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

You already have the stoichiometric equation, so you really have already answered this question........

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.

So to your problem:

$\text{Moles of iron required}$ $=$ $\frac{209.7 \cdot g}{55.8 \cdot g \cdot m o {l}^{-} 1} = 3.76 \cdot m o l$.

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.........

I have taken classes to iron foundries where ore is still smelted, and iron metal is produced. It is getting hard to find a iron foundry these days, especially one that is willing to show a bunch of students around a (dangerous!) operation. If you ever see a working foundry you feel very small when you hear the roar of the blast furnace, and feel its incredible heat. These days most of the iron and steel produced is in China, and India.