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

$\begin{array}{1 1}(a)\;0.84L& (b)\;2.24L \\(c)\;4.06L&(d)\;1.12L\end{array}$

$BaCO_3\rightarrow BaO+CO_2$
$\therefore$ 197g of $BaCO_3$ gives 22.4L of $CO_2$ at STP.
$9.85g$ $BaCO_3$ gives =$\large\frac{22.4\times 9.85}{197}$
$\Rightarrow 1.122$
Hence (d) is the correct answer.