What would be the heat released when an aqueous solution containing `0.5 mol` if `HNO_(3)` is mixed with `0.3` mol of `OH^(-1)` (enthalpy of neutralisation is `-57.1 kJ`)

Options

A. `28.5 kJ`
B. `17.1 kJ`
C. `45.7 kJ`
D. `1.7 kJ`

Answer

B

Explanation

`0.5 mol HNO_(3)-=0.5 molH^(+)`
`H^(+)+OH^(-)toH_(2)O`, `DeltaH=-57.1 kJ`
Here `OH^(-)=0.3 mol`
Thus, `OH^(-)` is the limiting reactant
When `1` mol of `OH^(-)` ions are neutralised heat released` = 57.1 kJ`
`0.3` mol of `OH^(-)` ions are neutralised heat released `=57.1xx0.3=17.13 kJ`