Q22 / 75 · JEE Main 2025 (2 Apr Shift 1) · Chemistry
\(\mathrm{CaCO}_3(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{CO}_2(\mathrm{~g}) \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)
Consider the above reaction, what mass of \(\mathrm{CaCl}_2\) will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of \(\mathrm{CaCO}_3\) ?
(Given : Molar mass of \(\mathrm{Ca}, \mathrm{C}, \mathrm{O}, \mathrm{H}\) and Cl are 40, \(12,16,1\) and \(35.5 \mathrm{~g} \mathrm{~mol}^{-1}\), respectively)
Consider the above reaction, what mass of \(\mathrm{CaCl}_2\) will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of \(\mathrm{CaCO}_3\) ?
(Given : Molar mass of \(\mathrm{Ca}, \mathrm{C}, \mathrm{O}, \mathrm{H}\) and Cl are 40, \(12,16,1\) and \(35.5 \mathrm{~g} \mathrm{~mol}^{-1}\), respectively)
- A3.908 g
- B2.636 g
- Correct answerC10.545 g
- D5.272 g
Solution
\(\begin{aligned}<br />& \mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \\<br>& \text { Moles of } \mathrm{CaCO}_3=\frac{1000}{100}=10<br />\end{aligned}\)
Moles of \(\mathrm{HCl}=0.76 \times \frac{250}{1000}=0.19\) (L.R.)
Moles of \(\mathrm{CaCl}_2\) formed \(=\frac{0.19}{2}\)
Mass of \(\mathrm{CaCl}_2=\frac{0.19}{2} \times 111=10.545 \mathrm{gm}\)
Moles of \(\mathrm{HCl}=0.76 \times \frac{250}{1000}=0.19\) (L.R.)
Moles of \(\mathrm{CaCl}_2\) formed \(=\frac{0.19}{2}\)
Mass of \(\mathrm{CaCl}_2=\frac{0.19}{2} \times 111=10.545 \mathrm{gm}\)
Hence, the correct answer is option: 'C'.