Question

1. write the balanced equation for the decomposition reaction of solid potassium chlorate into solid potassium chloride and oxygen gas. a. find the theoretical yield of oxygen gas if you start with 15.32 g of potassium chlorate. b. calculate the percent yield if 5.75 g are actually produced in lab.

Explanation:

Step1: Write the balanced chemical equation

The decomposition of potassium chlorate ($KClO_3$) is given by $2KClO_3(s)
ightarrow 2KCl(s)+3O_2(g)$.

Step2: Calculate the molar mass of $KClO_3$

The molar mass of $K$ is $39.10\ g/mol$, $Cl$ is $35.45\ g/mol$ and $O$ is $16.00\ g/mol$. So, $M_{KClO_3}=39.10 + 35.45+3\times16.00=122.55\ g/mol$.

Step3: Calculate the number of moles of $KClO_3$

$n_{KClO_3}=\frac{m}{M}=\frac{15.32\ g}{122.55\ g/mol}=0.125\ mol$.

Step4: Determine the mole - ratio of $KClO_3$ to $O_2$

From the balanced equation, the mole - ratio of $KClO_3$ to $O_2$ is $2:3$. So, for $0.125\ mol$ of $KClO_3$, the number of moles of $O_2$ produced is $n_{O_2}=\frac{3}{2}\times0.125\ mol = 0.1875\ mol$.

Step5: Calculate the molar mass of $O_2$

The molar mass of $O_2$ is $M_{O_2}=2\times16.00\ g/mol = 32.00\ g/mol$.

Step6: Calculate the theoretical yield of $O_2$

$m_{O_2}=n\times M=0.1875\ mol\times32.00\ g/mol = 6.00\ g$.

Step7: Calculate the percent yield

The percent yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$. Given actual yield = $5.75\ g$ and theoretical yield = $6.00\ g$. So, $\text{Percent Yield}=\frac{5.75\ g}{6.00\ g}\times100\% = 95.83\%$.

Answer:

a. $6.00\ g$
b. $95.83\%$