To neutralize 7.6 g of formic mixture. Calculation of the mass fraction of acetic acid in a mixture with formic acid

1. Determine the mass of Mg 3 N 2 that has completely decomposed with water, if 150 ml of a 4% hydrochloric acid solution with a density of 1.02 g/ml was required for salt formation with hydrolysis products.

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   1) Mg 3 N 2 + 6H 2 O → 3Mg(OH) 2 + 2NH 3 2) Mg(OH) 2 + 2HCl → MgCl 2 + 2H 2 O 3) NH 3 +HCl → NH 4 Cl n (HCl) = 150 * 1.02 * 0.04/ 36.5 = 0.168 mol Let x moles of Mg 3 N 2 enter into the reaction. According to equation 1, 3x mol of Mg(OH) 2 and 2x mol of NH 3 were formed. The neutralization of 3x mol of Mg(OH) 2 required 6x mol of HCl (according to equation 2), and the neutralization of 2x mol of NH 3 required 2x mol of HCl (according to equation 3), for a total of 8x mol of HCl. 8 x = 0.168 mol, X = 0.021 mol, n(Mg 3 N 2) = 0.021 mol, m(Mg 3 N 2) = M*n = 100 * 0.021 = 2.1 g. Answer: 2.1 g

2. Determine the mass fraction of sodium carbonate in the solution obtained by boiling 150 g of 8.4% sodium bicarbonate solution. What volume of a 15.6% barium chloride solution (density 1.11 g/ml) will react with the resulting sodium carbonate? The evaporation of water can be neglected.
   

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   1) 2NaHCO 3 - t → Na 2 CO 3 + H 2 O + CO 2 2) Na 2 CO 3 + BaCl 2 → BaCO 3 + 2 NaCl n(NaHCO 3) = 150 * 0.084/84 = 0.15 mol From equation (1) n(NaHCO 3) : n(Na ​​2 CO 3) = 2:1 => n(Na ​​2 CO 3) = 0.075 mol. m(Na 2 CO 3) = 0.075∙106 = 7.95 g n(CO 2) = 0.075 mol, m(CO 2) = 0.075∙44 = 3.3 g m(solution) = 150 – 3.3 = 146.7 g ω(Na 2 CO 3) = m(Na 2 CO 3)/m(solution) = 7.95/146.7 = 0.0542 or 5.42% From equation (2) n(Na ​​2 CO 3) : n(BaCl 2) = 1: 1 => n(BaCl 2) = 0.075 mol. m(BaCl 2) =n*M = 0.075 * 208 = 15.6 g. m(solution) = m(BaCl 2)/ω = 15.6/0.156 = 100 g V(solution) = m(solution)/ρ = 100/1.11 = 90.1 ml. Answer: 5.42%, 90.1 ml.

3. In what mass ratios should 10% solutions of sodium hydroxide and sulfuric acid be mixed to obtain a neutral solution of sodium sulfate? What is the mass fraction of salt in such a solution?

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   2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O Let the mass of the NaOH solution be 100 g, m(NaOH) = 10 g, n(NaOH)=0.25 mol, n(H 2 SO 4) = 0.125 mol, m(H 2 SO 4) = 12.25 g, m(H 2 SO 4 solution) = 122.5 g Ratio m(NaOH solution) : m(H 2 SO 4 solution) = 1:1.2 n(Na 2 SO 4) = 0.125 mol, m(Na 2 SO 4) = 17.75 g, m(solution) = 100 + 122.5 g = 222.5 g, w(Na 2 SO 4)=7.98%

4. How many liters of chlorine (n.o.) will be released if 26.1 g of manganese (IV) oxide is added to 200 ml of 35% hydrochloric acid (density 1.17 g/ml) when heated? How many grams of sodium hydroxide in a cold solution will react with this amount of chlorine?

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   1) MnO 2 + 4 HCl → MnCl 2 + 2 H 2 O + Cl 2 2) 2NaOH + Cl 2 → NaCl + NaClO + H 2 O n(HCl) = 200*1.17* 0.35/ 36.5= 2.24 mol – in excess n(MnO 2) = 26.1/ 87 = 0.3 mol – in deficiency According to equation (1) n(Cl 2) = 0.3 mol V(Cl 2) = 6.72 l According to equation (2) n(NaOH)= 0.6 mol, m(NaOH)=24 g Answer: 6.72 l, 24 g

5. In what volume of water should 11.2 liters of sulfur (IV) oxide (n.s.) be dissolved to obtain a solution of sulfurous acid with a mass fraction of 1%? What color will litmus acquire when added to the resulting solution?
6. What mass of lithium hydride must be dissolved in 100 ml of water to obtain a solution with a mass fraction of hydroxide of 5%? What color will litmus acquire when added to the resulting solution?

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   LiH + H 2 O → LiOH + H 2 Let m(LiH)= x g, then m (LiOH)=x*24/8 = 3x g. m (solution) = m(H 2 O) + m(LiH)– m(H 2) m (r-ra) = x+100 – x/4 = 0.75x+100 w = m (in-va)*100% / m (solution) 3x/(0.75x+100) = 0.05 3x=0.038x+5 2.96x = 5 x=1.7 g

7. In what mass of solution with a mass fraction of Na 2 SO 4 10% should 200 g of Na 2 SO 4 × 10H 2 O be dissolved in order to obtain a solution with a mass fraction of sodium sulfate of 16%? What kind of environment will the resulting solution have?
   

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   Let the mass of the solution be x g. It contains 0.1 x g Na 2 SO 4. 200 g of crystalline hydrate were added, in which the mass of sodium sulfate was 200 * 142/322 = 88.2 g. (0.1x +88.2)/(x+200) =0.16 0.1x +88.2 = 0.16x + 32 0.06x = 56.2 x = 937 Answer: 937 g, neutral.

8. Ammonia gas, released when 160 g of a 7% solution of potassium hydroxide was boiled with 9.0 g of ammonium chloride, was dissolved in 75 g of water. Determine the mass fraction of ammonia in the resulting solution.
   
9. Ammonia, released when 80 g of a 14% solution of potassium hydroxide was boiled with 8.03 g of ammonium chloride, was dissolved in water. Calculate how many milliliters of 5% nitric acid with a density of 1.02 g/ml will be used to neutralize the resulting ammonia solution.
   

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   KOH + NH 4 Cl → NH 3 + H 2 O + KCl NH 3 + HNO 3 → NH 4 NO 3 n (NH 4 Cl) = 8.03/ 53.5 = 0.15 mol m (KOH) = 80* 0.14 = 11.2 g n (KOH) = 11.2 / 56 = 0.2 mol KOH in excess. We carry out further calculations based on the shortage n (NH 4 Cl) = n (NH 3) = 0.15 mol n (HNO 3) = 0.15 mol m (HNO 3) = n*M = 0.15 * 63 = 9.45 g m (HNO 3 solution) = m (amount) * 100% / w = 9.45 / 0.05 = 189 g m = V*ρ V= m / ρ= 189 / 1.02 = 185.3 ml Answer: 185.3 ml

10. Calcium carbide is treated with excess water. The released gas occupied a volume of 4.48 l (n.s.). Calculate what volume of 20% hydrochloric acid with a density of 1.10 g/ml will be used to completely neutralize the alkali formed from calcium carbide.
    

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    1) CaC 2 + 2H 2 O = Ca(OH) 2 + C 2 H 2 n(C 2 H 2) = V/ V m = 4.48/22.4 = 0.2 mol From equation (1) => n(C 2 H 2) = n(Ca(OH) 2) n(Ca(OH) 2) = 0.2 mol 2) Ca(OH) 2 + 2HCl = CaCl 2 + 2H 2 O From equation (2) => n(Ca(OH) 2) : n(HCl) = 1:2 => n(HCl) = 0.4 mol m(HCl) =n*M = 0.4*36.5 = 14.6g m(HCl solution) = 14.6/0.2 = 73g V(HCl solution) = 73/1.1 = 66.4 ml

11. Calculate what volume of a 10% hydrogen chloride solution with a density of 1.05 g/ml will be used to completely neutralize the calcium hydroxide formed during the hydrolysis of calcium carbide, if the gas released during hydrolysis occupied a volume of 8.96 l (n.s.).
    

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    CaC 2 + 2 H 2 O → Ca(OH) 2 + C 2 H 2 Ca(OH) 2 + 2 HCl → CaCl 2 + 2 H 2 O n(C 2 H 2)= V / V no. = 8.96 / 22.4 = 0.4 mol n (C 2 H 2) = n Ca(OH) 2 = 0.4 mol n(HCl) = 0.4 * 2 = 0.8 mol m(HCl ingredients) = n*M = 0.8 * 36.5= 29.2 g w = m (in-va) * 100% / m (solution) m (solution) = m (in-va)*100% / w = 29.2 / 0.1 = 292 g m = V* ρ V= m / ρ = 292 / 1.05 = 278ml Answer: 278 ml

12. Aluminum carbide is treated with 200 g of a 30% sulfuric acid solution. The methane released in this case occupied a volume of 4.48 liters (n.s.). Calculate the mass fraction of sulfuric acid in the resulting solution.
    

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    Al 4 C 3 + 6H 2 SO 4 → 2Al 2 (SO 4) 3 + 3CH 4 n(CH 4) =V/Vm = 4.48 /22.4 = 0.2 mol m(CH 4) = m*M = 0.2 * 16 = 3.2 g m(Al 4 C 3) = 1 / 3 * 0.2 * 144 = 9.6 g Reacted according to the equation n(H 2 SO 4) = 0.4 mol, m(H 2 SO 4) = 0.4 * 98 = 39.2 g. Initially added m(H 2 SO 4) = m(solution) * ω = 200 g * 0.3 = 60 g. Remaining m(H 2 SO 4) = 60 - 39.2 = 20.8 g. m(H 2 SO 4) = 0.21 * 98 = 20.8 g m(p-pa) = m(Al 4 C 3) + m(p-pa H 2 SO 4) – m(CH 4) m(p-pa) = 9.6 g + 200 g – 3.2 g = 206.4 g ω(H 2 SO 4) = m(H 2 SO 4)/m(solution) = 20.8 / 206.4 * 100% = 10%

13. When aluminum carbide was treated with a solution of hydrochloric acid, the mass of which was 320 g and the mass fraction of HCl was 22%, 6.72 l (n.o.) of methane was released. Calculate the mass fraction of hydrochloric acid in the resulting solution.
    

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    Al 4 C 3 + 12HCl → 4AlCl 3 + 3CH 4 n(CH 4) = V/Vm = b.72/22.4 = 0.3 mol; According to the equation n(HCl) = 4 n(CH 4) = 1.2 mol m(HCl) = m(solution)*ω = 320 · 0.22 = 70.4 g; m(HCl) = 1.2 · 36.5 = 43.8 g entered into the reaction. Remaining m(HCl) = 70.4 – 43.8 = 26.6 g. m(p-pa) = 320 g + m(Al 4 C 3) – m(CH 4), According to the equation n(Al 4 C 3) = 1/3 n(CH 4) = 0.1 mol; m(Al 4 C 3) = 0.1 144 = 14.4 g, m(CH 4) = 0.3 16 = 4.8 g, m(p-pa) = 320 g + 14.4 g – 4.8 g = 329.6 g. ω(HCl) = 26.6 / 329.6 100% = 8.07%

14. Calcium hydride was added to an excess of hydrochloric acid solution (mass of acid solution 150 g, mass fraction of HCl 20%). In this case, 6.72 liters (n.s.) of hydrogen were released. Calculate the mass fraction of calcium chloride in the resulting solution.
    

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    n(CaCl 2) = ½ n(H 2) = 0.15 mol m(CaCl 2) = 111 * 0.15 = 16.65 g W(CaCl 2) = m v-va / m r-ra = 16.65/155.7 = 0.1069 or 10.69% Answer: W(CaCl 2) = 10.69%

15. Mixed 125 ml of a 5% lithium hydroxide solution (r = 1.05 g/ml) and 100 ml of a 5% nitric acid solution (ρ = 1.03 g/ml). Determine the medium of the resulting solution and the mass fraction of lithium nitrate in it.
    

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    LiOH + HNO₃ = LiNO₃ + H₂O m (LiOH solution) = V × ρ = 125 ml × 1.05 g/ml = 131.25 g m(LiOH) = 131.25 g × 0.05 = 6.563 g n(LiOH) = m/ M = 6.563/ 24 = 0.273 mol m (HNO₃ solution:) = V × ρ = 100 ml × 1.03 g/ml = 103 g m(HNO₃) = 103 g × 0.05 = 5.15 g n(HNO₃) = 5.15 /63 = 0.0817 mol LiOH is given in excess, the calculation is carried out using acid. n(LiNO₃) = 0.0817 mol m(LiNO₃) = n × M = 0.0817 × 69 = 5.64 g m(resulting solution) = m(LiOH solution) + m(HNO₃ solution) = 131.25 g + 103 g = 234.25 g ω(LiNO₃) = 5.64 / 234.25 × 100% = 2.4% Answer: alkaline, 2.4%;

16. Phosphorus (V) oxide weighing 1.42 g was dissolved in 60 g of 8.2% orthophosphoric acid and the resulting solution was boiled. What salt and in what quantity is formed if 3.92 g of potassium hydroxide is added to the resulting solution?
17. Sulfur (VI) oxide weighing 8 g was dissolved in 110 g of 8% sulfuric acid. What salt and in what quantity is formed if 10.6 g of potassium hydroxide is added to the resulting solution?

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    SO 3 + H 2 O = H 2 SO 4 n = m/MM(SO 3) = 80 g/mol, n(SO 3)=8/80=0.1 mol. According to equation (1) n(H 2 SO 4)=n(SO 3)=0.1 mol, n(KOH)=10.6/56=0.19 mol. In the original solution n(H 2 SO 4) = 110 * 0.08/98 = 0.09 mol. After adding sulfur oxide n(H 2 SO 4) = 0.09 + 0.1 = 0.19 mol. The amounts of alkali and acid substances are related as 1:1, which means an acid salt is formed H 2 SO 4 + KOH = KHSO 4 + H 2 O n(H 2 SO 4) = n(KOH) = n(KHSO 4) = 0.19 mol Answer: KHSO 4, 0.19 mol.

18. Ammonia, released when 107 g of a 20% ammonium chloride solution reacted with 150 g of an 18% sodium hydroxide solution, completely reacted with 60% phosphoric acid to form ammonium dihydrogen phosphate. Determine the mass fraction of sodium chloride in the solution and the required mass of a 60% phosphoric acid solution.
    

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    NH 4 Cl + NaOH = NaCl + NH 3 + H 2 O m(NH 4 Cl) = 107 g ∙ 0.2 = 21.4 g n(NH 4 Cl) = 21.4 g / 53.5 g/mol = 0.4 mol m(NaOH) = 150 g ∙ 0.18 = 27 g n(NaOH) = 27 g / 40 g/mol = 0.675 mol, therefore NaOH is in excess n(NaCl) = n(NH4Cl) = 0.4 mol m(NaCl) = 0.4 ∙ 58.5 = 23.4 g n(NH 3) = n(NH 4 Cl) = 0.4 mol m(NH 3) = 0.4 ∙ 17 = 6.8 g m(solution) = m(solution NH 4 Cl) + m(solution NaOH) − m(NH 3) = 107 + 150 − 6.8 = 250.2 g w(NaCl) = 23.4 / 250.2 = 0.094 or 9.4% NH 3 + H 3 PO 4 = NH 4 H 2 PO 4 n(NH 3) = n(H 3 PO 4) = 0.4 mol m(H 3 PO 4) = 98 ∙ 0.4 = 39.2 g m(solution H 3 PO 4) = 39.2 / 0.6 = 65.3 g

19. Hydrogen sulfide, released during the interaction of excess concentrated sulfuric acid with 1.44 g of magnesium, was passed through 160 g of a 1.5% bromine solution. Determine the mass of the precipitate that formed and the mass fraction of acid in the resulting solution.
    

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    4Mg + 5H 2 SO 4 = 4MgSO 4 + H 2 S + 4H 2 O H 2 S + Br 2 = 2HBr + S↓ n(Mg) = m/M = 1.44 g: 24 g/mol = 0.06 mol n(H 2 S) = ¼ n(Mg) = 0.015 mol m(H 2 S) = n * M = 0.015 mol * 34 g/mol = 0.51 mol m(ingredients Br 2) = 160 g * 0.015 = 2.4 g n(Br 2) = m/M = 2.4 g: 160 g/mol = 0.015 mol n(HBr)= 2n(Br 2) = 0.03 mol m(HBr) = n * M = 0.03 mol * 81 g/mol = 2.43 g n(S) = n(Br 2) = 0.015 mol m(S) = n * M = 0.015 mol * 32 g/mol = 0.48 g m(solution) = m(H 2 S)+ m(Br 2 solution) -m(S) = 0.51 g + 160 g - 0.48 = 160.03 g W(HBr) = m(HBr)/ m(solution) = 2.43 g / 160.03 g = 0.015 or 1.5% Answer: m (S) = 0.48 g, w (HBr) = 1.5%

20. Chlorine reacted without residue with 228.58 ml of 5% NaOH solution (density 1.05 g/ml) at elevated temperature. Determine the composition of the resulting solution and calculate the mass fractions of substances in this solution.
    

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    6NaOH + 3Cl 2 = 5NaCl + NaClO 3 + 3H 2 O (at t) m solution = 228.58 ∙ 1.05 = 240g; m(NaOH) = 240 ∙ 0.05 = 12g. n(NaOH) = 12/ 40 = 0.3 mol; n(Cl 2) = 0.15 mol; n(NaCl) = 0.25 mol; n(NaClO 3) = 0.05 mol m(NaCl) = 58.5 ∙ 0.25 = 14.625g; m(NaClO 3) = 106.5 ∙ 0.05 = 5.325g: m solution = 240 + m(Cl 2) = 240 + 71 ∙ 0.15 = 240 + 10.65 = 250.65g W(NaCl) = 14.625 / 250.65 = 0.0583 or 5.83% W(NaClO 3) = 5.325 / 250.65 = 0.0212 or 2.12%

21. Copper weighing 6.4 g was treated with 100 ml of 30% nitric acid (ρ = 1.153 g/ml). To completely bind the products, 200 g of sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkali in the used solution.
    

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    3Cu + 8HNO 3 = 3Cu(NO 3) 2 + 2NO + 4H 2 O m(HNO 3) = 100 ∙ 0.3 ∙ 1.153 = 34.59 g n(HNO 3) = 34.59/ 63 = 0.55 mol, n(Cu) = 6.4/ 64 = 0.1 mol n(HNO 3) g = 0.55 – 8/3 ∙ 0.1 = 0.28 mol Cu(NO 3) 2 + 2 NaOH = Cu(OH) 2 + 2 NaNO 3 HNO 3 + NaOH = NaNO 3 + H 2 O n(NaOH) = n(HNO 3) ex + 2n(Cu(NO 3) 2) = 0.28 + 0.1 ∙ 2 = 0.48 mol m(NaOH) = 0.48 ∙ 40 = 19.2 g. W(NaOH) = 19.2/ 200 = 0.096 or 9.6%

22. 2.84 g of phosphorus (V) oxide was dissolved in 60 g of 18% orthophosphoric acid and the resulting solution was boiled. What salt and in what quantity is formed if 30 g of sodium hydroxide is added to the resulting solution?
    

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    1) 3H 2 O + P 2 O 5 → 2H 3 PO 4 In the original solution m(H 3 PO 4) = m(solution)*ω = 60*0.18 = 10.8 g. n(P 2 O 5) = m/M = 2.84/142 = 0.02 mol As a result of the reaction, m(H 3 PO 4) = 0.04*98 = 3.92 g was formed Total m(H 3 PO 4) = 3.92 + 10.8 = 14.72 g. n(H 3 PO 4) = m/M = 14.72/98 = 0.15 mol n(NaOH) = m/M = 30/40 = 0.75 mol – in excess, salt is average. 2) 3NaOH +H 3 PO 4 → Na 3 PO 4 + 3H 2 O According to equation (2) n(Na ​​3 PO 4) = n(H 3 PO 4) = 0.15 mol m(Na 3 PO 4) = 0.15*164 = 24.6 g Answer: 24.6 g

23. Ammonia with a volume of 4.48 l (n.o.) was passed through 200 g of a 4.9% solution of orthophosphoric acid. Name the salt formed as a result of the reaction and determine its mass.
    
24. 5.6 l (n.s.) of hydrogen sulfide reacted without a residue with 59.02 ml of a 20% KOH solution (density 1.186 g/ml). Determine the mass of salt obtained as a result of this chemical reaction.
    

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    m(KOH solution)= V*ρ=1.186 *59.02 = 70g, m(KOH)= m(p-raKOH)*ω =70g * 0.2 = 14g, n(KOH) =m/M = 14/56 = 0.25 mol, n(H 2 S) = V/Vm = 5.6/22.4 = 0.25 mol. The amount of hydrogen sulfide is equal to the amount of alkali, therefore, an acidic salt is formed - hydrosulfide according to the reaction: H 2 S + KOH = KНS + H 2 O According to the equation n(KHS) = 0.25 mol, m(KHS) = M*n = 72 · 0.25 = 18 g. Answer: 18

25. To neutralize 7.6 g of a mixture of formic and acetic acids, 35 ml of a 20% potassium hydroxide solution (density 1.20 g/ml) was used. Calculate the mass of acetic acid and its mass fraction in the original mixture of acids.

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    COOH + KOH = HCOOC + H 2 O CH 3 COOH + KOH = CH 3 COOC + H 2 O m(p-paKOH) = V(p-pa)*ρ = 35*1.2 = 42 g. m(KOH) = m(p-pa)*ω(KOH) = 42*0.2 = 8.4 g n(KOH) = m(KOH)/M(KOH) = 8.4/56 = 0.15 mol Let n(HCOOH) = x mol, and n(CH 3 COOH) = y mol. m(HCOOH) = n(HCOOH)*M(HCOOH) = x*46 g m(CH 3 COOH) = n(CH 3 COOH)*M(CH 3 COOH) = y*60 g Let's create a system of equations: x + y = 0.15 60y + 46x = 7.6 Let's solve the system: x = 0.1 mol, y = 0.05 mol m(CH 3 COOH) = n(CH 3 COOH)*M(CH 3 COOH) = 0.05*60 = 3 g. ω(CH 3 COOH) = m(CH 3 COOH)/m(mixture) = 3/7.6 = 0.395 or 39.5%. Answer: 39.5%

26. Mixed 100 ml of a 30% perchloric acid solution (r = 1.11 g/ml) and 300 ml of a 20% sodium hydroxide solution (r = 1.10 g/ml). How many milliliters of water should be added to the resulting mixture so that the mass fraction of sodium perchlorate in it is 8%?
    

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    HClO 4 + NaOH = NaClO 4 + H 2 O m(NaOH solution) = V*ρ = 300*1.10 = 330 g. n(NaOH) = m(NaOH solution) *ω/ M = 330*0.2/40= 1.65 mol – in excess. m(HClO 4 solution) = V*ρ = 100*1.11 =111 g. n(HClO 4) = 111*0.3/100.5 = 0.331 mol, According to the equation n(HClO 4) = n(NaClO 4) = 0.331 mol, m (NaClO 4) =n*M = 0.331*122.5 = 40.5 g. Let the mass of added water be x g. 40.5/(111+330+ x) = 0.08 whence x = 65.3 g. V(H 2 O)=65.3 ml. Answer: 65.3 ml

27. 6.4 g of calcium carbide was added to 100 ml of a 5% hydrochloric acid solution (density 1.02 g/ml). How many milliliters of 15% nitric acid (density 1.08 g/ml) should be added to the resulting mixture to completely neutralize it?
    

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    1) CaC 2 + 2HCl = CaCl 2 + C 2 H 2 2) CaC 2 + 2H 2 O = Ca(OH) 2 + C 2 H 2 3) Ca(OH) 2 + 2HNO 3 = Ca(NO 3) 2 + 2H 2 O n(HCl) = m(HCl)/M(HCl) = Vp-ra(HCl)*(HCl)*p(HCl)/M(HCl) = 100*0.05*1.02/36.5 = 0.14 mol, n(CaC 2) = m/M = 6.4/64 = 0.1 mol. According to equation (1) n(CaC 2) : n(HCl) = 1: 2 => CaC 2 - in excess. n(CaC 2) = n(HCl)/2 = 0.07 mol reacted. What remains is n(CaC 2) = 0.1 - 0.07 = 0.03 mol. According to equation (2) n(CaC 2) = n(Ca(OH) 2) = 0.03 mol. According to equation (3) n(Ca(OH)2) : n(HNO3) = 1: 2 => n(HNO 3) = 2n(Ca(OH) 2) = 0.06 mol. Vp-ra(HNO 3) = m(p-ra)/ρ m(solution) = m(HNO 3) /ω = 0.06*63/0.15 = 25.2 g, V(p-raHNO 3) = 25.2/1.08 = 23.3 ml. Answer: 23.3 ml

28. Sodium nitrite weighing 13.8 g was added while heating to 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (n.s.) of nitrogen will be released in this case and what is the mass fraction of ammonium chloride in the resulting solution?

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    NaNO 2 + NH 4 Cl = N 2 + NaCl + 2H 2 O n(NaNO 2) = 13.8/69 = 0.2 mol n(NH 4 Cl) = 220 0.1/53.5 = 0.41 mol NH 4 Cl - in excess n(N 2) = n(NaNO 2) = 0.2 mol V(N 2) = 0.2 mol 22.4 l/mol = 4.48 l Let's calculate the mass of ammonium chloride remaining in excess: n(NH 4 Cl)g = 0.41 − 0.2 = 0.21 mol m(NH 4 Cl) g = 0.21 · 53.5 = 11.2 g Calculate the mass fraction of ammonium chloride: m(p-pa) = 13.8 + 220 − 0.2 28 = 228.2 g ω(NH 4 Cl) = 11.2/228.2 = 0.049 or 4.9%Answer: V(N 2) =4.48 l ω(NH 4 Cl) =4.9%

29. Potassium nitrite weighing 8.5 g was added while heating to 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (n.s.) of nitrogen will be released in this case and what is the mass fraction of ammonium bromide in the resulting solution?
    

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    KNO 2 + NH 4 Br = N 2 + KBr + 2H 2 O m(NH 4 Br) = 270 g * 0.12 = 32.4 g n(NH 4 Br) = m/M = 32.4 g: 98 g/mol = 0.33 mol n(KNO 2) = m/M = 8.5 g: 85 g/mol = 0.1 mol n(NH 4), react. with KNO 2 = 0.33 mol – 0.1 mol = 0.23 mol (since n(KNO 2) : n(NH 4 Br) = 1:1) m(NH 4 Br) remaining in the final solution = n * M = 0.23 mol * 98 g/mol = 22.54 g n(N 2) = n(KNO 2) = 0.1 mol (N 2) = n * M = 0.1 mol * 28 g/mol = 2.8 g V(N 2) = n * M = 0.1 mol * 22.4 l/mol = 2.24 l m(final solution) = m(KNO 2) + m(NH 4 Br solution) – m(N 2) = 8.5 g + 270 g – 2.8 g = 275.7 g W(NH 4 Br in final solution) = 22.54 g: 275.7 g = 8% Answer: V(N 2) = 2.24 l; W(NH 4 Br) = 8%

30. Mixed 300 ml of sulfuric acid solution with a mass fraction of 10% (density 1.05 g/ml) and 200 ml of potassium hydroxide solution with a mass fraction of 20% (density 1.10 g/ml). How many milliliters of water should be added to the resulting mixture so that the mass fraction of salt in it is 7%?
31. 12.8 g of calcium carbide was added to 120 ml of nitric acid solution with a mass fraction of 7% (density 1.03 g/ml). How many milliliters of 20% hydrochloric acid (density 1.10 g/ml) should be added to the resulting mixture to completely neutralize it?
    

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    1) CaC 2 + 2HNO 3 = Ca(NO 3) 2 + C 2 H 2. n(CaC 2) = 12.8/ 64 = 0.2 mol n(HNO 3) = (0.07 1.03 120) /63 = 0.137 mol CaC 2 – excess. 2) CaC 2 + 2H 2 O = Ca(OH) 2 + C 2 H 2. n(Ca(OH) 2) = 0.2 - 0.137/ 2 = 0.13 mol3) Ca(OH) 2 + 2HCl = CaCl 2 + 2H 2 O. n(HCl) = 0.13 2 = 0.26 mol m(solution) = m(HCl)/ w =(0.26 · 36.5)/ 0.2 = 47.45 g V(HCl solution) = m(solution) /ρ = 47.45/1.10 = 43.1 ml. Answer: 43.1 ml.

32. To the solution obtained by adding 4 g of potassium hydride to 100 ml of water, 100 ml of a 39% solution of nitric acid (r = 1.24 g/ml) was added. Determine the mass fractions of all substances (including water) in the final solution.
    

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    KH + H 2 O = KOH + H 2 n(KH) = 4 g: 40 g/mol = 0.1 mol n(H 2 O) = 100 g: 18 g/mol = 5.6 mol KH is in short supply, then n(KOH) = n(KH) = 0.1 molKOH + H 2 NO 3 = KNO 3 + H 2 O a) m(KOH) = 0.1 * 56 = 5.6 b) m substance (HNO 2) = 100 ml * 1.24 g/ml * 0.39 = 48.36 g n(HNO 3) = 48.36 g: 63 g/mol = 0.77 mol HNO 3 in excess, n excess (HNO 3) = 0.77 – 0.1 = 0.67 mol m(HNO 3) = 0.67 * 63 = 42.21 g m(solution) = 4g + 100g + 124g – 0.2g = 227.8g 3) W(KNO 3) = m(KNO 3) : m(p-pa) = (0.1 mol * 101 g/mol) 227.8 g * 100% = 4.4% W(HNO 3) = 42.21: 227.8 * 100% = 18.5 W(H 2 O) = 100% - (W(KNO 3) + W(HNO 3)) = 77.1% Answer: W(KNO 3) = 4.4% W(HNO 3) =18.5% W(H2O) = 77.1%

    1) 2Na 2 O 2 + 2H 2 O = 4NaOH + O 2
    2) 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O
    m(H 2 SO 4 solution) = 300 * 1.08 = 324 g
    m(H 2 SO 4) = 0.1*324 = 32.4 g
    n(H 2 SO 4) = 32.4/98 = 0.33 mol
    n(NaOH) : n(H 2 SO 4) = 2:1 => n(NaOH) = 0.33 * 2 = 0.66 mol
    n(Na 2 O 2): n(NaOH) = 1:2 => n(Na ​​2 O 2) = 0.66/2 = 0.33 mol
    m(Na 2 O 2) = n * M = 0.33 * 78 = 25.7 g
    n(Na 2 O 2): n(O 2) = 2:1 => n(O 2) = 0.33/2 = 0.165 mol
    V(O 2) = 0.165 * 22.4 = 3.7 l
    Answer: m(Na 2 O 2) = 25.7 g; V(O 2) = 3.7 l

33. When heated, potassium bicarbonate turns into carbonate. Calculate the mass fraction of potassium bicarbonate in the initial solution, by heating which you can obtain an 8% solution of potassium carbonate.
34. When 17.4 g of manganese dioxide reacted with 58 g of potassium bromide in a sulfuric acid medium, bromine was released in a 77% yield. What volume (no.) of propene can react with the resulting amount of bromine?
35. Carbon dioxide with a volume of 5.6 liters (n.s.) was passed through 164 ml of a 20% sodium hydroxide solution (ρ = 1.22 g/ml). Determine the composition and mass fractions of substances in the resulting solution.
36. Aluminum carbide was dissolved in a 15% solution of sulfuric acid weighing 300 g. The methane released in this case occupied a volume of 2.24 liters (n.s.). Calculate the mass fraction of sulfuric acid in the resulting solution.
37. 8 g of sulfur was burned in excess oxygen. The resulting gas was passed through 200 g of an 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution.
38. A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, and 8.96 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters (n.s.) of hydrogen will be released. Calculate the mass fraction of iron in the initial mixture.
39. A mixture of magnesium and zinc filings was treated with an excess of dilute sulfuric acid, and 22.4 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (n.s.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.
    

    2KHCO 3 + H 2 SO 4 = K 2 SO 4 + 2CO 2 + 2H 2 O (2)

    Na 2 CO 3 + 2HCl = NaCl + CO 2 +H 2 O NaHCO 3 +HCl = NaCl + CO 2 +H 2 O NaHCO 3 + NaOH = Na 2 CO 3 + H 2 O 1) we'll findquantityNaHCO3 n(NaOH)=(NaHCO 3)=m(NaOH) . ω (NaOH)/M(NaOH) = 80* 0.1/40 = 0.2 mol 2) we'll findmassNaHCO3 m(NaHCO 3)=n(NaHCO 3)*M(NaHCO 3)=0.2. 84=16.8 g 3) let's find the amount of Na 2 CO 3 n(HCl) = m(HCl) . ω (HCl)/M(HCl) = 73* 0.2/36.5 = 0.4 mol n(Na 2 CO 3) = (n(HCl) - n(NaOH))/2 = 0.1 mol 4) let's find the massNa 2 CO 3 m(Na 2 CO 3)=n. M = 0.1. 106 = 10.6 g 5) let's find the mass fractionNa 2 CO 3 ω (Na 2 CO 3)=m(Na 2 CO 3)/m(mixture). 100%=38.7% m(mixture).=m(Na 2 CO 3)+m(NaHCO 3)=27.4 g Answer: ω (Na 2 CO 3) = 38.7%

    1) 2Na + 2H 2 O → 2NaOH + H 2 2) Na 2 O + H 2 O → 2NaOH 1) let's find the quantityNa n(H 2) = V/Vm = 4.48 / 22.4 = 0.2 mol According to equation (1) n(Na) = n 1 (NaОH) = 2. n(H2) = 0.4 mol 2) find the massNa m(Na) = n. M = 0.4. 23 = 9.2 g 3) let's find the quantityNaOH n(NaOH) = m(p-p) . ω/M(NaOH) = 240. 0.1/40 = 0.6 mol 4) let's find the quantityNa 2 O n(Na 2 O) = (n(NaOH) - n 1 (NaOH))/2=0.1 mol m(Na 2 O)= n. M = 0.1. 62 = 6.2g 5) we'll findman ace shareNa ω (Na) = m(Na)/m(mixtures) . 100%=59.7% m(mixture)=m(Na)+m(Na 2 O)=15.4 Answer: ω(Na)=59,7%

    H 2 SO 4 + Na 2 CO 3 → Na 2 SO 4 + CO 2 + H 2
    1) Let's find the total quantityH2SO4
    n 1 (H 2 SO 4) = m. ω/M=490. 0.4/98=2 mol
    M(H 2 SO 4)=2+32+64=98 g/mol
    2) We'll findH2SO4reactedWithNa 2 CO 3
    n(Na 2 CO 3) = n(Na ​​2 CO 3 . 10H 2 O) = m/M = 143/286 = 0.5 mol
    M(Na 2 CO 3 . 10H 2 O) = (46 + 12 + 48) + (10 . 18) = 286 g/mol
    According to the equation n(H 2 SO 4) = n(Na ​​2 CO 3) = 0.5 mol
    3) we'll findH 2 SO 4 reacted withNaOH
    n(H 2 SO 4)=n 1 -n=2-0.5=1.5 mol
    4) find the mass of NaOH
    n(NaOH)=2n(H 2 SO 4)=2. 1.5=3 mol
    m(NaOH)=n. M =3. 40=120 g
    M(NaOH)=23+16+1=18 g/mol
    5) let's find the mass fractionNaOH
    ω (NaOH)= m(substance)/ . m(p-p)*100%=120/1200 . 100%=10%
    Answer: m(NaOH)=120g; ω (NaOH)=10%
    
    48) A mixture of magnesium and aluminum filings was treated with an excess of dilute hydrochloric acid, and 11.2 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of potassium hydroxide solution, then 6.72 liters (no.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.
    49) Calcium carbide weighing 6.4 g was dissolved in 87 ml of hydrobromic acid (ρ = 1.12 g/ml) with a mass fraction of 20%. What is the mass fraction of hydrogen bromide in the resulting solution?

1. Determine the mass of Mg 3 N 2 that has completely decomposed with water, if 150 ml of a 4% hydrochloric acid solution with a density of 1.02 g/ml was required for salt formation with hydrolysis products.

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   1) Mg 3 N 2 + 6H 2 O → 3Mg(OH) 2 + 2NH 3 2) Mg(OH) 2 + 2HCl → MgCl 2 + 2H 2 O 3) NH 3 +HCl → NH 4 Cl n (HCl) = 150 * 1.02 * 0.04/ 36.5 = 0.168 mol Let x moles of Mg 3 N 2 enter into the reaction. According to equation 1, 3x mol of Mg(OH) 2 and 2x mol of NH 3 were formed. The neutralization of 3x mol of Mg(OH) 2 required 6x mol of HCl (according to equation 2), and the neutralization of 2x mol of NH 3 required 2x mol of HCl (according to equation 3), for a total of 8x mol of HCl. 8 x = 0.168 mol, X = 0.021 mol, n(Mg 3 N 2) = 0.021 mol, m(Mg 3 N 2) = M*n = 100 * 0.021 = 2.1 g. Answer: 2.1 g

2. Determine the mass fraction of sodium carbonate in the solution obtained by boiling 150 g of 8.4% sodium bicarbonate solution. What volume of a 15.6% barium chloride solution (density 1.11 g/ml) will react with the resulting sodium carbonate? The evaporation of water can be neglected.
   

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   1) 2NaHCO 3 - t → Na 2 CO 3 + H 2 O + CO 2 2) Na 2 CO 3 + BaCl 2 → BaCO 3 + 2 NaCl n(NaHCO 3) = 150 * 0.084/84 = 0.15 mol From equation (1) n(NaHCO 3) : n(Na ​​2 CO 3) = 2:1 => n(Na ​​2 CO 3) = 0.075 mol. m(Na 2 CO 3) = 0.075∙106 = 7.95 g n(CO 2) = 0.075 mol, m(CO 2) = 0.075∙44 = 3.3 g m(solution) = 150 – 3.3 = 146.7 g ω(Na 2 CO 3) = m(Na 2 CO 3)/m(solution) = 7.95/146.7 = 0.0542 or 5.42% From equation (2) n(Na ​​2 CO 3) : n(BaCl 2) = 1: 1 => n(BaCl 2) = 0.075 mol. m(BaCl 2) =n*M = 0.075 * 208 = 15.6 g. m(solution) = m(BaCl 2)/ω = 15.6/0.156 = 100 g V(solution) = m(solution)/ρ = 100/1.11 = 90.1 ml. Answer: 5.42%, 90.1 ml.

3. In what mass ratios should 10% solutions of sodium hydroxide and sulfuric acid be mixed to obtain a neutral solution of sodium sulfate? What is the mass fraction of salt in such a solution?

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   2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O Let the mass of the NaOH solution be 100 g, m(NaOH) = 10 g, n(NaOH)=0.25 mol, n(H 2 SO 4) = 0.125 mol, m(H 2 SO 4) = 12.25 g, m(H 2 SO 4 solution) = 122.5 g Ratio m(NaOH solution) : m(H 2 SO 4 solution) = 1:1.2 n(Na 2 SO 4) = 0.125 mol, m(Na 2 SO 4) = 17.75 g, m(solution) = 100 + 122.5 g = 222.5 g, w(Na 2 SO 4)=7.98%

4. How many liters of chlorine (n.o.) will be released if 26.1 g of manganese (IV) oxide is added to 200 ml of 35% hydrochloric acid (density 1.17 g/ml) when heated? How many grams of sodium hydroxide in a cold solution will react with this amount of chlorine?

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   1) MnO 2 + 4 HCl → MnCl 2 + 2 H 2 O + Cl 2 2) 2NaOH + Cl 2 → NaCl + NaClO + H 2 O n(HCl) = 200*1.17* 0.35/ 36.5= 2.24 mol – in excess n(MnO 2) = 26.1/ 87 = 0.3 mol – in deficiency According to equation (1) n(Cl 2) = 0.3 mol V(Cl 2) = 6.72 l According to equation (2) n(NaOH)= 0.6 mol, m(NaOH)=24 g Answer: 6.72 l, 24 g

5. In what volume of water should 11.2 liters of sulfur (IV) oxide (n.s.) be dissolved to obtain a solution of sulfurous acid with a mass fraction of 1%? What color will litmus acquire when added to the resulting solution?
6. What mass of lithium hydride must be dissolved in 100 ml of water to obtain a solution with a mass fraction of hydroxide of 5%? What color will litmus acquire when added to the resulting solution?

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   LiH + H 2 O → LiOH + H 2 Let m(LiH)= x g, then m (LiOH)=x*24/8 = 3x g. m (solution) = m(H 2 O) + m(LiH)– m(H 2) m (r-ra) = x+100 – x/4 = 0.75x+100 w = m (in-va)*100% / m (solution) 3x/(0.75x+100) = 0.05 3x=0.038x+5 2.96x = 5 x=1.7 g

7. In what mass of solution with a mass fraction of Na 2 SO 4 10% should 200 g of Na 2 SO 4 × 10H 2 O be dissolved in order to obtain a solution with a mass fraction of sodium sulfate of 16%? What kind of environment will the resulting solution have?
   

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   Let the mass of the solution be x g. It contains 0.1 x g Na 2 SO 4. 200 g of crystalline hydrate were added, in which the mass of sodium sulfate was 200 * 142/322 = 88.2 g. (0.1x +88.2)/(x+200) =0.16 0.1x +88.2 = 0.16x + 32 0.06x = 56.2 x = 937 Answer: 937 g, neutral.

8. Ammonia gas, released when 160 g of a 7% solution of potassium hydroxide was boiled with 9.0 g of ammonium chloride, was dissolved in 75 g of water. Determine the mass fraction of ammonia in the resulting solution.
   
9. Ammonia, released when 80 g of a 14% solution of potassium hydroxide was boiled with 8.03 g of ammonium chloride, was dissolved in water. Calculate how many milliliters of 5% nitric acid with a density of 1.02 g/ml will be used to neutralize the resulting ammonia solution.
   

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   KOH + NH 4 Cl → NH 3 + H 2 O + KCl NH 3 + HNO 3 → NH 4 NO 3 n (NH 4 Cl) = 8.03/ 53.5 = 0.15 mol m (KOH) = 80* 0.14 = 11.2 g n (KOH) = 11.2 / 56 = 0.2 mol KOH in excess. We carry out further calculations based on the shortage n (NH 4 Cl) = n (NH 3) = 0.15 mol n (HNO 3) = 0.15 mol m (HNO 3) = n*M = 0.15 * 63 = 9.45 g m (HNO 3 solution) = m (amount) * 100% / w = 9.45 / 0.05 = 189 g m = V*ρ V= m / ρ= 189 / 1.02 = 185.3 ml Answer: 185.3 ml

10. Calcium carbide is treated with excess water. The released gas occupied a volume of 4.48 l (n.s.). Calculate what volume of 20% hydrochloric acid with a density of 1.10 g/ml will be used to completely neutralize the alkali formed from calcium carbide.
    

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    1) CaC 2 + 2H 2 O = Ca(OH) 2 + C 2 H 2 n(C 2 H 2) = V/ V m = 4.48/22.4 = 0.2 mol From equation (1) => n(C 2 H 2) = n(Ca(OH) 2) n(Ca(OH) 2) = 0.2 mol 2) Ca(OH) 2 + 2HCl = CaCl 2 + 2H 2 O From equation (2) => n(Ca(OH) 2) : n(HCl) = 1:2 => n(HCl) = 0.4 mol m(HCl) =n*M = 0.4*36.5 = 14.6g m(HCl solution) = 14.6/0.2 = 73g V(HCl solution) = 73/1.1 = 66.4 ml

11. Calculate what volume of a 10% hydrogen chloride solution with a density of 1.05 g/ml will be used to completely neutralize the calcium hydroxide formed during the hydrolysis of calcium carbide, if the gas released during hydrolysis occupied a volume of 8.96 l (n.s.).
    

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    CaC 2 + 2 H 2 O → Ca(OH) 2 + C 2 H 2 Ca(OH) 2 + 2 HCl → CaCl 2 + 2 H 2 O n(C 2 H 2)= V / V no. = 8.96 / 22.4 = 0.4 mol n (C 2 H 2) = n Ca(OH) 2 = 0.4 mol n(HCl) = 0.4 * 2 = 0.8 mol m(HCl ingredients) = n*M = 0.8 * 36.5= 29.2 g w = m (in-va) * 100% / m (solution) m (solution) = m (in-va)*100% / w = 29.2 / 0.1 = 292 g m = V* ρ V= m / ρ = 292 / 1.05 = 278ml Answer: 278 ml

12. Aluminum carbide is treated with 200 g of a 30% sulfuric acid solution. The methane released in this case occupied a volume of 4.48 liters (n.s.). Calculate the mass fraction of sulfuric acid in the resulting solution.
    

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    Al 4 C 3 + 6H 2 SO 4 → 2Al 2 (SO 4) 3 + 3CH 4 n(CH 4) =V/Vm = 4.48 /22.4 = 0.2 mol m(CH 4) = m*M = 0.2 * 16 = 3.2 g m(Al 4 C 3) = 1 / 3 * 0.2 * 144 = 9.6 g Reacted according to the equation n(H 2 SO 4) = 0.4 mol, m(H 2 SO 4) = 0.4 * 98 = 39.2 g. Initially added m(H 2 SO 4) = m(solution) * ω = 200 g * 0.3 = 60 g. Remaining m(H 2 SO 4) = 60 - 39.2 = 20.8 g. m(H 2 SO 4) = 0.21 * 98 = 20.8 g m(p-pa) = m(Al 4 C 3) + m(p-pa H 2 SO 4) – m(CH 4) m(p-pa) = 9.6 g + 200 g – 3.2 g = 206.4 g ω(H 2 SO 4) = m(H 2 SO 4)/m(solution) = 20.8 / 206.4 * 100% = 10%

13. When aluminum carbide was treated with a solution of hydrochloric acid, the mass of which was 320 g and the mass fraction of HCl was 22%, 6.72 l (n.o.) of methane was released. Calculate the mass fraction of hydrochloric acid in the resulting solution.
    

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    Al 4 C 3 + 12HCl → 4AlCl 3 + 3CH 4 n(CH 4) = V/Vm = b.72/22.4 = 0.3 mol; According to the equation n(HCl) = 4 n(CH 4) = 1.2 mol m(HCl) = m(solution)*ω = 320 · 0.22 = 70.4 g; m(HCl) = 1.2 · 36.5 = 43.8 g entered into the reaction. Remaining m(HCl) = 70.4 – 43.8 = 26.6 g. m(p-pa) = 320 g + m(Al 4 C 3) – m(CH 4), According to the equation n(Al 4 C 3) = 1/3 n(CH 4) = 0.1 mol; m(Al 4 C 3) = 0.1 144 = 14.4 g, m(CH 4) = 0.3 16 = 4.8 g, m(p-pa) = 320 g + 14.4 g – 4.8 g = 329.6 g. ω(HCl) = 26.6 / 329.6 100% = 8.07%

14. Calcium hydride was added to an excess of hydrochloric acid solution (mass of acid solution 150 g, mass fraction of HCl 20%). In this case, 6.72 liters (n.s.) of hydrogen were released. Calculate the mass fraction of calcium chloride in the resulting solution.
    

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    n(CaCl 2) = ½ n(H 2) = 0.15 mol m(CaCl 2) = 111 * 0.15 = 16.65 g W(CaCl 2) = m v-va / m r-ra = 16.65/155.7 = 0.1069 or 10.69% Answer: W(CaCl 2) = 10.69%

15. Mixed 125 ml of a 5% lithium hydroxide solution (r = 1.05 g/ml) and 100 ml of a 5% nitric acid solution (ρ = 1.03 g/ml). Determine the medium of the resulting solution and the mass fraction of lithium nitrate in it.
    

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    LiOH + HNO₃ = LiNO₃ + H₂O m (LiOH solution) = V × ρ = 125 ml × 1.05 g/ml = 131.25 g m(LiOH) = 131.25 g × 0.05 = 6.563 g n(LiOH) = m/ M = 6.563/ 24 = 0.273 mol m (HNO₃ solution:) = V × ρ = 100 ml × 1.03 g/ml = 103 g m(HNO₃) = 103 g × 0.05 = 5.15 g n(HNO₃) = 5.15 /63 = 0.0817 mol LiOH is given in excess, the calculation is carried out using acid. n(LiNO₃) = 0.0817 mol m(LiNO₃) = n × M = 0.0817 × 69 = 5.64 g m(resulting solution) = m(LiOH solution) + m(HNO₃ solution) = 131.25 g + 103 g = 234.25 g ω(LiNO₃) = 5.64 / 234.25 × 100% = 2.4% Answer: alkaline, 2.4%;

16. Phosphorus (V) oxide weighing 1.42 g was dissolved in 60 g of 8.2% orthophosphoric acid and the resulting solution was boiled. What salt and in what quantity is formed if 3.92 g of potassium hydroxide is added to the resulting solution?
17. Sulfur (VI) oxide weighing 8 g was dissolved in 110 g of 8% sulfuric acid. What salt and in what quantity is formed if 10.6 g of potassium hydroxide is added to the resulting solution?

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    SO 3 + H 2 O = H 2 SO 4 n = m/MM(SO 3) = 80 g/mol, n(SO 3)=8/80=0.1 mol. According to equation (1) n(H 2 SO 4)=n(SO 3)=0.1 mol, n(KOH)=10.6/56=0.19 mol. In the original solution n(H 2 SO 4) = 110 * 0.08/98 = 0.09 mol. After adding sulfur oxide n(H 2 SO 4) = 0.09 + 0.1 = 0.19 mol. The amounts of alkali and acid substances are related as 1:1, which means an acid salt is formed H 2 SO 4 + KOH = KHSO 4 + H 2 O n(H 2 SO 4) = n(KOH) = n(KHSO 4) = 0.19 mol Answer: KHSO 4, 0.19 mol.

18. Ammonia, released when 107 g of a 20% ammonium chloride solution reacted with 150 g of an 18% sodium hydroxide solution, completely reacted with 60% phosphoric acid to form ammonium dihydrogen phosphate. Determine the mass fraction of sodium chloride in the solution and the required mass of a 60% phosphoric acid solution.
    

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    NH 4 Cl + NaOH = NaCl + NH 3 + H 2 O m(NH 4 Cl) = 107 g ∙ 0.2 = 21.4 g n(NH 4 Cl) = 21.4 g / 53.5 g/mol = 0.4 mol m(NaOH) = 150 g ∙ 0.18 = 27 g n(NaOH) = 27 g / 40 g/mol = 0.675 mol, therefore NaOH is in excess n(NaCl) = n(NH4Cl) = 0.4 mol m(NaCl) = 0.4 ∙ 58.5 = 23.4 g n(NH 3) = n(NH 4 Cl) = 0.4 mol m(NH 3) = 0.4 ∙ 17 = 6.8 g m(solution) = m(solution NH 4 Cl) + m(solution NaOH) − m(NH 3) = 107 + 150 − 6.8 = 250.2 g w(NaCl) = 23.4 / 250.2 = 0.094 or 9.4% NH 3 + H 3 PO 4 = NH 4 H 2 PO 4 n(NH 3) = n(H 3 PO 4) = 0.4 mol m(H 3 PO 4) = 98 ∙ 0.4 = 39.2 g m(solution H 3 PO 4) = 39.2 / 0.6 = 65.3 g

19. Hydrogen sulfide, released during the interaction of excess concentrated sulfuric acid with 1.44 g of magnesium, was passed through 160 g of a 1.5% bromine solution. Determine the mass of the precipitate that formed and the mass fraction of acid in the resulting solution.
    

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    4Mg + 5H 2 SO 4 = 4MgSO 4 + H 2 S + 4H 2 O H 2 S + Br 2 = 2HBr + S↓ n(Mg) = m/M = 1.44 g: 24 g/mol = 0.06 mol n(H 2 S) = ¼ n(Mg) = 0.015 mol m(H 2 S) = n * M = 0.015 mol * 34 g/mol = 0.51 mol m(ingredients Br 2) = 160 g * 0.015 = 2.4 g n(Br 2) = m/M = 2.4 g: 160 g/mol = 0.015 mol n(HBr)= 2n(Br 2) = 0.03 mol m(HBr) = n * M = 0.03 mol * 81 g/mol = 2.43 g n(S) = n(Br 2) = 0.015 mol m(S) = n * M = 0.015 mol * 32 g/mol = 0.48 g m(solution) = m(H 2 S)+ m(Br 2 solution) -m(S) = 0.51 g + 160 g - 0.48 = 160.03 g W(HBr) = m(HBr)/ m(solution) = 2.43 g / 160.03 g = 0.015 or 1.5% Answer: m (S) = 0.48 g, w (HBr) = 1.5%

20. Chlorine reacted without a residue with 228.58 ml of a 5% NaOH solution (density 1.05 g/ml) at elevated temperature. Determine the composition of the resulting solution and calculate the mass fractions of substances in this solution.
    

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    6NaOH + 3Cl 2 = 5NaCl + NaClO 3 + 3H 2 O (at t) m solution = 228.58 ∙ 1.05 = 240g; m(NaOH) = 240 ∙ 0.05 = 12g. n(NaOH) = 12/ 40 = 0.3 mol; n(Cl 2) = 0.15 mol; n(NaCl) = 0.25 mol; n(NaClO 3) = 0.05 mol m(NaCl) = 58.5 ∙ 0.25 = 14.625g; m(NaClO 3) = 106.5 ∙ 0.05 = 5.325g: m solution = 240 + m(Cl 2) = 240 + 71 ∙ 0.15 = 240 + 10.65 = 250.65g W(NaCl) = 14.625 / 250.65 = 0.0583 or 5.83% W(NaClO 3) = 5.325 / 250.65 = 0.0212 or 2.12%

21. Copper weighing 6.4 g was treated with 100 ml of 30% nitric acid (ρ = 1.153 g/ml). To completely bind the products, 200 g of sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkali in the used solution.
    

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    3Cu + 8HNO 3 = 3Cu(NO 3) 2 + 2NO + 4H 2 O m(HNO 3) = 100 ∙ 0.3 ∙ 1.153 = 34.59 g n(HNO 3) = 34.59/ 63 = 0.55 mol, n(Cu) = 6.4/ 64 = 0.1 mol n(HNO 3) g = 0.55 – 8/3 ∙ 0.1 = 0.28 mol Cu(NO 3) 2 + 2 NaOH = Cu(OH) 2 + 2 NaNO 3 HNO 3 + NaOH = NaNO 3 + H 2 O n(NaOH) = n(HNO 3) ex + 2n(Cu(NO 3) 2) = 0.28 + 0.1 ∙ 2 = 0.48 mol m(NaOH) = 0.48 ∙ 40 = 19.2 g. W(NaOH) = 19.2/ 200 = 0.096 or 9.6%

22. 2.84 g of phosphorus (V) oxide was dissolved in 60 g of 18% orthophosphoric acid and the resulting solution was boiled. What salt and in what quantity is formed if 30 g of sodium hydroxide is added to the resulting solution?
    

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    1) 3H 2 O + P 2 O 5 → 2H 3 PO 4 In the original solution m(H 3 PO 4) = m(solution)*ω = 60*0.18 = 10.8 g. n(P 2 O 5) = m/M = 2.84/142 = 0.02 mol As a result of the reaction, m(H 3 PO 4) = 0.04*98 = 3.92 g was formed Total m(H 3 PO 4) = 3.92 + 10.8 = 14.72 g. n(H 3 PO 4) = m/M = 14.72/98 = 0.15 mol n(NaOH) = m/M = 30/40 = 0.75 mol – in excess, salt is average. 2) 3NaOH +H 3 PO 4 → Na 3 PO 4 + 3H 2 O According to equation (2) n(Na ​​3 PO 4) = n(H 3 PO 4) = 0.15 mol m(Na 3 PO 4) = 0.15*164 = 24.6 g Answer: 24.6 g

23. Ammonia with a volume of 4.48 l (n.o.) was passed through 200 g of a 4.9% solution of orthophosphoric acid. Name the salt formed as a result of the reaction and determine its mass.
    
24. 5.6 l (n.s.) of hydrogen sulfide reacted without a residue with 59.02 ml of a 20% KOH solution (density 1.186 g/ml). Determine the mass of salt obtained as a result of this chemical reaction.
    

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    m(KOH solution)= V*ρ=1.186 *59.02 = 70g, m(KOH)= m(p-raKOH)*ω =70g * 0.2 = 14g, n(KOH) =m/M = 14/56 = 0.25 mol, n(H 2 S) = V/Vm = 5.6/22.4 = 0.25 mol. The amount of hydrogen sulfide is equal to the amount of alkali, therefore, an acidic salt is formed - hydrosulfide according to the reaction: H 2 S + KOH = KНS + H 2 O According to the equation n(KHS) = 0.25 mol, m(KHS) = M*n = 72 · 0.25 = 18 g. Answer: 18

25. To neutralize 7.6 g of a mixture of formic and acetic acids, 35 ml of a 20% potassium hydroxide solution (density 1.20 g/ml) was used. Calculate the mass of acetic acid and its mass fraction in the original mixture of acids.

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    COOH + KOH = HCOOC + H 2 O CH 3 COOH + KOH = CH 3 COOC + H 2 O m(p-paKOH) = V(p-pa)*ρ = 35*1.2 = 42 g. m(KOH) = m(p-pa)*ω(KOH) = 42*0.2 = 8.4 g n(KOH) = m(KOH)/M(KOH) = 8.4/56 = 0.15 mol Let n(HCOOH) = x mol, and n(CH 3 COOH) = y mol. m(HCOOH) = n(HCOOH)*M(HCOOH) = x*46 g m(CH 3 COOH) = n(CH 3 COOH)*M(CH 3 COOH) = y*60 g Let's create a system of equations: x + y = 0.15 60y + 46x = 7.6 Let's solve the system: x = 0.1 mol, y = 0.05 mol m(CH 3 COOH) = n(CH 3 COOH)*M(CH 3 COOH) = 0.05*60 = 3 g. ω(CH 3 COOH) = m(CH 3 COOH)/m(mixture) = 3/7.6 = 0.395 or 39.5%. Answer: 39.5%

26. Mixed 100 ml of a 30% perchloric acid solution (r = 1.11 g/ml) and 300 ml of a 20% sodium hydroxide solution (r = 1.10 g/ml). How many milliliters of water should be added to the resulting mixture so that the mass fraction of sodium perchlorate in it is 8%?
    

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    HClO 4 + NaOH = NaClO 4 + H 2 O m(NaOH solution) = V*ρ = 300*1.10 = 330 g. n(NaOH) = m(NaOH solution) *ω/ M = 330*0.2/40= 1.65 mol – in excess. m(HClO 4 solution) = V*ρ = 100*1.11 =111 g. n(HClO 4) = 111*0.3/100.5 = 0.331 mol, According to the equation n(HClO 4) = n(NaClO 4) = 0.331 mol, m (NaClO 4) =n*M = 0.331*122.5 = 40.5 g. Let the mass of added water be x g. 40.5/(111+330+ x) = 0.08 whence x = 65.3 g. V(H 2 O)=65.3 ml. Answer: 65.3 ml

27. 6.4 g of calcium carbide was added to 100 ml of a 5% hydrochloric acid solution (density 1.02 g/ml). How many milliliters of 15% nitric acid (density 1.08 g/ml) should be added to the resulting mixture to completely neutralize it?
    

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    1) CaC 2 + 2HCl = CaCl 2 + C 2 H 2 2) CaC 2 + 2H 2 O = Ca(OH) 2 + C 2 H 2 3) Ca(OH) 2 + 2HNO 3 = Ca(NO 3) 2 + 2H 2 O n(HCl) = m(HCl)/M(HCl) = Vp-ra(HCl)*(HCl)*p(HCl)/M(HCl) = 100*0.05*1.02/36.5 = 0.14 mol, n(CaC 2) = m/M = 6.4/64 = 0.1 mol. According to equation (1) n(CaC 2) : n(HCl) = 1: 2 => CaC 2 - in excess. n(CaC 2) = n(HCl)/2 = 0.07 mol reacted. What remains is n(CaC 2) = 0.1 - 0.07 = 0.03 mol. According to equation (2) n(CaC 2) = n(Ca(OH) 2) = 0.03 mol. According to equation (3) n(Ca(OH)2) : n(HNO3) = 1: 2 => n(HNO 3) = 2n(Ca(OH) 2) = 0.06 mol. Vp-ra(HNO 3) = m(p-ra)/ρ m(solution) = m(HNO 3) /ω = 0.06*63/0.15 = 25.2 g, V(p-raHNO 3) = 25.2/1.08 = 23.3 ml. Answer: 23.3 ml

28. Sodium nitrite weighing 13.8 g was added while heating to 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (n.s.) of nitrogen will be released in this case and what is the mass fraction of ammonium chloride in the resulting solution?

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    NaNO 2 + NH 4 Cl = N 2 + NaCl + 2H 2 O n(NaNO 2) = 13.8/69 = 0.2 mol n(NH 4 Cl) = 220 0.1/53.5 = 0.41 mol NH 4 Cl - in excess n(N 2) = n(NaNO 2) = 0.2 mol V(N 2) = 0.2 mol 22.4 l/mol = 4.48 l Let's calculate the mass of ammonium chloride remaining in excess: n(NH 4 Cl)g = 0.41 − 0.2 = 0.21 mol m(NH 4 Cl) g = 0.21 · 53.5 = 11.2 g Calculate the mass fraction of ammonium chloride: m(p-pa) = 13.8 + 220 − 0.2 28 = 228.2 g ω(NH 4 Cl) = 11.2/228.2 = 0.049 or 4.9%Answer: V(N 2) =4.48 l ω(NH 4 Cl) =4.9%

29. Potassium nitrite weighing 8.5 g was added while heating to 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (n.s.) of nitrogen will be released in this case and what is the mass fraction of ammonium bromide in the resulting solution?
    

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    KNO 2 + NH 4 Br = N 2 + KBr + 2H 2 O m(NH 4 Br) = 270 g * 0.12 = 32.4 g n(NH 4 Br) = m/M = 32.4 g: 98 g/mol = 0.33 mol n(KNO 2) = m/M = 8.5 g: 85 g/mol = 0.1 mol n(NH 4), react. with KNO 2 = 0.33 mol – 0.1 mol = 0.23 mol (since n(KNO 2) : n(NH 4 Br) = 1:1) m(NH 4 Br) remaining in the final solution = n * M = 0.23 mol * 98 g/mol = 22.54 g n(N 2) = n(KNO 2) = 0.1 mol (N 2) = n * M = 0.1 mol * 28 g/mol = 2.8 g V(N 2) = n * M = 0.1 mol * 22.4 l/mol = 2.24 l m(final solution) = m(KNO 2) + m(NH 4 Br solution) – m(N 2) = 8.5 g + 270 g – 2.8 g = 275.7 g W(NH 4 Br in final solution) = 22.54 g: 275.7 g = 8% Answer: V(N 2) = 2.24 l; W(NH 4 Br) = 8%

30. Mixed 300 ml of sulfuric acid solution with a mass fraction of 10% (density 1.05 g/ml) and 200 ml of potassium hydroxide solution with a mass fraction of 20% (density 1.10 g/ml). How many milliliters of water should be added to the resulting mixture so that the mass fraction of salt in it is 7%?
31. 12.8 g of calcium carbide was added to 120 ml of nitric acid solution with a mass fraction of 7% (density 1.03 g/ml). How many milliliters of 20% hydrochloric acid (density 1.10 g/ml) should be added to the resulting mixture to completely neutralize it?
    

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    1) CaC 2 + 2HNO 3 = Ca(NO 3) 2 + C 2 H 2. n(CaC 2) = 12.8/ 64 = 0.2 mol n(HNO 3) = (0.07 1.03 120) /63 = 0.137 mol CaC 2 – excess. 2) CaC 2 + 2H 2 O = Ca(OH) 2 + C 2 H 2. n(Ca(OH) 2) = 0.2 - 0.137/ 2 = 0.13 mol3) Ca(OH) 2 + 2HCl = CaCl 2 + 2H 2 O. n(HCl) = 0.13 2 = 0.26 mol m(solution) = m(HCl)/ w =(0.26 · 36.5)/ 0.2 = 47.45 g V(HCl solution) = m(solution) /ρ = 47.45/1.10 = 43.1 ml. Answer: 43.1 ml.

32. To the solution obtained by adding 4 g of potassium hydride to 100 ml of water, 100 ml of a 39% solution of nitric acid (r = 1.24 g/ml) was added. Determine the mass fractions of all substances (including water) in the final solution.
    

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    KH + H 2 O = KOH + H 2 n(KH) = 4 g: 40 g/mol = 0.1 mol n(H 2 O) = 100 g: 18 g/mol = 5.6 mol KH is in short supply, then n(KOH) = n(KH) = 0.1 molKOH + H 2 NO 3 = KNO 3 + H 2 O a) m(KOH) = 0.1 * 56 = 5.6 b) m substance (HNO 2) = 100 ml * 1.24 g/ml * 0.39 = 48.36 g n(HNO 3) = 48.36 g: 63 g/mol = 0.77 mol HNO 3 in excess, n excess (HNO 3) = 0.77 – 0.1 = 0.67 mol m(HNO 3) = 0.67 * 63 = 42.21 g m(solution) = 4g + 100g + 124g – 0.2g = 227.8g 3) W(KNO 3) = m(KNO 3) : m(p-pa) = (0.1 mol * 101 g/mol) 227.8 g * 100% = 4.4% W(HNO 3) = 42.21: 227.8 * 100% = 18.5 W(H 2 O) = 100% - (W(KNO 3) + W(HNO 3)) = 77.1% Answer: W(KNO 3) = 4.4% W(HNO 3) =18.5% W(H2O) = 77.1%

    1) 2Na 2 O 2 + 2H 2 O = 4NaOH + O 2
    2) 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O
    m(H 2 SO 4 solution) = 300 * 1.08 = 324 g
    m(H 2 SO 4) = 0.1*324 = 32.4 g
    n(H 2 SO 4) = 32.4/98 = 0.33 mol
    n(NaOH) : n(H 2 SO 4) = 2:1 => n(NaOH) = 0.33 * 2 = 0.66 mol
    n(Na 2 O 2): n(NaOH) = 1:2 => n(Na ​​2 O 2) = 0.66/2 = 0.33 mol
    m(Na 2 O 2) = n * M = 0.33 * 78 = 25.7 g
    n(Na 2 O 2): n(O 2) = 2:1 => n(O 2) = 0.33/2 = 0.165 mol
    V(O 2) = 0.165 * 22.4 = 3.7 l
    Answer: m(Na 2 O 2) = 25.7 g; V(O 2) = 3.7 l

33. When heated, potassium bicarbonate turns into carbonate. Calculate the mass fraction of potassium bicarbonate in the initial solution, by heating which you can obtain an 8% solution of potassium carbonate.
34. When 17.4 g of manganese dioxide reacted with 58 g of potassium bromide in a sulfuric acid medium, bromine was released in a 77% yield. What volume (no.) of propene can react with the resulting amount of bromine?
35. Carbon dioxide with a volume of 5.6 liters (n.s.) was passed through 164 ml of a 20% sodium hydroxide solution (ρ = 1.22 g/ml). Determine the composition and mass fractions of substances in the resulting solution.
36. Aluminum carbide was dissolved in a 15% solution of sulfuric acid weighing 300 g. The methane released in this case occupied a volume of 2.24 liters (n.s.). Calculate the mass fraction of sulfuric acid in the resulting solution.
37. 8 g of sulfur was burned in excess oxygen. The resulting gas was passed through 200 g of an 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution.
38. A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, and 8.96 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters (n.s.) of hydrogen will be released. Calculate the mass fraction of iron in the initial mixture.
39. A mixture of magnesium and zinc filings was treated with an excess of dilute sulfuric acid, and 22.4 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (n.s.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.
    

    2KHCO 3 + H 2 SO 4 = K 2 SO 4 + 2CO 2 + 2H 2 O (2)

    Na 2 CO 3 + 2HCl = NaCl + CO 2 +H 2 O NaHCO 3 +HCl = NaCl + CO 2 +H 2 O NaHCO 3 + NaOH = Na 2 CO 3 + H 2 O 1) we'll findquantityNaHCO3 n(NaOH)=(NaHCO 3)=m(NaOH) . ω (NaOH)/M(NaOH) = 80* 0.1/40 = 0.2 mol 2) we'll findmassNaHCO3 m(NaHCO 3)=n(NaHCO 3)*M(NaHCO 3)=0.2. 84=16.8 g 3) let's find the amount of Na 2 CO 3 n(HCl) = m(HCl) . ω (HCl)/M(HCl) = 73* 0.2/36.5 = 0.4 mol n(Na 2 CO 3) = (n(HCl) - n(NaOH))/2 = 0.1 mol 4) let's find the massNa 2 CO 3 m(Na 2 CO 3)=n. M = 0.1. 106 = 10.6 g 5) let's find the mass fractionNa 2 CO 3 ω (Na 2 CO 3)=m(Na 2 CO 3)/m(mixture). 100%=38.7% m(mixture).=m(Na 2 CO 3)+m(NaHCO 3)=27.4 g Answer: ω (Na 2 CO 3) = 38.7%

    1) 2Na + 2H 2 O → 2NaOH + H 2 2) Na 2 O + H 2 O → 2NaOH 1) let's find the quantityNa n(H 2) = V/Vm = 4.48 / 22.4 = 0.2 mol According to equation (1) n(Na) = n 1 (NaОH) = 2. n(H2) = 0.4 mol 2) find the massNa m(Na) = n. M = 0.4. 23 = 9.2 g 3) let's find the quantityNaOH n(NaOH) = m(p-p) . ω/M(NaOH) = 240. 0.1/40 = 0.6 mol 4) let's find the quantityNa 2 O n(Na 2 O) = (n(NaOH) - n 1 (NaOH))/2=0.1 mol m(Na 2 O)= n. M = 0.1. 62 = 6.2g 5) we'll findman ace shareNa ω (Na) = m(Na)/m(mixtures) . 100%=59.7% m(mixture)=m(Na)+m(Na 2 O)=15.4 Answer: ω(Na)=59,7%

    H 2 SO 4 + Na 2 CO 3 → Na 2 SO 4 + CO 2 + H 2
    1) Let's find the total quantityH2SO4
    n 1 (H 2 SO 4) = m. ω/M=490. 0.4/98=2 mol
    M(H 2 SO 4)=2+32+64=98 g/mol
    2) We'll findH2SO4reactedWithNa 2 CO 3
    n(Na 2 CO 3) = n(Na ​​2 CO 3 . 10H 2 O) = m/M = 143/286 = 0.5 mol
    M(Na 2 CO 3 . 10H 2 O) = (46 + 12 + 48) + (10 . 18) = 286 g/mol
    According to the equation n(H 2 SO 4) = n(Na ​​2 CO 3) = 0.5 mol
    3) we'll findH 2 SO 4 reacted withNaOH
    n(H 2 SO 4)=n 1 -n=2-0.5=1.5 mol
    4) find the mass of NaOH
    n(NaOH)=2n(H 2 SO 4)=2. 1.5=3 mol
    m(NaOH)=n. M =3. 40=120 g
    M(NaOH)=23+16+1=18 g/mol
    5) let's find the mass fractionNaOH
    ω (NaOH)= m(substance)/ . m(p-p)*100%=120/1200 . 100%=10%
    Answer: m(NaOH)=120g; ω (NaOH)=10%
    
    48) A mixture of magnesium and aluminum filings was treated with an excess of dilute hydrochloric acid, and 11.2 liters (n.s.) of hydrogen were released. If the same mass of the mixture is treated with an excess of potassium hydroxide solution, then 6.72 liters (no.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.
    49) Calcium carbide weighing 6.4 g was dissolved in 87 ml of hydrobromic acid (ρ = 1.12 g/ml) with a mass fraction of 20%. What is the mass fraction of hydrogen bromide in the resulting solution?

In this problem there are two parallel reactions involving a mixture of substances.

Problem 3.9.
To neutralize 7.6 g of a mixture of formic and acetic acids, 35 ml of a 20% solution of potassium hydroxide (p = 1.2 g/ml) was used. Calculate the mass of acetic acid and its mass fraction in the original mixture of acids.
Given:
mass of acid mixture: m (acid mixture) = 7.6 g;
volume of KOH solution: V solution (KOH) = 35 ml;
mass fraction of KOH in the original solution: (KOH) in the original solution. r-re = ​​20%;
density of the initial KOH solution: p ref. solution (KOH) = 1.2 g/ml.
Find: mass of acetic acid: m(CH 3 COOH);
mass fraction of CH 3 COOH in the initial mixture: (CH 3 COOH) = ?
Solution:

When a mixture of acids interacts with potassium hydroxide, two reactions occur simultaneously:

CH 3 COOH + KOH → CH 3 COOC + H 2 O
HCOOH + KOH → HCOOOK + H 2 O

The solution to the problem is possible using a counter-algorithm with the compilation of a mathematical equation. The solution scheme can be presented as follows:

1. Let us denote the mass of acetic acid in the initial mixture by the value “a”:

m(CH 3 COOH) = a g.

Then the mass of formic acid can be determined by the difference:

m(HCOOH) = m(mixtures of acids) – m(CH 3 COOH) = (7.6 - a) g.

2. Using the equation for the reaction of acetic acid with an alkali solution, we determine the mass of KOH that was consumed in this interaction.

3. Similarly, using the equation for the reaction with formic acid, we find the mass of KOH, which was consumed in the second reaction.

4. Find the total mass of KOH, which was consumed in two reactions:

m(KOH) = m(KOH) in a district with uks toy + m(KOH) in a district with mur toy =
= 0.933. a + (9.25 – 1.217. a) = (9.25 – 0.284. a) g.

5. Find the mass of KOH contained in 200 ml of the original solution:

6. We equate the expression obtained in step 4 to the value of the KOH mass from step 5:

9.25 – 0.284. a = 8.4.

We got a mathematical equation. Solving it, we find the value of “a”:

We used the value “a” to denote the mass of acetic acid:

m(CH 3 COOH) = 3 g.

7. Find the mass fraction of CH 3 COOH in the initial mixture of acids

Answer: m(CH 3 COOH) = 3 g; (CH 3 COOH) = 39.5%.

Types of problems Problems on mixtures The problem statement contains the words: “mixture”, “technical”, “impurity”, names of minerals or alloys Problems on solutions The problem statement contains the words: “solution”, “mass fraction of dissolved substance” Problems on excess disadvantage The problem statement contains information about both reagents Product yield problems The problem statement contains the words: “substance yield”, “mass fraction of product yield” Problems in which the products of one reaction are used to carry out another reaction.

Here's the problem... To react 6.3 g of a mixture of aluminum and magnesium with sulfuric acid, 275.8 ml of a 10% sulfuric acid solution (density 1.066 g/ml) is required. Determine what mass of a 20% barium chloride solution is required to completely precipitate metal sulfates from the resulting solution.

Let's determine the amount of sulfuric acid in the solution: m (solution) = V (solution). = 278.8. 1.066 = 294 (g) m (H 2 SO 4) = m (solution). w(H 2 SO 4) = ,1 = 29.4 g n (H 2 SO 4) = 29.4 / 98 = 0.3 mol Another 1 point!

Let's determine the mass of the solution: n (BaCl2) = 0.3 mol m (BaCl2) = .3 = 62.4 (g) m (BaCl2 solution) = 62.4: 0.2 = 312 (g) Answer: m (BaCl 2 solution) = 312 g And 1 more point Total - 4 points Total - 4 points

Elements of solving the problem and evaluating the results 1. The equations of all reactions are written correctly. 2. The “amount of substance” (volume, mass) of the starting substances was found 3. A system of equations was compiled to find the amount of substances in the mixture. 4.The final data asked for in the problem has been calculated. All elements were performed correctly - 4 points One mistake was made - 3 points Two errors were made - 2 points Three errors were made - 1 point All elements were performed incorrectly - 0 points

Problems for independent solution A mixture of zinc and zinc carbonate was treated with an excess of hydrochloric acid solution, and 13.44 liters of gas (n.o.) were released. The gas was burned, the combustion products were cooled to the previous temperature, and the volume of the gas decreased to 8.96 liters. What was the composition of the initial mixture of substances?

Homework 1 Sodium nitrite weighing 13.8 g was added while heating to 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (n.s.) of nitrogen will be released in this case and what is the mass fraction of ammonium chloride in the resulting solution? Answer: w(NH 4 Cl) = 4.9%

Homework 2 Potassium nitrite weighing 8.5 g was added when heated to 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (n.s.) of nitrogen will be released in this case and what is the mass fraction of ammonium bromide in the resulting solution? Answer: V(N 2) = 2.24 l, w(NH 4 Br) = 8.2%

Homework 3 Mixed 300 ml of sulfuric acid solution with a mass fraction of 10% (density 1.05 g/ml) and 200 ml of potassium hydroxide solution with a mass fraction of 20% (density 1.10 g/ml). What volume of water should be added to the resulting mixture so that the mass fraction of salt in it is 7%? Answer: V = 262.9 l

Homework 4 12.8 g of calcium carbide was added to 120 ml of nitric acid solution with a mass fraction of 7% (density 1.03 g/ml). What volume of 20% hydrochloric acid (density 1.10 g/ml) should be added to the resulting mixture to completely neutralize it? Answer: V = 43.1 ml

Homework 5 When 8.7 g of manganese dioxide reacted with 22.4 g of potassium bromide in a sulfuric acid medium, bromine was released, the practical yield of which was 88%. What volume (no.) of ethylene can react with the resulting amount of bromine? Answer: V = 1.86 l