45 g of ethanoic acid is warmed with 46 g of ethanol and a few drops of concentrated sulphuric acid.

Step 1. Moles. Acid: $45\ \text{g} \div 60\ \text{g/mol} = 0.75\ \text{mol}$. Ethanol: $46\ \text{g} \div 46\ \text{g/mol} = 1.0\ \text{mol}$.

Step 2. Limit. The acid runs out first, so $0.75\ \text{mol}$ of ester forms: $0.75 \times 88 = 66\ \text{g}$.

Step 3. Left over. $1.0 - 0.75 = 0.25\ \text{mol}$ of ethanol, which is $0.25 \times 46 = 11.5\ \text{g}$.