CaCO_3 reacts with HCl to produce CaCl_2, CO_ | vedclass

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Question

(C) The balanced chemical equation is: $CaCO_3 + 2HCl \longrightarrow CaCl_2 + CO_2 + H_2O$ Moles of $HCl = 	ext{Molarity} 	imes 	ext{Volume (in L)

Vedclass · Accepted Answer

$0.94$

Vedclass · Answer

(C) The balanced chemical equation is: $CaCO_3 + 2HCl \longrightarrow CaCl_2 + CO_2 + H_2O$ Moles of $HCl = 	ext{Molarity} 	imes 	ext{Volume (in L)} = 0.75 \ M 	imes 0.025 \ L = 0.01875 \ mol$. According to the stoichiometry of the reaction, $1 \ mol$ of $CaCO_3$ reacts with $2 \ mol$ of $HCl$. Therefore, moles of $CaCO_3$ required $= \frac{0.01875}{2} = 0.009375 \ mol$. Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g/mol$. Mass of $CaCO_3 = 0.009375 \ mol 	imes 100 \ g/mol = 0.9375 \ g \approx 0.94 \ g$.

$CaCO_3$ reacts with $HCl$ to produce $CaCl_2, CO_2$ and $H_2O$. The approximate mass (in $g$) of $CaCO_3$ required to react completely with $25 \ mL$ of $0.75 \ M \ HCl$ is (atomic mass of $Ca=40, C=12, O=16, Cl=35.5$ and $H=1$).

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