Answer:
b) CO is the limiting reactant.
Step-by-step explanation:
1. Convert the given masses of reactants to moles using their molar masses.
2. Determine the limiting reactant.
3. Calculate the theoretical yield using the stoichiometry of the reaction.
4. Calculate the percent yield.
1. Convert masses to moles
Molar masses:
- Fe2O3: 2 x 55.85 (for Fe) + 3 x 16.00 (for O) = 159.70 g/mol
- CO: 12.01 (for C) + 16.00 (for O) = 28.01 g/mol
Given masses:
- Fe2O3: 167 g
- CO: 85.8 g
Moles of Fe2O3 = 167 / 159.70
Moles of CO = 85.8 / 28.01
The moles of each reactant are:
- Fe2O3: 1.0457 moles
- CO: 3.0632 moles
2. Determine the limiting reactant
From the balanced chemical equation, 1 mole of Fe2O3 reacts with 3 moles of CO.
Thus, the amount of CO required to completely react with 1.0457 moles of Fe2O3 is:
Required CO = 1.0457 x 3
The amount of CO required to completely react with 1.0457 moles of Fe2O3 is 3.1371 moles. Since we only have 3.0632 moles of CO, CO is the limiting reactant.
3. Calculate the theoretical yield using the stoichiometry of the reaction
From the balanced equation, 1 mole of Fe2O3 produces 2 moles of Fe.
The molar mass of Fe is 55.85 g/mol.
Given that CO is the limiting reactant, the theoretical yield of Fe is:
Theoretical yield = 2 x 1.0457 moles x 55.85 g/mol
The theoretical yield of Fe is 116.81 g.
4. Calculate the percent yield
The percent yield is given by:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
Given:
Actual Yield of Fe = 72.3 g
Using the above formula:
Percent Yield = (72.3 g / 116.81 g) x 100
The percent yield of Fe is approximately 61.90%.
To summarise:
a) Fe2O3 is not the limiting reactant.
b) CO is the limiting reactant.
c) The theoretical yield is 116.81 g.
d) The percent yield is 61.90%.
Thus, from the provided options, only option b is correct.