Question

The charge required to deposit 40.5 g of Al from the fused Al2(SO4)3 is

Answer 1

Given:

$\boxed{\sf{2Al^{+3} + 3SO_{4}^{-2}\longrightarrow Al_{2}(SO_{4})_{3}}}$

To find:

Charge required to deposit 40.5 grams of Aluminium from the compound ?

Calculation:

The half Reaction for Aluminium :

$\boxed{\sf{{Al}^{+3} + 3{e}^{-1}\rightarrow Al}}$

From the above Reaction , we can say that:

$\rm \implies \: 27 \: g \: Al \: requires \: 3F \: charge$

$\rm \implies \: 40.5 \: g \: Al \: requires \: \bigg(\dfrac{3}{27} \times 40.5 \bigg) \: F \: charge$

$\rm \implies \: 40.5 \: g \: Al \: requires \: \bigg( \dfrac{40.5}{9} \bigg) \: F \: charge$

$\rm \implies \: 40.5 \: g \: Al \: requires \: 4.5\: F \: charge$

So, charge required to deposit 40.5 gram of Aluminium is 4.5 F.

$\star$ Hope It Helps.