What is the volume of 0.1 , N , HCl required | vedclass

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(D) The chemical reaction is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$. According to the law of equivalence,the number of equivalents of $HCl$

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) The chemical reaction is: $CaCO_3 + 2HCl ightarrow CaCl_2 + H_2O + CO_2$. According to the law of equivalence,the number of equivalents of $HCl$ must equal the number of equivalents of $CaCO_3$. Number of equivalents of $CaCO_3 = \frac{	ext{mass}}{	ext{equivalent weight}} = \frac{1.0 \, g}{50 \, g/eq} = 0.02 \, eq$. For $HCl$,the number of equivalents is given by $N 	imes V(L)$. $0.1 \, N 	imes V(L) = 0.02 \, eq$. $V(L) = \frac{0.02}{0.1} = 0.2 \, L$. Converting to $cm^3$: $0.2 \, L 	imes 1000 \, cm^3/L = 200 \, cm^3$.

What is the volume of $0.1 \, N \, HCl$ required to react completely with $1.0 \, g$ of pure calcium carbonate? (in $cm^3$)

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