2Al + 3Br2 -------------> 2AlBr3
3 g Al = 0.11 mol Al.
6 g Br2 = 0.0375 mole bromine (it is diatomic).
moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025.
Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g.
moles of Al left = 0.11 - 0.025 = 0.086
