# Iron (III) oxide reacts with carbon monoxide to form molten iron and carbon dioxide. If .18 g of iron (III) oxide reacts with 0.11 g of carbon monoxide, how many grams of iron would be produced?

$F {e}_{2} {O}_{3} \left(s\right) + 3 C O \left(g\right) \rightarrow 2 F e \left(l\right) + 3 C {O}_{2} \left(g\right)$
$\text{Moles of ferric oxide } = \frac{0.18 \cdot g}{159.69 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.13 \times {10}^{-} 3 \cdot m o l \text{ metal oxide}$.
$\text{Moles of carbon monoxide } = \frac{0.11 \cdot g}{28.0 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.93 \times {10}^{-} 3 \cdot m o l \text{ CO}$.
$F {e}_{2} {O}_{3}$ is the reagent in deficiency (why?), and thus $2 \times 1.13 \times {10}^{-} 3 \cdot m o l \times 55.85 \cdot g \cdot m o {l}^{-} 1 \cong 0.150 \cdot g$ iron metal are produced.