Question

Solid aluminum Al and oxygen O2 gas react to form solid aluminum oxide Al2O3. Suppose you have 11.0 mol of Al and 13.0 mol of O2 in a reactor. Calculate the largest amount of Al2O3 that could be produced.

Answer 1

Step-by-step explanation:

The balanced chemical equation for the reaction is:

4 Al (s) + 3 O2 (g) -> 2 Al2O3 (s)

From the balanced equation, we can see that 4 moles of Al react with 3 moles of O2 to produce 2 moles of Al2O3.

So, for 11.0 moles of Al, we need 11.0/4 x 3 = 8.25 moles of O2. However, we only have 13.0 moles of O2, which is more than enough to react with all the Al.

Therefore, the limiting reactant is Al, and the maximum amount of Al2O3 that can be produced is:

11.0 mol Al x (2 mol Al2O3 / 4 mol Al) = 5.5 mol Al2O3

So, the largest amount of Al2O3 that could be produced is 5.5 mol.