A student reduces 8.65 g of butanone to butan-2-ol with sodium borohydride in ethanol…

Step 1. Butanone. $\dfrac{8.65\ \text{g}}{72.107\ \text{g/mol}} = 0.1200\ \text{mol}$.

Step 2. Borohydride. Four ketone molecules per NaBH$_4$: $\dfrac{0.1200\ \text{mol}}{4} = 0.0300\ \text{mol}$. Mass: $0.0300 \times 37.832 = 1.135\ \text{g}$.

Step 3. Alcohol. Theoretical: $0.1200 \times 74.123 = 8.89\ \text{g}$. At $82\ \%$: $8.89 \times 0.82 = 7.29\ \text{g}$.