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Question

On heating a mixture of one mole each of Li2CO3,MgCO3 and Na2CO3, the volume of CO2 evolved under STP conditions will be:

A
67.2 L
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B
33.6 L
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C
44.8 L
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D
22.4 L
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Solution

The correct option is D 44.8 L

Na2CO3 does not decompose on heating while both Li2CO3 and MgCO3 give CO2

MgCO3MgO+CO2

1 mole

1 mole 1 mole

Li2O3Li2O+CO2

1 mole 1 mole 1 mole

Therefore 2 moles CO2 will be evolved. So, volume occupied by 2 mole gas at STP is

2×22.4=44.8 liters


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