Limiting Reagent Worksheet #1 with three chemistry problems on balancing equations and determining limiting reagents.

1. Given the following reaction: (Balance the equation first!)
C₃H₈ + O₂ → CO₂ + H₂O
Balanced equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

a) If you start with 14.8 g of C₃H₈ and 3.44 g of O₂, determine the limiting reagent
- Molar mass of C₃H₈ = 44.1 g/mol
- Molar mass of O₂ = 32.0 g/mol
- Moles of C₃H₈ = 14.8 g / 44.1 g/mol = 0.3356 mol
- Moles of O₂ = 3.44 g / 32.0 g/mol = 0.1075 mol
- From the balanced equation, 1 mol C₃H₈ requires 5 mol O₂.
- O₂ required for 0.3356 mol C₃H₈ = 0.3356 mol × 5 = 1.678 mol O₂
- But only 0.1075 mol O₂ is available, which is less than required.
- Therefore, O₂ is the limiting reagent.

b) determine the number of moles of carbon dioxide produced
- From the balanced equation, 5 mol O₂ produces 3 mol CO₂.
- Moles of CO₂ produced = (3 mol CO₂ / 5 mol O₂) × 0.1075 mol O₂ = 0.0645 mol CO₂

c) determine the number of grams of H₂O produced
- From the balanced equation, 5 mol O₂ produces 4 mol H₂O.
- Moles of H₂O produced = (4 mol H₂O / 5 mol O₂) × 0.1075 mol O₂ = 0.0860 mol H₂O
- Molar mass of H₂O = 18.0 g/mol
- Grams of H₂O produced = 0.0860 mol × 18.0 g/mol = 1.548 g H₂O

d) determine the number of grams of excess reagent left
- Moles of C₃H₈ used = (1 mol C₃H₈ / 5 mol O₂) × 0.1075 mol O₂ = 0.0215 mol C₃H₈
- Moles of C₃H₈ remaining = 0.3356 mol - 0.0215 mol = 0.3141 mol
- Grams of C₃H₈ remaining = 0.3141 mol × 44.1 g/mol = 13.85 g C₃H₈

2. Given the following equation:
Al₂(SO₄)₃ + 6 NaOH → 3 Na₂SO₄ + 2 Al(OH)₃

a) If 10.0 g of Al₂(SO₄)₃ is reacted with 10.0 g of NaOH, determine the limiting reagent
- Molar mass of Al₂(SO₄)₃ = 342.15 g/mol
- Molar mass of NaOH = 40.00 g/mol
- Moles of Al₂(SO₄)₃ = 10.0 g / 342.15 g/mol = 0.02923 mol
- Moles of NaOH = 10.0 g / 40.00 g/mol = 0.2500 mol
- From the balanced equation, 1 mol Al₂(SO₄)₃ requires 6 mol NaOH.
- NaOH required for 0.02923 mol Al₂(SO₄)₃ = 0.02923 mol × 6 = 0.1754 mol NaOH
- Since 0.2500 mol NaOH is available, which is more than required, Al₂(SO₄)₃ is the limiting reagent.

b) Determine the number of moles of Al(OH)₃ produced
- From the balanced equation, 1 mol Al₂(SO₄)₃ produces 2 mol Al(OH)₃.
- Moles of Al(OH)₃ produced = 2 × 0.02923 mol = 0.05846 mol Al(OH)₃

c) Determine the number of grams of Na₂SO₄ produced
- From the balanced equation, 1 mol Al₂(SO₄)₃ produces 3 mol Na₂SO₄.
- Moles of Na₂SO₄ produced = 3 × 0.02923 mol = 0.08769 mol Na₂SO₄
- Molar mass of Na₂SO₄ = 142.04 g/mol
- Grams of Na₂SO₄ produced = 0.08769 mol × 142.04 g/mol = 12.45 g Na₂SO₄

d) Determine the number of grams of excess reagent left over in the reaction
- Moles of NaOH used = 6 × 0.02923 mol = 0.1754 mol
- Moles of NaOH remaining = 0.2500 mol - 0.1754 mol = 0.0746 mol
- Grams of NaOH remaining = 0.0746 mol × 40.00 g/mol = 2.984 g NaOH

3. Given the following equation:
Al₂O₃ + Fe → Fe₃O₄ + Al

Balanced equation: 3Fe + 4Al₂O₃ → 3Fe₃O₄ + 8Al

a) If 25.4 g of Al₂O₃ is reacted with 10.2 g of Fe, determine the limiting reagent
- Molar mass of Al₂O₃ = 101.96 g/mol
- Molar mass of Fe = 55.85 g/mol
- Moles of Al₂O₃ = 25.4 g / 101.96 g/mol = 0.2490 mol
- Moles of Fe = 10.2 g / 55.85 g/mol = 0.1826 mol
- From the balanced equation, 4 mol Al₂O₃ requires 3 mol Fe.
- Fe required for 0.2490 mol Al₂O₃ = (3 mol Fe / 4 mol Al₂O₃) × 0.2490 mol = 0.1868 mol Fe
- Since only 0.1826 mol Fe is available, which is less than required, Fe is the limiting reagent.

b) Determine the number of moles of Al produced
- From the balanced equation, 3 mol Fe produces 8 mol Al.
- Moles of Al produced = (8 mol Al / 3 mol Fe) × 0.1826 mol Fe = 0.4869 mol Al

c) Determine the number of grams of Fe₃O₄ produced
- From the balanced equation, 3 mol Fe produces 3 mol Fe₃O₄.
- Moles of Fe₃O₄ produced = 0.1826 mol Fe × (3 mol Fe₃O₄ / 3 mol Fe) = 0.1826 mol Fe₃O₄
- Molar mass of Fe₃O₄ = 231.55 g/mol
- Grams of Fe₃O₄ produced = 0.1826 mol × 231.55 g/mol = 42.27 g Fe₃O₄

d) Determine the number of grams of excess reagent left over in the reaction
- Moles of Al₂O₃ used = (4 mol Al₂O₃ / 3 mol Fe) × 0.1826 mol Fe = 0.2435 mol Al₂O₃
- Moles of Al₂O₃ remaining = 0.2490 mol - 0.2435 mol = 0.0055 mol
- Grams of Al₂O₃ remaining = 0.0055 mol × 101.96 g/mol = 0.5608 g Al₂O₃