The mass of iron converted into Fe$$_3$$O$$_4$$ by the action of 18 g of steam is : (Given : Molar mass of H, O and Fe are 1, 16 and 56 g mol$$^{-1}$$ respectively) Assume iron is present in excess :

The reaction of iron metal with steam is

$$3\,Fe \;+\; 4\,H_2O(g) \;\longrightarrow\; Fe_3O_4 \;+\; 4\,H_2(g)$$

Step 1 : Calculate moles of steam available.
Mass of steam given = $$18 \text{ g}$$.
Molar mass of $$H_2O$$ = $$2(1) + 16 = 18 \text{ g mol}^{-1}$$.
Hence

$$n_{H_2O} = \frac{18}{18} = 1 \text{ mol}$$

Step 2 : Use stoichiometry of the balanced equation.
According to the equation, $$4$$ mol of $$H_2O$$ react with $$3$$ mol of $$Fe$$.
Therefore

$$1 \text{ mol } H_2O \; \text{will react with} \; \frac{3}{4} \text{ mol } Fe = 0.75 \text{ mol}$$

Step 3 : Convert moles of iron to mass.
Molar mass of $$Fe$$ = $$56 \text{ g mol}^{-1}$$.

$$m_{Fe} = 0.75 \times 56 = 42 \text{ g}$$

Thus, when iron is in excess, $$18 \text{ g}$$ of steam can convert $$42 \text{ g}$$ of iron into $$Fe_3O_4$$.

Option D which is: $$42$$ g