# To produce a mass of 279*g mass of iron metal, what quantities of aluminum, and ferric oxide are required?

Sep 25, 2016

#### Answer:

Approx. $135 \cdot g$ of aluminum metal, and $400 \cdot g$ of iron oxide are required.

#### Explanation:

We need (i) a stoichometrically balanced equation:

$F {e}_{2} {O}_{3} + 2 A l \rightarrow A {l}_{2} {O}_{3} + 2 F e$

And (ii) the molar quantity of iron metal produced:

$=$ $\frac{279 \cdot g}{55.85 \cdot g \cdot m o {l}^{-} 1}$ $=$ $5.00 \cdot m o l$.

Since a $5.00 \cdot m o l$ of iron metal were produced, the given stoiohiometry requires at least a $2.50 \cdot m o l$ quantity of $\text{ferric oxide}$, and a $5.00 \cdot m o l$ quantity of $\text{aluminum metal}$.

Given these molar quantities, we calculate equivalent masses of $2.50 \cdot m o l \times 159.69 \cdot g \cdot m o {l}^{-} 1$ $=$ ??*g of iron oxide.

And $5.00 \cdot m o l \times 26.98 \cdot g \cdot m o {l}^{-} 1$ $=$ ??*g of aluminum metal.

This would be an expensive way to produce iron.