46.4 g of iron oxide was reduced with hydrogen. In this case, 33.6 g of iron were obtained.

46.4 g of iron oxide was reduced with hydrogen. In this case, 33.6 g of iron were obtained. Set the formula for the starting iron oxide.

Given:
m (FexOy) = 46.4 g
m (Fe) = 33.6 g

To find:
FexOy -?

Decision:
1) FexOy + H2 -> Fe + H2O;
2) m (O in FexOy) = m (FexOy) – m (Fe) = 46.4 – 33.6 = 12.8 g;
3) n (O in FexOy) = m (O in FexOy) / M (O) = 12.8 / 16 = 0.8 mol;
4) n (Fe) = m (Fe) / M (Fe) = 33.6 / 56 = 0.6 mol;
5) n (Fe in FexOy) = n (Fe) = 0.6 mol;
6) x: y = n (Fe in FexOy): n (O in FexOy) = 0.6: 0.8 = 3: 4;
The formula of the initial oxide is Fe3O4 – iron oxide (II, III).

Answer: The formula of the initial oxide is Fe3O4 – iron oxide (II, III).