Question

calculate the amount of KCLO3 needed to supply sufficient oxygen for burning 112L of CO gas at NTP.

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  ANS

Answer 2

Answer:

Explanation:

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