Question

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

Answer 1

CO is a gas.  Assume it is ideal.
22.4 Liters of CO at STP 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

Answer 2

Answer:

204.6 g of KClO3 is needed to supply sufficient oxygen for burning 112L of CO at STP.

Explanation:

For explanation, please refer to the given attachments.

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