100 , g CaCO_3 reacts with 1 , L 1 , N HCl. O | vedclass

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(C) The balanced chemical equation is: $CaCO_3 + 2HCl 	o CaCl_2 + CO_2 + H_2O$Equivalent weight of $CaCO_3 = 50 \, g/eq$. So,$100 \, g$ of $CaCO_3$ c

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$22$

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(C) The balanced chemical equation is: $CaCO_3 + 2HCl 	o CaCl_2 + CO_2 + H_2O$Equivalent weight of $CaCO_3 = 50 \, g/eq$. So,$100 \, g$ of $CaCO_3$ corresponds to $2 \, equivalents$.Given $1 \, L$ of $1 \, N$ $HCl$ corresponds to $1 \, equivalent$ of $HCl$.Since $HCl$ is the limiting reagent,the amount of $CO_2$ produced depends on the amount of $HCl$ consumed.$2 \, equivalents$ of $HCl$ produce $1 \, mole$ $(44 \, g)$ of $CO_2$.Therefore,$1 \, equivalent$ of $HCl$ will produce $0.5 \, moles$ of $CO_2$,which is $0.5 	imes 44 \, g = 22 \, g$ of $CO_2$.

$100 \, g$ $CaCO_3$ reacts with $1 \, L$ $1 \, N$ $HCl$. On completion of reaction,how much weight of $CO_2$ will be obtained? (in $g$)

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