Question

What is the mass of iron that could be obtained from 67.0 grams of iron(III) oxide?

Answer 1

`67.0` `g` of iron(III) oxide represents `(67.0*cancelg)/(159.69*cancel(g)*mol^-1)`, approx. `0.4` `mol`.

Explanation:

So, if I smelt this quantity of ferric oxide I can get AT MOST 0.4 mol of iron metal, approx. 20 g of the stuff.

Of course, you will have to check the calculation yourself; I may have made a mistake.

In an iron foundry, these metal oxides are reduced on industrial scales. What do they use to reduce the iron from Fe(III) to Fe(O)?