Whenever you are given more than one reactants extent, you are faced with a limiting reactant problem:
1) Establish the limiting reactant: By stoichiometry you need 3 Br2/2Al while you are given 6 Br2/5Al.. You have a lower ratio than you need, so Br2 is the LR.
2) Assume all the LR reacts and use stoichiometry to find Al reacted and AlBr3 produced:
At the end, you will have 0 moles of Br2
you will have 5 moles - 6moles Br2(2 Al/3 Br2) = 1 moles Al
you will have 0 + 6 moles Br2(2 AlBr3/3 Br2) = 4 moles AlBr3
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