A student mixes 15.6 g of benzene (78.11 g/mol) with 7.85 g of ethanoyl chloride (78.50 g/mol) and…

Step 1. Moles. Benzene: $\dfrac{15.6}{78.11} = 0.200\ \text{mol}$. Ethanoyl chloride: $\dfrac{7.85}{78.50} = 0.100\ \text{mol}$. The acid chloride is limiting, so the limiting amount is $100\ \text{mmol}$.

Step 2. Theory. $0.100\ \text{mol} \times 120.15\ \text{g/mol} = 12.0\ \text{g}$.

Step 3. Actual. $12.0\ \text{g} \times 0.78 = 9.37\ \text{g}$.

Step 4. Aluminium chloride. $1.1 \times 0.100\ \text{mol} \times 133.34\ \text{g/mol} = 14.7\ \text{g}$.