8. write the balanced equation for the decomposition reaction of solid potassium chlorate into solid…

Explanation:

Step1: Write the balanced chemical equation

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

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. $95.83%$