18 g of H₂O reacts with x g of Fe to produce Fe₃O₄. Find x (Fe = 56 u)

The correct option is 42 g.

To find the mass of iron ($\mathrm{Fe}$) required to react with 18 g of water ($H_{2}O$) to produce magnetite (${\mathrm{Fe}}_3{O}_4$), we first write the balanced chemical equation for the reaction:
$3\mathrm{Fe}+4H_{2}O\to {\mathrm{Fe}}_3{O}_4+4H_2$

From the balanced equation, we can see that:
3 moles of $\mathrm{Fe}$ react with 4 moles of $H_{2}O$.

Let's calculate the molar masses of the substances involved:
Molar mass of $H_{2}O$ = $2\times 1+16=18\text{g/mol}$
Molar mass of $\mathrm{Fe}$ = $56\text{g/mol}$

Next, we determine the number of moles of $H_{2}O$ given in the question:
$\text{Moles of}{H}_{2}O=\frac{18\text{g}}{18\text{g/mol}}=1\text{mol}$

According to the stoichiometry of the balanced reaction:
4 moles of $H_{2}O$ require = 3 moles of $\mathrm{Fe}$
Therefore, 1 mole of $H_{2}O$ requires:
$\text{Moles of}\mathrm{Fe}=\frac{3}{4}\times 1=0.75\text{moles}$

Now, we calculate the mass $x$ of $\mathrm{Fe}$ corresponding to 0.75 moles:
$x=\text{Moles of}\mathrm{Fe}\times \text{Molar mass of}\mathrm{Fe}$
$x=0.75\times 56\text{g}=42\text{g}$

Thus, 42 g of iron is required for the reaction.