The amount of heat evolved when `500 cm^(3) 0.1 M HCl` is mixed with `200 cm^(3)` of `0.2 M NaOH` is

Options

A. `2.292 kJ`
B. `1.292 kJ`
C. `0.292 kJ`
D. `3.393 kJ`

Answer

A

Explanation

`500 cm^(3)` of `0.1 M HCl = (500xx0.1)/(1000)=0.05 mol`
`200 cm^(3) 0.2 M NaOH=(200xx0.2)/(1000)=0.04 mol`
`:. 0.04` moles of `NaOH` is neutralized by `0.04` mol of `HCl`.
`:. "Heat released"=57.3xx0.04=2.292 kJ`