Using formula 4Al+3O2 -------gt; 2Al2O3A.) What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
The first step in all of these problems, is to make sure that your chemical equation is balanced, because this will tell you the ratio of the compounds you're comparing. Now, you know that your limiting reagent is going to be the deciding factor for how much product you produce. You can see that you have a ratio of 4:3:2, For every 4 parts Al you use 3 parts O2 to make 2 parts Al2O3, so, using this ratio, which of your reactants is the limiting reactant? Let's look at Aluminum, the ratio of Al to Al2O3 is 4:2, so, we have a ratio of 1.4 moles of Al / (x) moles of Al2O3 = 4/2 (x) moles of Al2O3 = 1.4/2 (x) moles of Al2O3 = 0.7 With 1.4 moles of Al, you'll make 0.7 moles of Al2O3 Now let's look at Oxygen, the ratio of O to Al2O3 is 3:2, so, we have a ratio of 1.35 moles of O / (x) moles of Al2O3 = 3/2 (x) moles of Al2O3 = 1.35/1.5 (x) moles of Al2O3 = 0.9 With 1.35 Moles of O, you'll make 0.9 moles of Al2O3 NOW, this is where the limiting reagent comes into play again! You can only make as much of your product as your limiting reagent will allow! So while that 1.35 moles of oxygen is willing to make up to 0.9 moles of product, your 1.4 moles of Aluminum can only make 0.7 moles when it's giving everything it's got. SO.... your limiting reagent here is Aluminum, and when you react 1.4 moles of Al with 1.35 moles of O, you form 0.7 moles of Aluminum Oxide and have left over Oxygen (because not all of it is used up!)
