Question:
The mass of \(CaCO_3\) (in g) required to react completely with 25 mL of 0.75 M HCl is:
\[
(Ca=40,\; C=12,\; O=16)
\]
The mass of \(CaCO_3\) (in g) required to react completely with 25 mL of 0.75 M HCl is:
\[
(Ca=40,\; C=12,\; O=16)
\]
Show Hint
Always convert volume from mL to L before applying
\[
n=MV
\]
in molarity problems.
Updated On: Jun 12, 2026
- \(0.94\)
- \(0.47\)
- \(1.88\)
- \(0.79\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
The reaction is
\[
CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2
\]
The stoichiometric ratio is
\[
1\; \text{mol } CaCO_3
:
2\; \text{mol } HCl
\]
Step 1: Calculate moles of HCl. \[ M=\frac{n}{V} \] \[ n=MV \] \[ n=0.75\times\frac{25}{1000} \] \[ n=0.01875\;mol \]
Step 2: Find moles of \(CaCO_3\) required. From stoichiometry, \[ 2\;mol\;HCl \rightarrow 1\;mol\;CaCO_3 \] \[ n(CaCO_3) = \frac{0.01875}{2} = 0.009375\;mol \]
Step 3: Calculate mass. Molar mass of \(CaCO_3\): \[ 40+12+48=100 \] \[ m=nM \] \[ m=0.009375\times100 \] \[ m=0.9375g \] \[ \boxed{0.94g} \]
Step 1: Calculate moles of HCl. \[ M=\frac{n}{V} \] \[ n=MV \] \[ n=0.75\times\frac{25}{1000} \] \[ n=0.01875\;mol \]
Step 2: Find moles of \(CaCO_3\) required. From stoichiometry, \[ 2\;mol\;HCl \rightarrow 1\;mol\;CaCO_3 \] \[ n(CaCO_3) = \frac{0.01875}{2} = 0.009375\;mol \]
Step 3: Calculate mass. Molar mass of \(CaCO_3\): \[ 40+12+48=100 \] \[ m=nM \] \[ m=0.009375\times100 \] \[ m=0.9375g \] \[ \boxed{0.94g} \]
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