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  4. The mass of CaCO3 is required to react with 25 mL of 0.75 M HCl is

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Q. The mass of $CaCO_3$ is required to react with 25 mL of 0.75 M HCl is

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Solution:

$CaCO_{3}+2HCl \to CaCl_{2}+CO_{2}+H_{2}$
$25\, mL$ 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
$=9.375\times10^{-3}\,mol\times 100\,g\,mol^{-1}$
$=0.9375\,g=0.94\,g$