Question

On heating, $4.90\mathrm{g}$ of Potassium chlorate(${\mathrm{KClO}}_3$) shows a weight loss of $0.384\mathrm{g}$. What will be the percentage of the original decomposed Potassium chlorate(${\mathrm{KClO}}_3$)?

• A. $10\%$
• B. $15\%$
• C. $20\%$
• D. $30\%$

Answer 1

The correct option is C

$20\%$

Explanation of correct answer:

Option C $20\%$

Step 1: Number of moles of Oxygen

On heating, the Sodium chlorate decomposes to potassium chloride and oxygen gas.

$\underset{\mathrm{Sodium}\mathrm{chloride}}{2{\mathrm{KClO}}_3}\stackrel{∆}{\left(\mathrm{s}\right)\to }\underset{\mathrm{Pottassium}\mathrm{chloride}}{2\mathrm{KCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}\mathrm{gas}}{3{\mathrm{O}}_2\left(\mathrm{g}\right)}$

The gas Oxygen decomposes and it is the reason for the mass loss after heating of Potassium chlorate.

Mass loss is the mass of Oxygen formed which when divided by the molecular weight of the Oxygen gas gives the moles of Oxygen formed.

Mass loss is equal to $0.384\mathrm{g}$

That is, the number of moles of Oxygen formed is

$$\begin{gathered} \mathrm{moles}=\frac{\mathrm{givenmass}}{\mathrm{molecularmass}} \\ \begin{array}{r}\mathrm{moles}=\frac{0.384}{32}\\ \mathrm{moles}=0.012\end{array} \end{gathered}$$

Step 2: Amount of Sodium chlorate decomposed

Two moles of Potassium chlorate form three moles of Oxygen gas,

The ratio of the formation of moles of Potassium chlorate and Oxygen gas is going to be constant, therefore using the ratio and the calculated moles of Oxygen is,

$$\begin{gathered} \mathrm{moles}\mathrm{of}{\mathrm{KClO}}_3=\frac{2}{3}\times \mathrm{moles}\mathrm{of}{\mathrm{O}}_2 \\ \mathrm{moles}\mathrm{of}{\mathrm{KClO}}_3=\frac{2}{3}\times 0.012 \\ \mathrm{moles}\mathrm{of}{\mathrm{KClO}}_3=0.008 \end{gathered}$$

To, get the amount of Sodium chlorate, multiply the moles of Potassium chlorate decomposed by the molar mass of Potassium chlorate.

$$\begin{gathered} {\mathrm{KClO}}_3\mathrm{decomposed}=\mathrm{molar}\mathrm{mass}\mathrm{of}{\mathrm{KClO}}_3\times \mathrm{moles}\mathrm{of}{\mathrm{KClO}}_3\mathrm{decomposed} \\ {\mathrm{KClO}}_3\mathrm{decomposed}=122.5\times 0.008 \\ {\mathrm{KClO}}_3\mathrm{decomposed}=0.9804\mathrm{grams} \end{gathered}$$

Step 3: Percentage of Potassium chlorate decomposed.

The percentage decomposed out of the whole sample of $4.90\mathrm{g}$ of ${\mathrm{KClO}}_3$

$$\begin{gathered} \%of\mathrm{original}\mathrm{sample}=\frac{0.98}{4.90}\times 100 \\ \%of\mathrm{original}\mathrm{sample}=20\% \end{gathered}$$

Therefore, $20\%$ is the potassium chlorate decomposed in the reaction when $4.90\mathrm{g}$ of ${\mathrm{KClO}}_3$ is heated and there is a weight loss of 0.384 grams.

Hence option C is correct.