Question

Heat of neutralisation for the reaction,
NaOH + HCl-NaCl + H2O
is 57.1 kJ mol-'. The heat released when 0.25 mole of NaOH is
titrated against 0.25 molc of HCI will be:
(a) 22.5 kJ
(b) 57.1 kJ
(c) 28.6 kJ
(d) 14.3 kJ​

Answer 1

Answer:

57.1kJ

Explanation:

Heat of neutralization for the reaction, is 57.1kJ mol^-1 . The heat released when 0.25 mole of NaOH is titrated against 0.25 mole HCl will be: NaOH + HCl→ NaCl + H2O