Question

Question #7580c

Answer 1

The yield of Fe₂O₃ is 79.3 %.

This is a three-part question. "They" gave you the masses of two reactants, so the first problem is

1. Determine the limiting reactant. Then
2. Calculate the theoretical yield. Finally
3. Calculate the percentage yield.

Determine the limiting reactant.

Write the balanced equation.

4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

Calculate the amount of Fe₂O₃ formed from each reactant.

From FeS₂: 300 g FeS₂ × `(1" mol FeS₂")/(120.0" g FeS₂") × (2" mol Fe₂O₂")/(4" mol FeS₂")` = 1.250 mol Fe₂O₃

From O₂: 200 g O₂ × `(1" mol O₂")/(32.00" g O₂") × (2" mol Fe₂O₃")/(11" mol O₂")` = 1.136 mol Fe₂O₃

Since O₂ gives the smallest amount of Fe₂O₃, O₂ is the limiting reactant.

Calculate the Theoretical Yield

Theoretical yield = 1.136 mol Fe₂O₃ × `(158.7" g Fe₂O₃")/(1" mol Fe₂O₃")` = 180 g Fe₂O₃

Calculate the Percentage Yield

Percentage Yield =

`"Actual yield"/"Theoretical yield" × 100 % = (143" g")/(180" g") × 100 %` = 79.3 %

Hope this helps.