10 g of a piece of marble (CaCO_3) was put i | vedclass

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(C) The chemical equation for the reaction is: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$.According to the stoichiometry,$1 \ m

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$50$

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(C) The chemical equation for the reaction is: $CaCO_3(s) + 2HCl(aq) ightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$.According to the stoichiometry,$1 \ mol$ of $CaCO_3$ $(100 \ g)$ produces $22400 \ mL$ of $CO_2$ at $STP$.If the marble were $100 \%$ pure,$10 \ g$ of $CaCO_3$ would produce: $\frac{22400 \ mL}{100 \ g} 	imes 10 \ g = 2240 \ mL$ of $CO_2$.However,only $1120 \ mL$ of $CO_2$ was obtained.The percentage purity of $CaCO_3$ is calculated as: $\frac{	ext{Actual volume of } CO_2}{	ext{Theoretical volume of } CO_2} 	imes 100 = \frac{1120 \ mL}{2240 \ mL} 	imes 100 = 50 \%$.

$10 \ g$ of a piece of marble $(CaCO_3)$ was put into excess of dilute $HCl$ acid. When the reaction was complete,$1120 \ mL$ of $CO_2$ was obtained at $STP$. The percentage of pure $CaCO_3$ in the marble is $............ \%$.

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