Calculate the amount of KClO3 needed to supply sufficient oxygen for burning 112 l of CO gas at NTP?

Please elaborate nicely.

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Solution

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 2.52/3 = 5/3 moles of KClO3.

Molecular mass of KClO3 = 122.5 gms/mole

So amount of KClO3 required = 122.55/3 gms=204.166 gms

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Calculate the amount of KClO3 needed to supply sufficient oxygen for burning 112 l of CO gas at NTP? Please elaborate nicely.

Calculate the amount of KClO3 needed to supply sufficient oxygen for burning 112 l of CO gas at NTP?

Please elaborate nicely.