calculate the amount of potassium chlorate needed to supply sufficient oxygen for burning 112 litre of carbon monoxide gas at ntp
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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
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
Answered by
48
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×=
moles of KClO3.
Molecular mass of KClO3 = 122.5 gms/mole.
So amount of KClO3 required = 122.5×gms.
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