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Assuming fully decomposed,the volume of $CO_2$ released at STP,on heating 9.85g of $BaCO_3$ (at wt of Ba=137)will be:


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$BaCO_3\qquad\underrightarrow {\Delta}\qquad BaO+CO_2$
9.85g $BaCO_3$ produced $=\large\frac{22.4}{197}$$\times 9.85$
$\Rightarrow 1.12l\;CO_2$
Hence (c) is the correct answer.
answered Jan 8, 2014 by sreemathi.v

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