what is the volume of Co2 gas liberated on calcination of 50gr of CaCo3
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Given - Mass of CaCO3 - 50 gram
Find - Volume of carbon dioxide liberated.
Solution - The calcination reaction of CaCO3 - CaCO3 --> CaO + CO2.
100 gram of CaCO3 gives 44 gram of CO2
50 gram of CaCO3 will give - 44*50/100
22 gram of CO2 will release on calcination of 50 gram of CaCO3.
Number of moles of CO2 - 22/44 - 0.5 mole.
1 mole of gas occupies volume of 22.4 litre.
0.5 mole of gas will occupy - 22.4*0.5 - 11.2 litre.
Thus, volume of CO2 gas liberated will be 11.2 litre.
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