Aluminum reacts with oxygen gas to produce aluminum oxide. How many moles of aluminum oxide would you expect to produce if you reacted 14.5 moles of oxygen gas with excess aluminum?
4Al +3O2 --> 2Al2O3
14.5 mol O2 x 4 mol Al / 3 mol O2 = 19.3 moles
You have the right idea but the wrong numbers. You have calculated mols Al. You want mols Al2O3; the problem asks for mols Al2O3.
14.5 mol O2 x 2 mol Al2O3 / 3 mol O2 = 9.67 moles.
Correct?
yes
To solve this problem, we will use the balanced chemical equation provided:
4Al + 3O2 --> 2Al2O3
According to the equation, for every 3 moles of O2, we would expect to produce 2 moles of Al2O3.
Given that you have 14.5 moles of O2, we can use a proportion to find the number of moles of Al2O3 produced.
14.5 mol O2 x (2 mol Al2O3 / 3 mol O2) = 19.3 moles of Al2O3
Therefore, if you react 14.5 moles of oxygen gas with excess aluminum, you would expect to produce 19.3 moles of aluminum oxide.