1 mol of `H_(2) SO_(4)` in mixed with 2 mol of `NaIH`. The heat evolved will be

Options

A. `57.3 kJ`
B. `2 xx 57.3 KJ`
C. `57.3//2 kJ`
D. cannot be predicted

Answer

B

Explanation

During neutralization process,
`H^(+) (aq.) + OH^(-) (aq.) rarr H_(2) O(1)`
`57.3 kJ` energy (heat) is released for energy `1 mol H_(2) O` formation. When `1 mol` of `H_(2) SO_(4)` is mixed with 2 mol of `NaOH`,
we get 2 mol of `H_(2) O`:
`H_(2) SO_(4) + 2 NaOH rarr Na_(2) SO_(4) + 2 H_(2) O`
Thus, the amount of energy released is `2 xx 57.3 kJ`.