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

1 Answer
anor277
Jan 31, 2016

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)?

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