Example \(\PageIndex{1}\): Decomposition of Potassium Chlorate
Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below:
\[2 \ce{KClO_3} \left( s \right) \rightarrow 2 \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( g \right)\nonumber \]
In a certain experiment, \(40.0 \: \text{g} \: \ce{KClO_3}\) is heated until it completely decomposes. The experiment is performed and the oxygen gas is collected and its mass is found to be \(14.9 \: \text{g}\).
- What is the theoretical yield of oxygen gas?
- What is the percent yield for the reaction?
Solution
a. Calculation of theoretical yield
First, we will calculate the theoretical yield based on the stoichiometry.
Given: Mass of \(\ce{KClO_3} = 40.0 \: \text{g}\)
Mass of O2 collected = 14.9g
Find: Theoretical yield, g O2
Step 2: List other known quantities and plan the problem.
1 mol KClO3 = 122.55 g/mol
1 mol O2 = 32.00 g/mol
Step 3: Apply stoichiometry to convert from the mass of a reactant to the mass of a product:
Step 4: Solve.
\[40.0 \: \cancel{\text{g} \: \ce{KClO_3}} \times \dfrac{1 \: \cancel{\text{mol} \: \ce{KClO_3}}}{122.55 \: \cancel{\text{g} \: \ce{KClO_3}}} \times \dfrac{3 \: \cancel{\text{mol} \: \ce{O_2}}}{2 \: \cancel{\text{mol} \: \ce{KClO_3}}} \times \dfrac{32.00 \: \text{g} \: \ce{O_2}}{1 \: \cancel{\text{mol} \: \ce{O_2}}} = 15.7 \: \text{g} \: \ce{O_2}\nonumber \]
The theoretical yield of \(\ce{O_2}\) is \(15.7 \: \text{g}\), 15.67 g unrounded.
Step 5: Think about your result.
The mass of oxygen gas must be less than the \(40.0 \: \text{g}\) of potassium chlorate that was decomposed.
b. Calculation of percent yield
Now we will use the actual yield and the theoretical yield to calculate the percent yield.
Given: Theoretical yield =15.67 g, use the un-rounded number for the calculation.
Actual yield = 14.9g
Find: Percent yield, % Yield
Step 2: List other known quantities and plan the problem.
No other quantities needed.
Step 3: Use the percent yield equation below.
\(\text{Percent Yield} = \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\)
Step 4: Solve.
\(\text{Percent Yield} = \dfrac{14.9 \: \text{g}}{15.\underline{6}7 \: \text{g}} \times 100\% = 94.9\%\)
Step 5: Think about your result.
Since the actual yield is slightly less than the theoretical yield, the percent yield is just under \(100\%\).