Question

$KCl{O}_3$ decomposes into $KCl$ and $O_2$. If the volume of $O_2$ obtained in this reaction is $1.12$ $L$ at $STP$, the weight of $KCl$ formed in the reaction is:

• A. $7.45g$
• B. $2.48g$
• C. $4.96g$
• D. $1.24g$

Answer 1

The correct option is B $2.48g$

$2KCl{O}_3\to 2KCl+3O_2$

Here if the volume of $O_2$ at STP is $1.12$ L, then at STP $1$ mol gas will have volume $22.4$ L, so $1.12$ L must

have $\frac{1}{22.4}\times 1.12=\frac{1.12}{22.40}=\frac{112}{2240}=\frac{1}{20}$ moles.

It can be seen from the reaction that for $3$ mole $O_2\to 2$ moles of $KCl$ produced.

So if 1 mole $O_2$ is produced $\to \frac{2}{3}$ mole of $KCl$ must be produced along with.

$\therefore$ If $O_2$ yield is $\frac{1}{20}$ mole during the reaction, then $KCl$ must be $\frac{2}{3}\times \frac{1}{20}$ moles.

Hence $KCl$ moles $=\frac{2}{60}$ moles

As Molar mass of $KCl=39+35.5=74.5$ g.

so $\frac{2}{60}$ moles of $KCl$ must have weight $=2/60\times 74.5=2.48$ g.