Question

Consider the reaction below.

If you start with 0.39 moles of C ₃H ₈ (propane) and 7.0 moles of O ₂, what is the percent yield if 0.51 moles of carbon dioxide is produced?
C ₃H ₈(g) + 5 O ₂(g) → 3 CO ₂(g) + 4 H ₂O(g)

Answer 1

The required answer is 43.59

EXPLANATION -

The reaction is

We have 0.39 mol propane and 7.0 mol O₂.

From the stoichiometric reaction, we can see that

1 mol C₃H₈ reacts with 5 mol O₂.

So 0.39 mol C₃H₈ reacts with 5 * 0.39 mol O₂.

= 1.95 mol O₂.

But we have 7.0 mol O₂. So O₂ is present in excess and C₃H₈ is the limiting reagent.

The yield of the products will depend on the amount of limiting reagent present.

From the stoichiometric reaction, we can see that

1 mol C₃H₈ produces 3 mol CO₂.

So 0.39 mol C₃H₈ should produce 3 * 0.39 mol CO₂.

= 1.17 mol CO₂.

This is the theoretical yield of the product. The actual yield is usually less than the theoretical yield. Sometimes they may be equal (very rare cases) but actual yield can never be more than the theoretical yield.

Actual yield of CO₂ is 0.51 mol (given).

So

  

= 43.59

The required answer is 43.59