Consider the reaction 4Fe(s) + 3O2(g) --> 2Fe2O3(s) How many grams of oxygen are needed to produce 12.5 g of Fe2O3 (iron(III) oxide)?

3.76 g of O₂ are needed to produced 12.5 g of Fe₂O₃

Explanation:

The reaction is:  4Fe (s) + 3O₂ (g)  →  2Fe₂O₃ (s)

4 moles of iron react to 3 moles of oxygen in order to produce 2 moles of iron (III) oxide.

Let's determine the moles of the produced product.

12.5 g . 1mol/ 159.69g = 0.0783 moles

If we assume Iron in excess, we work with the oxygen.

2 moles of Fe₂O₃ are produced by 3 moles of oxygen

Then, 0.0783 moles of Fe₂O₃ might be produced by (0.0783 . 3)/2

0.117 moles.

We convert the moles to mass → 0.117 mol . 32 g/1mol = 3.76 g

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