Question

Calcium carbonate reacts with aqueous $HCl$ to give $CaCl_2$ and $CO_2$ according to the reaction,
$CaCO_{3(s)} + 2HCl_{(aq)} \to CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
What mass of $CaCO_3$ is required to react completely with $25\, mL$ of $0.75\, M \,HCl$ ?

• A. 5 g
• B. 10 g
• C. 0.94 g
• D. 8.5 g

Answer 1

Answer: (C)

$CaCO_{3(s)} + 2HCl_{(aq)} \to CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
No. of moles of $HCl$ given $= M_{HCl} V_{HCl}$
$= 0.75 \,mol \,L^{-1} \times 25 \times 10^{-3 }\,L = 0.0188 \,mol$
$2$ moles of $HCl$ requires $1$ mole $CaCO_3$
$\therefore 0.0188$ mole of $HCl$ will require $= \frac{0.0188}{2}$
$= 0.0094$ mole of $CaCO_3$
Molar mass of $CaCO_3 = 100 \,g/mol$
$\therefore$ Mass of $CaCO_3$ required
$= 100 \times 0.0094 = 0.94 \,g$