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Performing Calculations involving Limiting and Excess Reagents.

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1 Performing Calculations involving Limiting and Excess Reagents

2 __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 1.You have a sealed vessel that contains 623 mL of oxygen and 111g aluminum. How much aluminum oxide can I produce? 2.How much of your excess reactant is left over? 3.If you collected 1.06g of aluminum oxide, what was the percent yield of the reaction? 234

3 1.You have a sealed vessel that contains 623 mL of oxygen and 111g aluminum. How much aluminum oxide can I produce? __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 234 WHAT IS YOUR LIMITING REACTANT?? 623 mL O 2 x 1 mol O L 1 L O mL x = 111g Al x 1 mol Al 26.98g = mol O mol Al HAVE AVAILABLE x 4 mol Al 3 mol O 2 = mol Al In order to use up all the oxygen available, you would need mol Al. So you should ask yourself, it is going to be possible to use up all the O 2 ?? YES! Therefore, oxygen is limiting and aluminum is in excess.

4 1.You have a sealed vessel that contains 623 mL of oxygen and 111g aluminum. How much aluminum oxide can I produce? __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 234 Oxygen is limiting and aluminum is in excess. The amount of product you can form is going to be dependent on the reactant that will get used up! Therefore, it is dependent on your limiting reactant mol O 2 x 2 mol Al 2 O 3 3 mol O 2 = x g Al 2 O 3 1 mol Al 2 O g Al 2 O 3 formed

5 2. How much of your excess reactant is left over? __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 234 The amount of excess reactant left over is dependent upon how much we had to start and how much was used during the reaction mol O 2 x 4 mol Al 3 mol O 2 = mol Al We already know that oxygen in our limiting reactant. Therefore, when it is completely used in the reaction, mol of aluminum is also used. According to our original calculations, we originally had 4.11 mol of aluminum mol Al – mol Al = 3.74 mol Al remains unused 3.74 mol Al 26.98g 1 mol Al x = 101g Al remaining

6 3. If you collected 1.06g of aluminum oxide, what was the percent yield of the reaction? __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 234 In question 1, you’ve already determined that the amount of aluminum oxide produced was 1.89g. This is your THEORETICAL YIELD Percent Yield = Actual yield Theoretical yield x 100% Percent Yield = 1.06 g 1.89g x 100% = 56.1%


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