The amount of heat evolved when 500 text{ cm} | vedclass

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(A) The neutralization reaction is: $HCl + NaOH ightarrow NaCl + H_{2}O$Initial moles of $HCl = \frac{500 	imes 0.1}{1000} = 0.05 	ext{ mol}$Initi

Vedclass · Accepted Answer

$2.292$

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(A) The neutralization reaction is: $HCl + NaOH ightarrow NaCl + H_{2}O$Initial moles of $HCl = \frac{500 	imes 0.1}{1000} = 0.05 	ext{ mol}$Initial moles of $NaOH = \frac{200 	imes 0.2}{1000} = 0.04 	ext{ mol}$Since $NaOH$ is the limiting reagent,$0.04 	ext{ mol}$ of $HCl$ will react with $0.04 	ext{ mol}$ of $NaOH$.The heat of neutralization for $1 	ext{ mole}$ of a strong acid and a strong base is $57.3 	ext{ kJ/mol}$.Therefore,the heat evolved for $0.04 	ext{ mol}$ is $57.3 	imes 0.04 	ext{ kJ} = 2.292 	ext{ kJ}$.

The amount of heat evolved when $500 \text{ cm}^{3}$ of $0.1 \text{ M } HCl$ is mixed with $200 \text{ cm}^{3}$ of $0.2 \text{ M } NaOH$ is (in $\text{ kJ}$)

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