When 4.0 mol H2O react with 4.0 mol Al in the reaction below, 2.7 mol of Al are used up in the reaction. How many moles of Al are left over unreacted?

2 Al + 3 H2O → Al2O3 + 3H2 (C?)

2.7 mol Al left over

2.0 mol Al left over

1.3 mol Al left over

0.7 mol Al left over

From the amounts given and the reaction, the limiting reactant would be H2O and the excess reactant would be Al. Only a part of the initial amount of Al is being used up in the reaction while all of the H2O is reacted. The amount of Al left unreacted is 4.0 mol - 2.7 mol = 1.3 mol Al. The correct answer is the third option.