Chemistry - Thermochemistry Question with Solution | TestHub

Heat of neutralization $(\Delta H)=57\mathrm{kJ}$, mole of ${\mathrm{HNO}}_3=0.5$ mole and mole of $\mathrm{NaOH}=0.2$ mole. When ${\mathrm{HNO}}_3$ solution is added to $\mathrm{NaOH}$ solution, then 0.2 mole of ${\mathrm{HNO}}_3$ solution will combine with 0.2 mole of ${\mathrm{OH}}^{-}$ions of $\mathrm{NaOH}$ solution. $\therefore$ Heat released $=\Delta H\times 0.2=57\times 0.2=11.4\mathrm{kJ}$.