A student warms 4.65 g of aniline (93 g/mol) with 6.12 g of acetic anhydride (102 g/mol) and a…

Step 1. $n(\text{aniline}) = \dfrac{4.65}{93} = 0.0500\ \text{mol}$ and $n(\text{anhydride}) = \dfrac{6.12}{102} = 0.0600\ \text{mol}$. Aniline is the limiting reagent: $0.0500\ \text{mol}$.

Step 2. Acetic anhydride is in excess by $0.0600 - 0.0500 = 0.0100\ \text{mol}$.

Step 3. Theoretical mass: $0.0500\ \text{mol} \times 135\ \text{g/mol} = 6.75\ \text{g}$.

Step 4. $\%\ \text{yield} = \dfrac{5.40}{6.75} \times 100 = 80.0\ \%$.