Question

(1.) Consider the equation: 4Fe + 3O2 yields 2Fe2O3
- What is the theoretical yield of Fe2O3 when 12.0g of iron rusts?
- If 14.21 g of Iron (III) oxide is actually produced what is the percent yield ?

(2.) what is the molarity of a 126.81 L solution that contains 5.1 moles of KCl ?

Answer 1

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

1a. How to determine the theoretical yield of Fe₂O₃

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

1b. How to determine the percentage yield

• 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%

2. How to determine the molarity of the KCl solution

• 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

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