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(D) The balanced chemical equation for the decomposition of $BaCO_3$ is:$BaCO_3(s) 	o BaO(s) + CO_2(g)$The molar mass of $BaCO_3 = 137 + 12 + (3 	im

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$1.12$

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(D) The balanced chemical equation for the decomposition of $BaCO_3$ is:$BaCO_3(s) 	o BaO(s) + CO_2(g)$The molar mass of $BaCO_3 = 137 + 12 + (3 	imes 16) = 197 \ g/mol$.According to the stoichiometry of the reaction,$1 \ mol$ of $BaCO_3$ $(197 \ g)$ produces $1 \ mol$ of $CO_2$ gas,which occupies $22.4 \ L$ at $STP$.Therefore,the volume of $CO_2$ produced from $9.85 \ g$ of $BaCO_3$ is:$V = \frac{22.4 \ L}{197 \ g} 	imes 9.85 \ g = 1.12 \ L$.

Assuming complete decomposition,the volume of $CO_2$ released at $STP$ on heating $9.85 \ g$ of $BaCO_3$ (Atomic mass of $Ba = 137$) will be ................ $L$.

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