Question

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

Answer 1

We know that 22.4 L of CO at stp is equal to 1 Mole

So 112 L of CO is equal to 5 moles.

        Now, 2 CO + O2  ==> 2 CO2

So,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 is as follows:
        2 K Cl O3  => 2 KCl + 3 O2

Hence,3 moles of Oxygen are obtained from 2 moles of KClO3.

 And we get 2.5 moles of Oxygen from  2.5*2/3 = 5/3 moles of KClO3.

As, Molecular mass of KClO3 = 122.5 gms/mole 

Therefore,the amount of KClO3 required = 122.5*5/3  gms

Hope it helps!