Question

How many seconds are required to produce 4.00 g of aluminum metal from the electrolysis of molten alcl3 with an electrical current of 12.0 a?

Answer 1

Answer: 3618 seconds

Step-by-step explanation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=(4g)/(27g/mol)=0.15moles`

According to mole concept:

1 mole of an atom contains
`6.022* 10^(23)` number of particles.

We know that:

Charge on 1 electron =
`1.6* 10^(-19)C`

Charge on 1 mole of electrons =
`1.6* 10^(-19)* 6.022* 10^(23)=96500C`

`AlCl_3\rightarrow Al^(3+)+3Cl^-`

At cathode:
`Al^(3+)+3e^-\rightarrow Al`

1 mole of aluminium is deposited by =
`3* 96500=289500C`

Thus 0.15 moles of aluminium is deposited by =
`(289500C)/(1)* 0.15=43425C`

To calculate the time required, we use the equation:

`I=(q)/(t)`

where,

I = current passed =12.0 A

q = total charge =
`43425C`

t = time required in seconds = ?

Putting values in above equation, we get:

`12.0A=(43425C)/(t)\\\\t=(43425C)/(12.0A)=3618s`

Hence, the amount of time required to produce 4.00 g of aluminum metal from the electrolysis of molten
`AlCl_3` with an electrical current of 12.0 A is 3618 seconds