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?

a. 0.193 moles of Fe, actual yield

b. 0.640 moles of Fe, the theoretical yield

c. 30 % yield

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%

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .