Question

What mass of $CaC{O}_3$ is required to react completely with 25 ml of 0.75 M HCl according to the reaction $CaC{O}_3\left(s\right)+2HCl\left(aq\right)\to CaC{l}_2\left(aq\right)+C{O}_2\left(g\right)+H_{2}O\left(l\right)$

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

Answer 1

Answer: (D)

Given
${\left(\text{CaCO}\right)}_{\text{3}}\text{(s)}\text{+}\text{2HCl(aq)}\to {\left(\text{CaCl}\right)}_{\text{2}}\text{(aq)}\text{+}{\left(\text{CO}\right)}_{\text{2}}\text{(g)}\text{+}{\left(\text{H}\right)}_{\text{2}}\text{O(l)}$
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 ${\text{CaCO}}_{\text{3}}\text{=}\frac{\text{100}}{\text{2}}\times \text{0}\text{.01875}$
$=0.9375\text{g}$