When 10.0 grams of iron react with oxygen, the balanced chemical equation is:
4Fe + 3O2 → 2Fe2O3
This means that for every 4 moles of iron that react, we produce 2 moles of iron (III) oxide.
First, we need to convert the mass of iron to moles:
10.0 g Fe × (1 mol Fe / 55.85 g Fe) = 0.179 mol Fe
Now we can use the mole ratio from the balanced equation to find the number of moles of iron (III) oxide produced:
0.179 mol Fe × (2 mol Fe2O3 / 4 mol Fe) = 0.0895 mol Fe2O3
Finally, we can convert the number of moles of iron (III) oxide to grams:
0.0895 mol Fe2O3 × (159.69 g Fe2O3 / 1 mol Fe2O3) = 14.3 g Fe2O3
Therefore, 10.0 grams of iron reacting with oxygen produces 14.3 grams of iron (III) oxide.