Question

How is aluminum oxidized under basic conditions? What is the reduction product?

What volume of this product is collected under standard conditions when a `5.4*g` mass of this metal is oxidized in a solution prepared from a `20*g` mass of sodium hydroxide?

Answer 1

Approx. `7*L`.

Explanation:

We interrogate the oxidation reaction:

`Al +4HO^(-) rarr Al(OH)_4^(-) + 3e^(-)` `(i)`

And water is reduced to dihydrogen gas:

`H_2O +e^(-)rarr 1/2H_2(g) +HO^(-)` `(ii)`

And so, we take `(i) +3xx(ii):`

`Al +HO^(-) +3H_2O rarr Al(OH)_4^(-) + 3/2H_2(g)`

We have the starting masses, and thus:

`"Moles of aluminum"=(5.4*g)/(26.98*g*mol^-1)=0.200*mol`.

`"Moles of NaOH"=(20.0*g)/(40.00*g*mol^-1)=0.500*mol`.

Clearly, hydroxide is the reagent in excess...........as required.

And given the stoichiometry, there are `0.300*mol` dihydrogen gas evolved. At `"STP"`, the molar volume is `22.4*L*mol^-1`.

And thus the dihydrogen gas evolved has a volume of `0.300*molxx22.4*L*mol^-1=??L`.

Aluminum is amphoteric and can be oxidized under acid, or under basic conditions (as here). I take it this is 1st year undergrad?