# Combining 0.265 mol of Fe_2O_3 with excess carbon produced 14.7 g of Fe. Fe_2O_3 + 3C -> 2Fe +3CO. What is the actual yield of iron in moles?

Jan 18, 2017

Approx. 50%

#### Explanation:

You have the chemical equation, which shows that each mole of ferric oxide should yield 2 moles of iron metal:

$F {e}_{2} {O}_{3} \left(s\right) + 3 C \left(s\right) \rightarrow 2 F e \left(l\right) + 3 C O \left(g\right)$

$\text{Moles of iron}$ $=$ $\frac{14.7 \cdot g}{55.8 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.263 \cdot m o l$

Because stoichiometric reaction would have yielded $2 \times 0.265 \cdot m o l$ of iron metal, the yield of the reaction is just under 50%.