Answer:
a. Theoretical yield of Si = <u>72.58 kg</u>
b. Percent Yield =<u> 91.2 %</u>
Explanation:
The chemical reaction involved is: SiO₂ (l) + 2C (s) → Si (l) + 2CO (g)
The stoichiometric ratio = 1 mol SiO₂ ÷ 2 mol C = 0.5 mol SiO₂ ÷ 1 mol C
Given: mass of SiO₂: w₁ = 155.3kg = 155.3 × 10³g, mass of C: w₂ = 79.3kg = 79.3 × 10³g, mass of Si: w₃ = 66.2kg = 66.2 × 10³g (∵ 1 kg =1000g)
Molar mass of SiO₂: m₁ = 60.08 g/mol, atomic mass of C: m₂ = 12.01 g/mol, atomic mass of Si: m₃ = 28.08 g/mol
Number of moles of SiO₂ taken: n₁ = w₁ ÷ m₁ = 155300 g ÷ 60.08 g/mol = 2584.88 mol
Number of moles of C taken: n₂ = w₂ ÷ m₂ = 79300 g ÷ 12.01 g/mol = 6602.83 mol
Actual ratio = n₁ ÷ n₂ = 2584.88 mol SiO₂ ÷ 6602.83 mol C = 0.39 mol SiO₂ ÷ 1 mol C
Since the stoichiometric ratio > actual ratio
<u>Therefore, SiO₂ is the limiting reagent.</u>
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Now, the mass of SiO₂ in 1 mol = number of moles × molar mass = 1 mol × 60.08 g/mol = 60.08 g
the mass of Si in 1 mol = number of moles × molar mass = 1 mol × 28.08 g/mol = 28.08 g
So 28.08 g of Si is produced from 60.08 g SiO₂
Therefore, the amount of Si produced from 155300 g SiO₂ = 155300 g × 28.08 g ÷ 60.08 g = 72583.62 g = 72.58 kg
<u>Therefore, the </u><u>theoretical yield</u><u> of Si = 72.58 kg</u>
<u>Actual yield</u><u> of Si = 66.2 kg</u>
Therefore,<u> </u><u>Percent Yield</u><u> </u>= Actual yield ÷ Theoretical yield × 100 = 66.2 kg ÷ 72.58 kg × 100 =<u> 91.2 %</u>
