Solving moles-to-moles limiting reactant problems

Question

Solid aluminum and oxygengas react to form solid aluminum oxide. Suppose you have 5.0 mol of Al and 9.0 mol of O2 in a reactor.

Suppose as much as possible of the Al reacts. How much will be left? Round your answer to the nearest 0.1 mol.

Answer 1

First, we need to determine the reaction: 4Al (s) + 3O2 (g) --> 2Al2O3

Remember that Al has a +3 charge while oxygen has a -2 charge.

From this, we need to determine how many moles of the product (Al2O3) each reactant can form.

(5.0 mol Al / 4) x 2 = 2.5 mol Al2O3

(9.0 mol O2 / 3) x 2 = 6.0 mol Al2O3

Because oxygen gas can produce more moles of Al2O3, Al (s) is the limiting reactant. This also tells that only 2.5 moles of Al2O3 can be formed, and there will be no Al (s) remaining. Only oxygen gas will remain.

To calculate the amount of oxygen used, we can use: (5.0 mol Al / 4) x 3 = 3.75 mol oxygen gas used

Finally, subtract the amount used from initial amount: 9.0 mol - 3.75 mol = 5.25 mol oxygen gas remaining.

This rounds to 5.3 moles of oxygen gas leftover.