What will be the volume of CO_2 at S.T.P. obt | vedclass

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(A) The chemical equation for the decomposition of $BaCO_3$ is:$BaCO_3(s) 	o BaO(s) + CO_2(g)$Calculate the molar mass of $BaCO_3$:$M(BaCO_3) = 137 +

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$1.12$

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(A) The chemical equation for the decomposition of $BaCO_3$ is:$BaCO_3(s) 	o BaO(s) + CO_2(g)$Calculate the molar mass of $BaCO_3$:$M(BaCO_3) = 137 + 12 + (3 	imes 16) = 137 + 12 + 48 = 197 \ g/mol$According to the stoichiometry of the reaction,$1 \ mol$ of $BaCO_3$ produces $1 \ mol$ of $CO_2$.At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.Therefore,$197 \ g$ of $BaCO_3$ produces $22.4 \ L$ of $CO_2$ at $S.T.P.$For $9.85 \ g$ of $BaCO_3$,the volume of $CO_2$ produced is:$V = \frac{22.4 \ L}{197 \ g} 	imes 9.85 \ g$$V = \frac{22.4 	imes 9.85}{197} = 1.12 \ L$Thus,the correct option is $A$.

What will be the volume of $CO_2$ at $S.T.P.$ obtained by heating $9.85 \ g$ of $BaCO_3$? (Atomic mass of $Ba = 137$)

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