Q26 / 170 · Manipal (MET) 2017 · Chemistry

The mass of $\mathrm{CaCO}_3$ required to react with 25 mL of 0.75 M HCl is

  • Correct answer
    A
    0.94 g
  • B
    9.4 g
  • C
    0.094 g
  • D
    0.49 g

Solution

$\begin{gathered} \mathrm{CaCO}_3+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \\ 25 \mathrm{~mL} \text { of } 0.75 \mathrm{M} \mathrm{HCl}=\frac{25}{1000} \times 0.75=0.01875 \mathrm{~mol} \end{gathered}$ $\begin{aligned} & \text { Moles of } \mathrm{CaCO}_3 \text { required }=\frac{\text { moles of } \mathrm{HCl}}{2} \\ & \qquad=\frac{0.01875}{2}=9.375 \times 10^{-3} \mathrm{~mol} \end{aligned}$ $\begin{aligned} & \text { Mass of } \mathrm{CaCO}_3 \text { required } \\ & \quad=9.375 \times 10^{-3} \mathrm{~mol} \times 100 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \quad=0.9375 \mathrm{~g}=0.94 \mathrm{~g}\end{aligned}$

Hence, the correct answer is option: 'A'.

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