Moles of Benzene Formed from Reduction of Phenol with Zinc

When phenol ($\text{C}_6\text{H}_5\text{OH}$) is heated with zinc ($\text{Zn}$) dust, it undergoes reduction to form benzene ($\text{C}_6\text{H}_6$), which is product $\text{X}$, along with zinc oxide ($\text{ZnO}$).

The balanced chemical equation for the reduction reaction is: $\text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO}$

Step 1: Calculate the molar mass of phenol ($\text{C}_6\text{H}_5\text{OH}$): $\text{Molar mass of phenol} = (6 \times 12) + (6 \times 1) + 16 = 72 + 6 + 16 = 94 \text{ g mol}^{-1}$

Step 2: Calculate the initial number of moles of phenol: $\text{Initial moles of phenol } (n_{\text{phenol}}) = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4.7 \text{ g}}{94 \text{ g mol}^{-1}} = 0.05 \text{ mol}$

Step 3: Calculate the moles of benzene ($\text{X}$) formed based on reaction completion: Given that the reaction goes to $60\%$ completion, the number of moles of phenol converted into product $\text{X}$ is: $n_{\text{reacted}} = 0.05 \text{ mol} \times \frac{60}{100} = 0.03 \text{ mol}$

From the stoichiometry of the reaction, $1 \text{ mole}$ of phenol gives $1 \text{ mole}$ of benzene ($\text{X}$). Therefore, the number of moles of product $\text{X}$ formed is: $n_{\text{X}} = 0.03 \text{ mol} = 3 \times 10^{-2} \text{ mol}$

Thus, the required value is 3.