What is the weight of 50% pure HCl required t | vedclass

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

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$146$

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(B) The chemical reaction is: $CaCO_3 + 2HCl ightarrow CaCl_2 + H_2O + CO_2$.Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g/mol$.Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.From the stoichiometry,$1 \ mol$ of $CaCO_3$ $(100 \ g)$ reacts with $2 \ mol$ of $HCl$ ($2 	imes 36.5 = 73 \ g$ of pure $HCl$).Since we have $100 \ g$ of $CaCO_3$,we need $73 \ g$ of pure $HCl$.The $HCl$ solution is $50\%$ pure by weight,meaning $50 \ g$ of pure $HCl$ is present in $100 \ g$ of the solution.Therefore,the weight of the $50\%$ $HCl$ solution required = $\frac{73 \ g}{0.50} = 146 \ g$.

What is the weight of $50\%$ pure $HCl$ required to react completely with $100 \ g$ of $100\%$ pure limestone $(CaCO_3)$ (in $g$)?

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