Passage of 3 Faraday of charge through aqueous solution of AgNO₃, CuSO₄, Al(NO₃)₃ and NaCl, respectively, will deposit the metals in the ratio (molar) 

$\begin{array}{rl} & {\mathrm{Ag}}^{+}+1\overline{\mathrm{e}}\to \mathrm{Ag}\\ & 3\mathrm{F}3\text{mole}\\ & {\mathrm{Cu}}^{+2}+2\overline{\mathrm{e}}\to \mathrm{Cu}\\ & 2\mathrm{F}3\text{mole}\\ & 3\mathrm{F}\frac{3}{2}\text{mole}\end{array}$
In case of aq Al(NO₂)₃ & aq  NaCl Metal are not discharged, H⁺ will discharge
$n_{Ag}:n_{Cu}:n_{Al}:n_{Na}=3:\frac{3}{2}:0:0$
= 6 :3 : 0 : 0