Heat of neutralization for the reaction NaOH + HCl ® NaCl + H O is -57.1kJ mole-1 . The heat released when 0.25 moles of NaOH is treated with 0.25 moles of HCl is

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Heat of neutralization for the reaction NaOH + HCl ® NaCl + H O is -57.1kJ mole^-¹ . The heat released when 0.25 moles of NaOH is treated with 0.25 moles of HCl is

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22.5 kJ

57.1 kJ

14.3 kJ

28.6 kJ

answer is C.

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NaOH + HCl ® NaCl + H₂O ; DH = -57.1kJ

Per 1mole $\to$ -57.1kJ

0.25 mole = -57.1x0.25 = 14.3kJ

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