For how long, a current of 1.5 amperes has to be passed through the electrolyte in order to deposit 1.5 g of Al, when the electrode reaction is

${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}⟶\mathrm{Al}$ (round off the answer to two digits after the decimal)

Reaction: $\underset{1 \mathrm{mol}}{{\mathrm{Al}}^{3+}}+{\underset{3 \mathrm{mol}}{3\mathrm{e}}}^{-}\to \underset{1 \mathrm{mol}\left(=27 \mathrm{g}\right)}{\mathrm{Al}}$

Charge on 3 moles  of electrons = 3 x 96500 coulombs

Total charge to deposit 27g Al = 3 x 96500 C

$\therefore$ Charge to deposit 1.5 $\mathrm{g} \mathrm{Al}=\frac{\left(3\times 96500\times 1.5\right)}{27}=16083.3 \mathrm{C}$

To Calculate time (t). Current = 1.5 ampere; time  $\left(\mathrm{t}\right)=$?

No. of coulombs = ampere x time (in seconds)

                           $=1.5 \times \mathrm{t}.\mathrm{C}$

1.5 t = 16083.3

$\mathrm{t}=\frac{16083.3}{1.5}\text{ Or }\frac{16083.3}{1.5\times 60\times 60}=2.978\approx 2.98 \mathrm{hours}$