Question

Refer to the experiment shown in which a sample of solid potassium chlorate $\left(KCl{O}_3\right)$ was heated in a test tube and decomposed by the following reaction:

$2KCl{O}_3\left(s\right)\to 2KCl\left(s\right)+3O_2\left(g\right)$

The oxygen produced was collected by displacement of water at ${22}^{\circ }C$ and pressure of $750mmHg$. The volume of the gas collected was $2.24L$. (The vapour pressure of water at ${22}^{\circ }C$ is $20.0mmHg$.)

If $61.0g$ of solid $KCl{O}_3(molarmass=122g/mol)$ were to be completely decomposed by this method, what mass of $KCl\left(s\right)$ would you expect to remain in the test tube?

• A. $111g$
• B. $74g$
• C. $56g$
• D. $37g$

Answer 1

The correct option is D $37g$

The equation is: $2KCl{O}_3$ $=$ $2KCl$ $+$ $3O_2$

Molecular weight of $KCl{O}_3$ is $122.5g/mol$.

Thus two moles of $KCl{O}_3$ weigh ($122.5$$\times$$2$), that is $245grams$.

Molecular weight of $KCl$ is $74.5g/mol$. Thus, two moles of $KCl$ weigh ($74.5\times 2$), that is $149grams$.

From the equation, it is seen that two moles of Potassium chlorate give two moles of Potassium chloride.

$245grams$ of $KCl{O}_3$ give $149grams$ of $KCl$

$1gram$ of $KCl{O}_3$ give $\frac{149}{245}$$gram$ of $KCl$

Thus, $61grams$ of $KCl{O}_3$ give $\frac{149}{245}$$\times$$61$ $grams$ of $KCl$, that is $37grams$ of $KCl$.

Thus, option $D$ is the correct answer.