What is the volume of #H_2# has formed at STP when 6.32g of #Al# reacts with excess #NaOH#?

1 Answer
anor277
Oct 27, 2016

#Al + NaOH + 3H_2O → Na^(+) + [Al(OH)_4]^(-) + 3/2 H_2 (g)uarr#

Approx, #8*L# of dihydrogen gas are evolved.

Explanation:

Aluminum metal will reduce the hydrogen in water under basic conditions to give the aluminate ion #Al(OH)_4^-#.

#"Moles of metal"# #=# #(6.32*g)/(26.98*g*mol^-1)# #=# #0.234*mol#.

Now given the stoichiometric equation, 3/2 equiv of dihydrogen gas evolve from each equiv metal, i.e. #0.351*mol#.

Since at #"STP"# #1# #mol# of ideal gas occupies #22.4*L*mol^-1#, we take the product, #0.351*molxx22.4*L*mol^-1=??L#

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