Answer:
The equation for the reaction between Al and Br2 is:
2Al + 3Br2 -> 2AlBr3
If there is an excess of Al and 6.0 mol of Br2, then the limiting reactant will be Br2, and the number of moles of AlBr3 formed will be equal to the number of moles of Br2 consumed, which is 6.0 mol.
Therefore, assuming 100% yield, 6.0 mol of AlBr3 will be formed.