Will 23.2 grams of iron oxide be enough to produce 16.8 grams of iron?

Given:
m (Fe3O4) = 23.2 g
m (Fe) = 16.8 g

To find:
m (Fe3O4) – enough?

Decision:
1) Fe3O4 + 4CO => 3Fe + 4CO2;
2) M (Fe) = Mr (Fe) = Ar (Fe) = 56 g / mol;
M (Fe3O4) = Mr (Fe3O4) = Ar (Fe) * N (Fe) + Ar (O) * N (O) = 56 * 3 + 16 * 4 = 232 g / mol;
3) n (Fe3O4) = m (Fe3O4) / M (Fe3O4) = 23.2 / 232 = 0.1 mol;
4) n (Fe) = m (Fe) / M (Fe) = 16.8 / 56 = 0.3 mol;
5) n must. (Fe3O4) = n (Fe) / 3 = 0.3 / 3 = 0.1 mol;
6) n (Fe3O4) = n (Fe3O4).

Answer: The mass of Fe3O4 (23.2 g) will be enough.