The amount of 50% (w/w) solution of hydrochl | vedclass

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(B) The balanced chemical equation is:$CaCO_3 + 2HCl ightarrow CaCl_2 + CO_2 + H_2O$Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g/mol$.M

Vedclass · Accepted Answer

$292$

Vedclass · Answer

(B) The balanced chemical equation is:$CaCO_3 + 2HCl ightarrow CaCl_2 + CO_2 + H_2O$Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g/mol$.Moles of $CaCO_3 = \frac{200 \ g}{100 \ g/mol} = 2 \ mol$.According to the stoichiometry, $1 \ mol$ of $CaCO_3$ reacts with $2 \ mol$ of $HCl$.Therefore, $2 \ mol$ of $CaCO_3$ requires $4 \ mol$ of $HCl$.Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.Mass of pure $HCl$ required $= 4 \ mol 	imes 36.5 \ g/mol = 146 \ g$.Given that the solution is $50\% \ (w/w)$, it means $50 \ g$ of $HCl$ is present in $100 \ g$ of solution.Mass of solution $= \frac{146 \ g}{0.50} = 292 \ g$.

The amount of $50\% \ (w/w)$ solution of hydrochloric acid required to react with $200 \ g$ of $CaCO_3$ would be (in $g$)

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