Question

How many grams of ca metal are produced by the electrolysis of molten cabr2 using a current of 30.0 amp for 8.0 hours?

Answer 1

`m=179.4g` of Ca

Step-by-step explanation:

First we are going to write down the balanced reaction:

`CaBr_(2)=Ca^(2+)+_(2)Br^(-1)`

The reduction for the
`Ca^(2+)` ion will be:

`Ca^(2+)+2e^(-)=Ca`

It means that 2 Faraday left for each mol of Ca.

Converting from Faraday to Coulombs:

`2Faraday*(96485Coulombs)/(1Faraday)= 192970Coulombs`

Then we can apply the Faraday´s law:

`m=(M*I*t)/(e^(-)*96485)`

where

m=mass in grams

M=molar mass

I=current

t=time in seconds

e-=number of electrons per mol

Replacing values:

`m=(40.078g*30.0A*28800s)/(2*96485C)`

`m=179.4g` of Ca