17kg of aluminum was produced from 51kg of aluminum oxide (Al2O3) by electrolysis. What was the percentage yield (AR, Al=27,O=16).

Question

Qb.in the manufacturer of fertilizer ammonium sulphate H2SO⁴ +NH3 =(NH4)2SO4.What is the maximum mass of ammonium sulphate that can be abstain from 2kg of sulphuric acid and 1kg of ammonia?
Qc.In the blast furnace,the overall reaction is :2Fe2O3(s) + 3C=3CO2(s)+4Fe(s
).What is the maximum mass of iron that can be obtained from 700 to one's of iron (iii)oxide and 70 tones of coke?(1tonne=1000kg)

Answer 1

a) Calculating the theoretical yield of aluminum:
Molar mass of Al2O3 = 2(27) + 3(16) = 54 + 48 = 102 g/mol
51 kg of Al2O3 = 51000 g
Number of moles of Al2O3 = 51000 g / 102 g/mol = 500 mol
From the balanced equation, 2 moles of Al are produced from 1 mole of Al2O3
Number of moles of Al produced = 500 mol * 2 = 1000 mol
Molar mass of Al = 27 g/mol
Mass of Al produced = 1000 mol * 27 g/mol = 27000 g = 27 kg

Percentage yield = (Actual yield / Theoretical yield) * 100
Percentage yield = (17 kg / 27 kg) * 100 = 62.96%

b) Calculating the maximum mass of ammonium sulphate:
1 kg of H2SO4 = 1000 g
Molar mass of H2SO4 = 1(1) + 32 + 4(16) = 98 g/mol
Number of moles of H2SO4 = 1000 g / 98 g/mol ≈ 10.20 mol
Molar mass of NH3 = 14 g/mol
Mass of NH3 = 1000 g
Number of moles of NH3 = 1000 g / 14 g/mol ≈ 71.43 mol
From the balanced equation, 1 mole of H2SO4 reacts with 2 moles of NH3 to produce 1 mole of (NH4)2SO4
Number of moles of (NH4)2SO4 produced = 10.20 mol / 1 * 71.43 mol / 2 ≈ 36.07 mol
Molar mass of (NH4)2SO4 = 18 + 2(1) + 32 + 4(16) = 132 g/mol
Max mass of (NH4)2SO4 = 36.07 mol * 132 g/mol ≈ 4762.40 g ≈ 4.76 kg

c) Calculating the maximum mass of iron:
700 tonnes of Fe2O3 = 700000 kg
Molar mass of Fe2O3 = 2(56) + 3(16) = 112 + 48 = 160 g/mol
Number of moles of Fe2O3 = 700000 kg / 160 g/mol = 4375 mol
From the balanced equation, 2 moles of Fe are produced from 2 moles of Fe2O3
Number of moles of Fe produced = 4375 mol * 2 = 8750 mol
Molar mass of Fe = 56 g/mol
Mass of Fe produced = 8750 mol * 56 g/mol = 490000 g = 490 kg

Therefore, the maximum mass of iron that can be obtained is 490 kg.