Answer
$26.8^{\circ}C$
Work Step by Step
$0.100\text{ L}*\frac{0.500\text{ moles $HCl$}}{1\text{ L}}=0.0500\text{ mol HCl}$
$0.300\text{ L}*\frac{0.100\text{ moles $Ba(OH)_2$}}{1\text{ L}}=0.0300\text{ mol $Ba(OH)_2$}$
$\frac{0.0300\text{ mol $Ba(OH)_2$}}{0.0500\text{ mol $HCl$}}>\frac{1\text{ mol $Ba(OH)_2$}}{2\text{ mol $HCl$}}$
Therefore $HCl$ is the limiting reactant.
$0.0500\text{ mol HCl}*\frac{-118 \text{ kJ}}{2\text{ mol HCl}}=-2.95\text{ kJ}$
$q=mc\Delta T$
$2950=400*4.18*(T-25.0)$
$T=26.8^{\circ}C$
