How many moles of electrons are required to
(i) reduce 1 mol of `MnO_(4)^(-)` to `Mn^(2+)`
(ii) produce 10.0 g of Al from molten `Al_(2) O_(3)` ?

Explanation

Moles of electrons required :
(i) `MnO_(4)^(-) + 5e^(-) to Mn^(2+)`
1 mol of `MnO_(4)^(-)` will be reduced by 5 mol of electrons
(ii) `Al^(3+) + 3e^(-) to Al`
3 mol of electrons are required to produce 1 mol or 27 g of Al
`therefore` 10 g of Al will be produced by `= (3)/(27) xx 10 = 1.11` mol electrons .