Question

On passing a current through molten kcl 19.5g of k is deposited. The amount of Al deposited by the same quantity of electricity if passed through molten Alcl3 is?

Answer 1

K+ + e = K

Al3+ + 3e = Al

Aluminium requires 3 times the amount of electricity to deposit one mole


Moles of K deposited = 19.5/39 = 0.5

Moles of Al deposited = 0.5/3 = 1/6

Mass of Al deposited = 27 x 1/6 

                                   = 4.5 g

Answer 2

Answer:

Explanation:

K+ + e = K

Al3+ + 3e = Al

Aluminium requires 3 times the amount of electricity to deposit one mole

Moles of K deposited = 19.5/39 = 0.5

Moles of Al deposited = 0.5/3 = 1/6

Mass of Al deposited = 27 x 1/6

= 4.5 g