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