1a. The theoretical yield of Fe₂O₃ obtained is 17.14 g
1b. The percentage yield of Fe₂O₃ obtained is 82.9%
2. The molarity of the KCl solution is 0.04 M
<h3>1a. How to determine the theoretical yield of Fe₂O₃ </h3>
Balanced equation
4Fe + 3O₂ —> 2Fe₂O₃
Molar mass of Fe = 56 g/mol
Mass of Fe from the balanced equation = 4 × 56 = 224 g
Molar mass of Fe₂O₃ = (56×2) + (16×3) = 160 g/mol
Mass of Fe₂O₃ from the balanced equation = 2 × 160 = 320 g
SUMMARY
From the balanced equation above,
224 g of Fe reacted to produce 320 g of Fe₂O₃
Therefore,
12 g of Fe will react to produce = (12 × 320) / 224 = 17.14 g of Fe₂O₃
Thus, the theoretical yield of Fe₂O₃ is 17.14 g
<h3>1b. How to determine the percentage yield </h3>
- Actual yield of Fe₂O₃ = 14.21 g
- Theoretical yield of Fe₂O₃ = 17.14 g
- Percentage yield of Fe₂O₃ =?
Percentage yield = (Actual / Theoretical) × 100
Percentage yield of Fe₂O₃ = (14.21 / 17.14) ×100
Percentage yield of Fe₂O₃ = 82.9%
<h3>2. How to determine the molarity of the KCl solution</h3>
- Mole of KCl = 5.1 moles
- Volume = 126.81 L
- Molarity of KCl =?
Molarity = mole / volume
Molarity of KCl = 5.1 / 126.81
Molarity of KCl = 0.04 M
Learn more about stoichiometry:
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