Assuming full decomposition, the volume of CO2 released at STP on heating 9.85g of BaCO3 (Atomic mass of Ba=137) will be
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 BaCO3→BaO+CO2↑
Molecular weight of BaCO3=137+12+3×16=197 ∵ 197gm produces 22.4L at S.T.P. ∴ 9.85gm produces 22.4197×9.85=1.12L at S.T.P.