Whenever an acid is neutralised by a base, the net reaction is : `H^(+)(aq) + OH^(-)(aq) to H_(2)O(l) + 57.1 kJ`
Calculate the heat evolved for the following experiments :
(i) 0.25 mol of HCI solution is neutralised by 0.25 mol of NaOH solution.
(ii) 0.60 mol of `HNO_(3)` solution is mixed with 0.30 mol of KOH solution.
(iii)`100 cm^(3)` of 0.2 M HCl is mixed with `200 cm^(3)` of 0.3M NaOH.
`400 cm^(3)` of 0.2 M `H_(2)SO_(4)` is mixed with `600 cm^(3)` of 0.1 M NaOH solution.

Explanation

To solve the problem, we will calculate the heat evolved for each of the given neutralization reactions step by step.

### Given Reaction:
The net reaction for the neutralization of an acid by a base is:
\[ H^+(aq) + OH^-(aq) \rightarrow H_2O(l) + 57.1 \, \text{kJ} \]

This means that for every mole of water produced, 57.1 kJ of heat is released.

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