KI reacts with`H_2SO_4` producing `I_2` and `H_2S`. The volume of 0.2 `H_2SO_4` required to produce 3.4g of `H_2S` is
(a). 2.5 L
(b). 3.8 L
(c). 4 L
(d). 5 L

Explanation

(a).
Since the n-factor of `H_2S` is not known, the problem is solved by mole concept by balancing the equation.
`2I^(ɵ)toI_2+cancel(2e^(-))]xx4`
`underline(8H^(o+)+cancel(8e^(-))+underset(x=4)(SO_4^(2-))tounderset(x=-2)(S^(2-))+4H_2O)`
`underline(8I^(ɵ)+8H^(o+)+SO_4^(2-)to4I_2+S^(2-)+4H_2O)`.
Add other ions, i.e., `2H^(o+),8K^(o+)` and `4SO_4^(2-)` to both sides to balance the equation. Net redox equation is . Net redox equation is
" mol of "`H_2S=(3.4)/(34)=0.1` mol
" mol of "`H_2SO_4=0.1xx5mol=0.5mol`
`0.2xxV=0.5`
`V=2.5L`