It is chemistry question

The reaction is a free radical substitution.
$\text{CH}_4\text{(g)} + \text{Cl}_2\text{(g)} \xrightarrow{\text{UV light}} \text{CH}_3\text{Cl(g)} + \text{HCl(g)}$

From the equation, 1 mole of $\text{CH}_4$ reacts with 1 mole of $\text{Cl}_2$. The mole ratio is 1:1.

Molar mass of $\text{CH}_4 = 12.01 + 4 \times 1.01 = 16.05 \text{ g/mol}$
Moles of $\text{CH}_4 = (32.1 \text{ g}) / (16.05 \text{ g/mol}) = 2.00 \text{ mol}$

Molar mass of $\text{Cl}_2 = 2 \times 35.45 = 70.90 \text{ g/mol}$
Moles of $\text{Cl}_2 = (70.9 \text{ g}) / (70.90 \text{ g/mol}) = 1.00 \text{ mol}$

Since the mole ratio is 1:1, and we have 2.00 mol of $\text{CH}_4$ and 1.00 mol of $\text{Cl}_2$, chlorine ($\text{Cl}_2$) is the limiting reactant.

From the balanced equation, 1 mole of $\text{Cl}_2$ produces 1 mole of $\text{CH}_3\text{Cl}$.
Therefore, 1.00 mol of $\text{Cl}_2$ will produce 1.00 mol of $\text{CH}_3\text{Cl}$.

Molar mass of $\text{CH}_3\text{Cl} = 12.01 + 3 \times 1.01 + 35.45 + 1.01 = 49.49 \text{ g/mol}$
Mass of $\text{CH}_3\text{Cl} = (\text{moles of } \text{CH}_3\text{Cl}) \times (\text{molar mass of } \text{CH}_3\text{Cl})$
Mass of $\text{CH}_3\text{Cl} = (1.00 \text{ mol}) \times (49.49 \text{ g/mol}) = 49.49 \text{ g}$

The mass of chloromethane produced is $49.49 \text{ g}$.