Question

Calcium carbonate reacts with aqueous $HCl$ to give $CaC{l}_2$ and $C{O}_2$ according to the reaction,
$CaC{O}_{3(s)}+2HC{l}_{(aq)}\to CaC{l}_{2(aq)}+C{O}_{2(g)}+H_2{O}_{(l)}$
What mass of $CaC{O}_3$ is required to react completely with $25mL$ of $0.75MHCl$ ?

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

Answer 1

Answer: (C)

$CaC{O}_{3(s)}+2HC{l}_{(aq)}\to CaC{l}_{2(aq)}+C{O}_{2(g)}+H_2{O}_{(l)}$
No. of moles of $HCl$ given $=M_{HCl}{V}_{HCl}$
$=0.75mol{L}^{-1}\times 25\times 10^{-3}L=0.0188mol$
$2$ moles of $HCl$ requires $1$ mole $CaC{O}_3$
$\therefore 0.0188$ mole of $HCl$ will require $=\frac{0.0188}{2}$
$=0.0094$ mole of $CaC{O}_3$
Molar mass of $CaC{O}_3=100g/mol$
$\therefore$ Mass of $CaC{O}_3$ required
$=100\times 0.0094=0.94g$