Question

Some quantity of electricity was passed through two cells, one containing $HCl$ and the other $Ag{NO}_3$ solution. What volume of $H_2$ will be liberated at NTP, if $1.08\times {10}^{-4}kg$ of Ag got deposited at the other electrode?
(Molar mass of $Ag=108g{mol}^{-1}$, $H=1g{mol}^{-1}$)

Answer 1

$2e^{-}+AgN{O}_3+HCl⟶Ag+H_2+N{O}_3^{-}+C{l}^{-}$

By stoichiometry, moles of $Ag=$ Moles of $H_2$ liberated.

Moles of $Ag=\frac{1.08\times {10}^{-4}\times {10}^3}{108}$ moles$={10}^{-3}$ moles.

$\therefore$ Moles of $H_2$ liberated $={10}^{-3}$ moles.

At NTP, $1$ mole of $H_2=22.414d{m}^3$

$\therefore$ ${10}^{-3}$ mole of $H_2=22.414\times {10}^{-3}d{m}^3=22.414m^3$