CaCO_{3(s)} + 2 HCl_{(aq)} rightarrow CaCl_{2 | vedclass

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Question

(C) The balanced chemical equation is: $CaCO_{3(s)} + 2 HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$Molar mass of $CaCO_3 = 40 + 12 +

Vedclass · Accepted Answer

$10.545$

Vedclass · Answer

(C) The balanced chemical equation is: $CaCO_{3(s)} + 2 HCl_{(aq)} ightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g \ mol^{-1}$.Moles of $CaCO_3 = \frac{1000 \ g}{100 \ g \ mol^{-1}} = 10 \ mol$.Moles of $HCl = Molarity 	imes Volume (L) = 0.76 \ mol \ L^{-1} 	imes 0.250 \ L = 0.19 \ mol$.According to the stoichiometry,$2 \ mol$ of $HCl$ reacts with $1 \ mol$ of $CaCO_3$.Since $0.19 \ mol$ of $HCl$ is present,it will react with $0.095 \ mol$ of $CaCO_3$.Thus,$HCl$ is the Limiting Reagent $(L.R.)$.Moles of $CaCl_2$ formed $= \frac{1}{2} 	imes 	ext{moles of } HCl = \frac{0.19}{2} = 0.095 \ mol$.Molar mass of $CaCl_2 = 40 + (2 	imes 35.5) = 111 \ g \ mol^{-1}$.Mass of $CaCl_2 = 0.095 \ mol 	imes 111 \ g \ mol^{-1} = 10.545 \ g$.

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