# How many grams of CO are needed to react wit an excess of Fe_2O_3 to produce 209.7 g Fe?

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

Apr 6, 2017

#### Explanation:

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

This equation UNEQUIVOCALLY tells you that $3 \cdot m o l$, i.e. $84 \cdot g$ carbon monoxide reduces $1 \cdot m o l$ $\text{ferric oxide}$ to give $2 \cdot m o l$, iron metal, i.e. $2 \cdot m o l \times 55.8 \cdot g \cdot m o {l}^{-} 1 = 111.6 \cdot g$. I hope you are happy with the mole concept; if not, raise a flag and someone will help you.

$\text{Moles of iron required}$ $=$ $\frac{209.7 \cdot g}{55.8 \cdot g \cdot m o {l}^{-} 1} = 3.76 \cdot m o l$.
Given the stoichiometry of the equation, we require AT LEAST $\frac{1}{2} \times 3.76 \cdot m o l$ $F {e}_{2} {O}_{3}$, i.e. $\frac{1}{2} \times 3.76 \cdot m o l \times 159.7 \cdot g \cdot m o {l}^{-} 1$
$= 300 \cdot g$ $\text{ferric oxide}$..........and THEREFORE we require $\frac{3}{2} \times 3.76 \cdot m o l \times 28.01 \cdot g \cdot m o {l}^{-} 1$
$= 158 \cdot g$ $\text{CO}$ as reductant.........