The heat of neutralisation of a strong base acid is $13.7 \mathrm{kcal}$. The heat released when 0.6 mole $\mathrm{HCl}$ solution is added to a 0.25 mole of NaOH is

Given, the heat of neutralisation of a strong base and a strong acid is $13.7 \mathrm{kcal}$. The reaction of neutralisation is as follows $$ \begin{array}{l} \mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} = -137kcal \\ 1 \mathrm{mol} \quad 1 \mathrm{mol} \end{array} $$ According to question, $$ \begin{array}{l} \mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \\ 0.6 \mathrm{mol} \quad 0.25 \mathrm{mol} \end{array} $$ In equation (i), NaOH acts as a limiting reagent. For 1 mole of NaOH and 1 mole of HCl, heat of neutralisation $=137 \mathrm{kcal}$ : For 0.25 mole of NaOH and 0.6 mole of $\mathrm{HCl}$, heat of neutralisation $=137 \times 0.25 \Rightarrow 3.425 \mathrm{kcal}$