## Chemistry 9th Edition

Published by Cengage Learning

# Chapter 6 - Thermochemistry - Exercises - Page 288: 66

#### 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. $0.0500\text{ mol HCl}*\frac{-118 \text{ kJ}}{2\text{ mol HCl}}=-2.95\text{ kJ}$ $q=mc\Delta T$ $2950=400*4.18*(T-25.0)$ $T=26.8^{\circ}C$

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