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?

1 Answer
Mar 16, 2017

Answer:

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?