Assuming full decomposition, the volume of `CO_(2)` released at STP on heating 9.85 g of `BaCO_(3)` (At mass Ba = 137) will be
A. 0.84 L
B. 0.24 L
C. 4.06 L
D. 1.12 L
A. 0.84 L
B. 0.24 L
C. 4.06 L
D. 1.12 L
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Correct Answer – D
`underset(197)(BaCO_(3))toBaO+underset(22.4L)(CO_(2))`
`V_(CO_(2))` released `= (22.4)/(197)xx9.85 = 1.12L`