Limiting Reactant and Percent Yield Worksheet Consider the following reaction: HCl + H2O -> H3O+ + Cl- If 21 moles of HCl react with 6.96 moles of H2O, what are the limiting reactants? Consider the following reaction: FeS2 + 10O2 -> Fe2O3 + 8SO2 When 26.62 moles of FeS2 reacts with 59.44 moles of O2, what are the limiting and excess reactants? Consider the following reaction: Si + 2N2 -> Si3N4 When 600 g of Si reacts with 500 g of N2, what are the limiting and excess reactants? Given the following equation: 2Al(SO3)3 + 6NaOH -> 3Na2SO3 + 2Al(OH)3 If 10.0 g of Al(SO3)3 reacted with 10.0 g of NaOH, determine the limiting and excess reactants
The Correct Answer and Explanation is:
Limiting Reactant and Percent Yield Worksheet Solutions
Problem 1:
Reaction: HCl + H₂O → H₃O⁺ + Cl⁻
Given: 21 moles of HCl and 6.96 moles of H₂O
This reaction has a 1:1 molar ratio between HCl and H₂O. That means 1 mole of HCl reacts with 1 mole of H₂O.
- Required amount of H₂O for 21 moles of HCl:
21 moles HCl × (1 mole H₂O / 1 mole HCl) = 21 moles H₂O
Available: 6.96 moles H₂O
Conclusion: H₂O is the limiting reactant because there is not enough to react with all the HCl.
Problem 2:
Reaction: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
Given: 26.62 moles FeS₂ and 59.44 moles O₂
- Required O₂ for 26.62 moles FeS₂:
26.62 mol FeS₂ × (11 mol O₂ / 4 mol FeS₂) = 73.19 moles O₂
Available: 59.44 moles O₂
Conclusion: O₂ is the limiting reactant. FeS₂ is the excess reactant.
Problem 3:
Reaction: 3Si + 2N₂ → Si₃N₄
Molar masses:
- Si = 28.09 g/mol
- N₂ = 28.02 g/mol
Moles of Si:
600 g ÷ 28.09 g/mol = 21.37 moles Si
Moles of N₂:
500 g ÷ 28.02 g/mol = 17.85 moles N₂
Required N₂ for 21.37 moles Si:
21.37 mol Si × (2 mol N₂ / 3 mol Si) = 14.25 moles N₂
Available: 17.85 moles N₂
Conclusion: Si is the limiting reactant because it runs out first. N₂ is in excess.
Problem 4:
Reaction: 2Al(SO₃)₃ + 6NaOH → 3Na₂SO₃ + 2Al(OH)₃
Molar masses:
- Al(SO₃)₃ = 342.15 g/mol
- NaOH = 40.00 g/mol
Moles of Al(SO₃)₃:
10.0 g ÷ 342.15 g/mol = 0.02923 moles
Moles of NaOH:
10.0 g ÷ 40.00 g/mol = 0.25 moles
Required NaOH for 0.02923 moles Al(SO₃)₃:
0.02923 mol × (6 mol NaOH / 2 mol Al(SO₃)₃) = 0.0877 moles NaOH
Available: 0.25 moles NaOH
Conclusion: Al(SO₃)₃ is the limiting reactant. NaOH is in excess.
Explanation Summary
The limiting reactant is the substance that gets completely consumed first during a chemical reaction, limiting the amount of product formed. To identify the limiting reactant, you compare the actual mole ratio of reactants with the required stoichiometric ratio from the balanced equation. The reactant that cannot meet the required amount is the limiting reactant. Once the limiting reactant is used up, the reaction stops, even if other reactants are still present. In each problem above, careful mole calculations and conversions help determine which substance is limiting and which is in excess, based on the balanced chemical equation and given quantities.
