What would be the amount of heat released when an aqueous solution containing 0.5 mole of ${\text{HNO}}_{\text{3}}$

is mixed with 0.3 mole of  ${\text{OH}}^{-}$ (enthalpy of neutralization is – 57.1 kJ) ?

${\text{HNO}}_{\text{3}}{\text{\hspace{0.17em}+\hspace{0.17em}OH}}^{-}\to {\text{\hspace{0.17em}H}}_{\text{2}}{\text{O\hspace{0.17em}+\hspace{0.17em}NO}}_{\text{3}}^{-}$

The enthalpy of neutralization is −57.1 kJ/mol.
0.3 mole of HNO_3​ will react with 0.3 moles of OH⁻. The heat released is:

$0.3 \mathrm{mol}\times -57.1 \mathrm{kJ}/\mathrm{mol}=17.1 \mathrm{kJ} \mathrm{or} 4.1 \mathrm{kcal} \mathrm{or} 4100 \mathrm{cal}$