Calculate the amount of heat released when heat of neutralization is `-57.0 kJ :`
`(a)` `0.5 mol e` of `HNO_(3)` in aqueous solution.
`(b)` `200mL` of `0.1 MH_(2)SO_(4)` is mixed with `150mL` of `0.2M KOH`.

Explanation

`(a)` 1 mole of `HNO_(3)` and 1 mole of `NaOH`
give heat `=57.0kJ`
`:. 0.3 mol e of HNO_(3)` and `0.3 mol e `of
`NaOH` give heat `=57xx0.3=17.1kJ`
`(b) Meq. ` of `H_(2)SO_(4)=200xx0.1xx2=40,`
`Meq ` of `KOH=150xx0.2xx1=30`
`:' 1000Meq.` of `H_(2)SO_(4)` and `1000Meq.` of `KOH` on mixing produce heat `=57.0kJ`
`:. 30 Meq.` of `H_(2)so_(4)` and `30Meq.` of `KOH` on mixing produce heat
`=(57xx30)/(1000)=1.71kJ`