Question

What mass of $CaCO_{3}$ is required to react completely with 25 ml of 0.75 M HCl according to the reaction $CaCO_{3}\left(s\right)+2HCl\left(aq\right) \rightarrow CaCl_{2}\left(aq\right)+CO_{2}\left(g\right)+H_{2}O\left(l\right)$

• A. 1 g
• B. 0.3 g
• C. 0.8 g
• D. 0.93 g

Answer 1

Answer: (D)

Given
$\left(\text{CaCO}\right)_{\text{3}} \text{(s)} \, \text{+} \, \text{2HCl(aq)} \rightarrow \left(\text{CaCl}\right)_{\text{2}} \text{(aq)} \, \text{+} \, \left(\text{CO}\right)_{\text{2}} \text{(g)} \, \text{+} \, \left(\text{H}\right)_{\text{2}} \text{O(l)}$
Moles present in $\text{25} \, \text{ml}$ of $0.75 \, \text{M} \, \text{HCl} = \frac{0.75}{1000} \times 25$
$=0.01875$ moles
Since 2 moles react completely with $\text{CaCO}_{\text{3}} \, \text{=} \, \text{100} \, \text{g}$
$\text{0} \text{.01875}$ moles react completely with $\text{CaCO}_{\text{3}} \, \text{=} \, \frac{\text{100}}{\text{2}} \, \times \, \text{0} \text{.01875}$
$= 0.9375 \, \text{g}$