Question

Combining 0.320 mol Fe2O3 with excess carbon produced 10.8 g Fe . Fe 2 O 3 + 3 C ⟶ 2 Fe + 3 CO a. What is the actual yield of iron in moles? b. What is the theoretical yield of iron in moles?c. What is the percent yield?

Answer 1

a. 0.193 moles of Fe, actual yield

b. 0.640 moles of Fe, the theoretical yield

c. 30 % yield

Step-by-step explanation:

The reaction is this: Fe₂O₃ + 3C → 2Fe + 3CO

First of all, we determine the moles of produced Fe.

10.8 g / 55.85 g/mol = 0.193 moles → Actual yield of iron in moles

We have 0.320 moles, so we can calculate the theoretical yield by a rule of three.

1 mol of Fe₂O₃ can produce 2 moles of Fe

0.320 moles of oxide, may produce (0.320 . 2) / 1 = 0.640 moles

As we determined the theoretical yield and the actual yield, we can know the percent yield → (Produced yield / Theoretical Yield) . 100 = % yield

(0.193 mol / 0.640 mol) . 100 = 30%