Answer:
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
