A student has the two beakers shown in the figure…

Step 1. Moles in each beaker. Acid: $0.300\ \text{mol L}^{-1} \times 0.200\ \text{L} = 0.0600\ \text{mol}$. Base: $0.500\ \text{mol L}^{-1} \times 0.300\ \text{L} = 0.1500\ \text{mol}$.

Step 2. The limiting reagent. The acid needs $2 \times 0.0600 = 0.1200\ \text{mol}$ of $\text{NaOH}$. There is $0.1500\ \text{mol}$, more than enough. So the acid is used up and sodium hydroxide is in excess. The friend is right this time, but for the wrong reason. Left: $0.1500 - 0.1200 = 0.0300\ \text{mol}$ of $\text{NaOH}$.

Step 3. Its molarity. $\dfrac{0.0300\ \text{mol}}{0.500\ \text{L}} = 0.0600\ \text{mol L}^{-1}$.

Step 4. Sodium sulfate. One mole of acid gives one mole of salt: $0.0600\ \text{mol}$. Molarity: $\dfrac{0.0600\ \text{mol}}{0.500\ \text{L}} = 0.120\ \text{mol L}^{-1}$.

Step 5. Sodium ions. All $0.1500\ \text{mol}$ of sodium from beaker B is still there, in the salt and in the leftover base. Molarity: $\dfrac{0.1500\ \text{mol}}{0.500\ \text{L}} = 0.300\ \text{mol L}^{-1}$.

Step 6. The dry solid. Sodium sulfate ($142\ \text{g mol}^{-1}$): $0.0600\ \text{mol} \times 142\ \text{g mol}^{-1} = 8.52\ \text{g}$. Sodium hydroxide ($40\ \text{g mol}^{-1}$): $0.0300\ \text{mol} \times 40\ \text{g mol}^{-1} = 1.20\ \text{g}$. Total: $9.72\ \text{g}$.