FINISHED_Final_Notebook_Jones

Limiting Reagents:
Example: What would be the limiting reagent if 10g of H were reacted with 15g of O ?
2
2
2H + O = 2H O
2 2 2
Step 1: Figure out how much of a product can be made with a reactant
10g H |1 moL |
2
| 2g H | 2 moL H O | 1 moL H O
2
2 moL H O | 18g H O
2 2
2
2
90g H O 2
Step 2: Go through the same process for the second reactant
15g O 2 | 1 moL O 2 | 2 moL H 2 O | 18g H O2
| 32g O | 1 moL O | 1 moL H O
2
2
2
20g H O 2
Step 3: Because oxygen produces less water than hydrogen, oxygen is the limiting reagent
Percent Yield / Using the Limiting Reagent:
Example: How much H O would be produced with a 69.6% yield if 10g of H were reacted with 15g of O ?
2
Step 1: Since, in this equation, the limiting reagent is oxygen, we use that to calculate the percent yield.
20g of water is the theoretical yield (as demonstrated previously), and we need to find 69.6% of that.
So, we multiply 69.6% (or .696) by 20 to get our answer.
0.696 x 20 = 13.92g of H O 2
So, 13.92g is our percent yield. If, instead of a percent yield, we were given the actual amount produced,
we would divide the actual amount by the theoretical amount to find the percent error, or percent yield.
For example:
13.92g of H 2 O
0.696 or 69.6%
20g of H 2 O
2
2
107