Question

The mass of $CaC{O}_3$ is required to react with 25 mL of 0.75 M HCl is

• A. 0.94 g
• B. 9.4 g
• C. 0.094 g
• D. 0.49 g

Answer 1

Answer: (A)

$CaC{O}_3+2HCl\to CaC{l}_2+C{O}_2+H_2$
$25mL$ of $0.75MHCl$
$=\frac{25}{100}L\times (0.75mol{L}^{-1})$
$=0.01875mol$
Moles of $CaC{O}_3$ required $=\frac{\text{Moles of HCl}}{2}$
$=\frac{0.01875}{2}=9.375\times 10^{-3}mol$
Mass of $CaC{O}_3$ required
$=9.375\times 10^{-3}mol\times 100gmo{l}^{-1}$
$=0.9375g=0.94g$