Question

According to the equation above, how many moles of potassium chlorate, KClO3, must be decomposed to generate 1.0 L of O2 gas at standard temperature and pressure?

Answer 1

Answer:

Moles of potassium chlorate = 0.02976 moles

Explanation:

At standard pressure and temperature,

22.4 L of a gas consists of 1 mole

Thus, given, volume of $O_2$ = 1.0 L

So,

1 L of a gas consists of $\frac{1}{22.4}$ mole

Moles of oxygen gas = 0.04464 moles

The reaction is shown below as:-

$2KClO_3\rightarrow 2KCl+3O_2$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

1 mole of oxygen gas are produced when $\frac{2}{3}$ moles of potassium chlorate undergoes reaction.

Thus,

0.04464 mole of oxygen gas are produced when $\frac{2}{3}\times 0.04464$ moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate = 0.02976 moles