1 mol of `H_(2) SO_(4)` in mixed with 2 mol of `NaIH`. The heat evolved will beA. `57.3 kJ`B. `2 xx 57.3 KJ`C. `57.3//2 kJ`D. cannot be predicted

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

1.	1 mol of `H_(2) SO_(4)` in mixed with 2 mol of `NaIH`. The heat evolved will beA. `57.3 kJ`B. `2 xx 57.3 KJ`C. `57.3//2 kJ`D. cannot be predicted
Answer» Correct Answer - B 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`.

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1 mol of `H_(2) SO_(4)` in mixed with 2 mol of `NaIH`. The heat evolved will beA. `57.3 kJ`B. `2 xx 57.3 KJ`C. `57.3//2 kJ`D. cannot be predicted

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