##### 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 O_{2} collected = 14.9g

Find: Theoretical yield, g O_{2}

*Step 2: List other known quantities and plan the problem.*

1 mol KClO_{3} = 122.55 g/mol

1 mol O_{2} = 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.__6__7 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.__6__7 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 \(100\%\).