Question

The mass of $CaCO_3$ 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)

0.94 g

Answer 2

$CaCO_{3}+2HCl \to CaCl_{2}+CO_{2}+H_{2}$ $25\, mL$ of $0.75\, M\, HCl$ $=\frac{25}{100}L\times (0.75\,mol\,L^{-1})$ $= 0.01875 \,mol$ Moles of $CaCO_{3}$ required $=\frac{\text{Moles of HCl}}{2}$ $=\frac{0.01875}{2}=9.375\times10^{-3}\,mol$ Mass of $CaCO_{3}$ required $=9.375\times10^{-3}\,mol\times 100\,g\,mol^{-1}$ $=0.9375\,g=0.94\,g$