Question

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

Answer 1

Approx. `135*g` of aluminum metal, and `400*g` of iron oxide are required.

Explanation:

We need (i) a stoichometrically balanced equation:

`Fe_2O_3 + 2Al rarr Al_2O_3 + 2Fe`

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

`=` `(279*g)/(55.85*g*mol^-1)` `=` `5.00*mol`.

Since a `5.00*mol` of iron metal were produced, the given stoiohiometry requires at least a `2.50*mol` quantity of `"ferric oxide"`, and a `5.00*mol` quantity of `"aluminum metal"`.

Given these molar quantities, we calculate equivalent masses of `2.50*molxx159.69*g*mol^-1` `=` `??*g` of iron oxide.

And `5.00*molxx26.98*g*mol^-1` `=` `??*g` of aluminum metal.

This would be an expensive way to produce iron.