Question

Calculate the amount of KClO3 needed to supply sufficient oxygen for burning 112 L of CO at NTP.

Answer 1

CO is a gas. Assume it is ideal.
22.4 Liters of CO at NTP is equal to 1 mole.
So 112 L of CO is equal to 5 moles.

2 CO + O2 ==> 2 CO2
We need 1 mole of Oxygen for 2 moles of CO to burn. So for 112L of CO we need 2.5 moles of O2.

Decomposition of Chlorate:
2 K Cl O3 => 2 KCl + 3 O2
So 3 moles of Oxygen are obtained from 2 moles of Potassium chlorate.
We get 2.5 moles of Oxygen from 2.5*2/3 = 5/3 moles of KClO3.

Molecular mass of KClO3 = 122.5 gms/mole

So amount of KClO3 required = 122.5*5/3 gms = 204.166 gms