To determine the amount of excess reactant left, we need to first identify which is the limiting and the excess reactant. We do as follows:
A + 2B → AB2
2 mol A ( 2 mol B / 1 mol A) = 4 mol B needed to be consumed completely
5 mol B ( 1 mol A / 2 mol B) = 2.5 mol A needed to be consumed completely
Therefore, the limiting would be reactant A and the excess would be reactant B. The amount of the excess would be:
5 mol B - 4 mol B = 1 mol B <----------LAST OPTION