Question

How many moles of potassium chlorate need to be heated to produce 11.2 litres oxygen at NTP?

• A. $\frac{1}{4}$ mol
• B. $\frac{1}{3}$ mol
• C. $\frac{2}{3}$ mol
• D. $\frac{1}{2}$ mol

Answer 1

The correct option is B $\frac{1}{3}$ mol

Determining required number of moles of potassium chlorate:

Given volume of $O_2$ is 11.2 L
We know, at NTP 1 mole of any gas occupies 22.4 L.

Moles of $O_2=\frac{Volume\left(inL\right)atNTP}{22.4L}$
= $\frac{11.2}{22.4}$
= $\frac{1}{2}$ mol
Chemical reaction involved
$\underset{PotassiumChlorate}{2KCl{O}_3}\to \underset{PotassiumChloride}{2KCl}+\underset{Oxygen}{3O_2}$

Thus, according to the reaction.
3 moles of $O_2$ are produced from 2 mol of $KCl{O}_3$
1 mol of $O_2$ will be produced from $\frac{2}{3}mol$ of $KCl{O}_3$
Thus,
$\frac{1}{2}molof{O}_2$ will be produced from
$\frac{2}{3}\times \frac{1}{2}molofKCl{O}_3$

Hence, $\frac{1}{3}molofKCl{O}_3$ are required
So, the correct answer is option (b).