Answer :
0.36 g of H2 would be produced.
Stoichiometric reactions
According to the equation of the reaction: 2Al + 6HBr → 2AlBr3 + 3H2
The mole ratio of Al to HBr is 1:3.
Mole of 15.8 g Al = 15.8/26.98
= 0.59 mole
Mole of 29.2 g HBr = 29.2/80.91
= 0.36 mole
Thus, HBr is the limiting reagent.
Mole ratio of HBr and H2 = 2:1
Equivalent mole of H2 = 0.36/2
= 0.18 mole
Mass of 0.18 mole H2 = 0.18 x 2
= 0.36 g
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