If the heat of neutralization for a strong acid - base reaction is `-57.1 kJ`, what would be the heat released when `350 cm^(3)` at `0.20 M H_(2) SO_(4)` is mixed with `650 cm^(3)` of `0.10 M NaOH`?

Options

A. `37.1 kJ`
B. `3.71 kJ`
C. `3.17 kJ`
D. `0.317 kJ`

Answer

B

Explanation

Amount of `H_(2)SO_(4)=MxxY`
`=0.20xx350=70 m mol`
`:. "Amount of" H^(+) ions=2xx70=140 m mol`
Amount of `NaOH=MxxV`
`=0.10xx650=65 m mol`
`:. "Amount of" OH^(-) ions=65 m mol`
Thus `NaOH` is the limiting reactant. As such `65 m mol` of `OH^(-)` ions will react with `65 m mol` of `H^(+)` ions to give `65 m mol` of water.
`1` mol of `H^(+)` ions react with `1` mol of `OH^(-)` ions to give `1` mol of `H_(2)O` and in the process `57.1 kJ` of heat is produced.
Thus heat produced `=57.1xx65xx10^(-3)`
`=3.71 kJ`