When 20.5 g of methane and 45.0 g of chlorine gas undergo a reaction that has a 75.0% yield, what mass (g) of chloromethane (CH3Cl) forms? Hydrogen chloride also forms.

• Molar mass of $\text{CH}_4 = 12.01 \text{ g/mol (C)} + (4 \times 1.008 \text{ g/mol (H)}) = 16.042 \text{ g/mol}$
• Molar mass of $\text{Cl}_2 = 2 \times 35.45 \text{ g/mol (Cl)} = 70.90 \text{ g/mol}$
• Molar mass of $\text{CH}_3\text{Cl} = 12.01 \text{ g/mol (C)} + (3 \times 1.008 \text{ g/mol (H)}) + 35.45 \text{ g/mol (Cl)} = 50.484 \text{ g/mol}$

From the balanced equation, $1 \text{ mol of CH}_4$ reacts with $1 \text{ mol of Cl}_2$.
To react all $1.2779 \text{ mol of CH}_4$, we would need $1.2779 \text{ mol of Cl}_2$.
Since we only have $0.63469 \text{ mol of Cl}_2$, chlorine ($\text{Cl}_2$) is the limiting reactant.

From the balanced equation, $1 \text{ mol of Cl}_2$ produces $1 \text{ mol of CH}_3\text{Cl}$.

Rounding to three significant figures: