What volume of 0.1 , M , HCl in cm^3 is requi | vedclass

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

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) The chemical reaction is: $CaCO_3 + 2HCl ightarrow CaCl_2 + H_2O + CO_2$. Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \, g/mol$. Moles of $CaCO_3 = \frac{1 \, g}{100 \, g/mol} = 0.01 \, mol$. According to the stoichiometry,$1 \, mol$ of $CaCO_3$ reacts with $2 \, mol$ of $HCl$. Therefore,$0.01 \, mol$ of $CaCO_3$ requires $0.02 \, mol$ of $HCl$. Using the formula $M = \frac{n}{V(L)}$,we have $V(L) = \frac{n}{M} = \frac{0.02 \, mol}{0.1 \, M} = 0.2 \, L$. Converting to $cm^3$ (or $mL$): $0.2 \, L 	imes 1000 \, cm^3/L = 200 \, cm^3$.

What volume of $0.1 \, M \, HCl$ in $cm^3$ is required to react completely with $1 \, g$ of pure calcium carbonate $(CaCO_3)$?

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