the balanced equation for the above reaction is as follows;
2Al + 3Br₂ --> 2AlBr₃
stoichiometry of Al to Br₂ is 2:3
there are 5.0 mol of Al and 6.0 mol of Br₂
if Al is the limiting reactant,
if 2 mol of Al reacts with 3 mol of Br₂
then 5.0 mol of Al reacts with - 3/2 x 5 = 7.5 mol of Br₂
but only 6 mol of Br₂ is present therefore Br₂ is the limiting reactant.
if 3 mol of Br₂ reacts with 2 mol of Al
then 6.0 mol of Br₂ reacts with - 2/3 x 6 = 4 mol of Al
the molar ratio of Al to AlBr₃ is 2:2
the number of moles of AlBr₃ formed is 4 mol of AlBr₃
amount of excess Al - 5.0 mol - 4.0 mol = 1 mol
therefore at the end of the reaction,
1 mol of Al and 4 mol of AlBr₃ in the reaction vessel
