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Question

Assuming fully decomposed, the volume of CO2 released at STP on heating 9.85 g of BaCO3 will be: (Atomic mass of Ba=137)

A
0.84 L
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B
2.24 L
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C
4.06 L
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D
1.12 L
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Solution

The correct option is D 1.12 L
BaCO3BaO+CO2
When 197 grams of BaCO3 is decomposed it forms 22.4 liters of CO2
If 9.85 grams of BaCO3 is decomposed it forms X liters of CO2.
X=9.85×22.4197=1.12.
Hence volume is 1.12 liters

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