Answer to Question #330207 in General Chemistry for Harry

Question

If 3.2443 moles of carbon dioxide are produced, how many grams of iron (III) oxalate were reacted if the reaction proceeds with a 99.6% yield?

Answer 1

Solution:

carbon dioxide = CO₂

iron(III) oxalate = Fe₂(C₂O₄)₃

Assume that the reaction is the decomposition of iron(III) oxalate

Then, the balanced chemical equation is:

Fe₂(C₂O₄)₃ → 2FeC₂O₄ + 2CO₂

According to stoichiometry:

1 mol of Fe₂(C₂O₄)₃ produces 2 mol of CO₂

X mol of Fe₂(C₂O₄)₃ produces 3.2443 mol of CO₂

Thus,

Moles of Fe₂(C₂O₄)₃ = X = (3.2443 mol CO₂) × (1 mol Fe₂(C₂O₄)₃ / 2 mol CO₂) = 1.62215 mol Fe₂(C₂O₄)₃

Given the yield of the reaction:

Moles of Fe₂(C₂O₄)₃ = (1.62215 mol × 100%) / (99.6%) = 1.628665 mol

The molar mass of Fe₂(C₂O₄)₃ is 375.747 g mol⁻¹

Therefore,

Mass of Fe₂(C₂O₄)₃ = (1.628665 mol) × (375.747 g mol⁻¹) = 611.966 g

Mass of Fe₂(C₂O₄)₃ = 611.966 g

Answer: 611.966 grams of iron(III) oxalate were reacted