# Ferric oxide may be reduced to pure iron either with carbon monoxide or with coke (pure carbon). Suppose that 150.0 lb of ferric oxide are available. How many pounds of carbon monoxide would be required to reduce the oxide?

## How many pounds of coke would be needed? In each case, how many pounds of pure iron would be produced?

##### 1 Answer
May 28, 2016

$C \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow C O \left(g\right)$
$F {e}_{2} {O}_{3} \left(s\right) + 3 C O \left(g\right) \rightarrow 2 F e \left(s\right) + 3 C {O}_{2} \left(g\right)$

#### Explanation:

The answer relates to the stoichiometric equation, which tells us that $159.7 \cdot g$ $F {e}_{2} {O}_{3}$ reacts with $84.0 \cdot g$ $C O$ to give $112 \cdot g$ $F e$.

You have $150 \cdot {\text{lbs"xx454*g*"lb}}^{-} 1$ $=$ $68 , 100 \cdot g$ $F {e}_{2} {O}_{3}$. Given the equations you should be able to tells us the quantities of steel produced, and the quantity of gas required. If you have difficulties, state them here, and someone will help you.