Question

A sample of mixture of $\text{CaC}{{\text{l}}_{\text{2}}}$ and $\text{NaCl}$ weighing $\text{5}\text{.55 g}$ was treated to precipitate all the $\text{Ca}$ as $\text{CaC}{{\text{O}}_{\text{3}}}$ which was then heated and quantitatively converted to $\text{1}\text{.68 g}$ of $\text{CaO}$. The mass percentage of $\text{CaC}{{\text{l}}_{\text{2}}}$ in the mixture is:
a. $\text{60 }\!\!%\!\!\text{ }$
b. $\text{70 }\!\!%\!\!\text{ }$
c. $\text{45 }\!\!%\!\!\text{ }$
d. $\text{56 }\!\!%\!\!\text{ }$

Answer 1

Hint: When all the carbon atoms are precipitated as calcium carbonate the number of moles formed of calcium carbonate should be the same as that of Calcium moles initially present.

Complete step by step answer:
As we know that calcium chloride $\text{CaC}{{\text{l}}_{\text{2}}}$ is and inorganic compound. It is a white colour crystalline solid and is highly soluble in water. It is used as De-icing agent and helps in freezing-point depression. Moreover, sodium chloride $\text{NaCl}$ commonly known as salt is an ionic white coloured crystalline compound. It is used in food and act as main ingredient in many process and industries.
Now as per question we got to know that combined mass of $\text{NaCl}$ and $\text{CaC}{{\text{l}}_{\text{2}}}$ is $\text{5}\text{.55 g}$. All the $\text{Ca}$ atoms are precipitated in mixture as $\text{CaC}{{\text{O}}_{\text{3}}}$. $\text{CaC}{{\text{O}}_{\text{3}}}$ is then decomposed to give $\text{CaO}$ and $\text{C}{{\text{O}}_{\text{2}}}$. The mass of $\text{CaO}$ given is $\text{1}\text{.68 g}$. Molecular mass of $\text{CaO}$ as we know is $\text{56 g}$. So, the number of moles of $\text{CaO}$ produced in reaction is $\dfrac{\text{1}\text{.68}}{\text{56}}\text{ = 0}\text{.03moles}$. The reaction takes place as $\text{CaC}{{\text{O}}_{\text{3}}}\to \text{CaO + C}{{\text{O}}_{\text{2}}}$. According to the equation, we got to know that $\text{1 mole}$of $\text{CaO}$ is formed by the decomposition of $\text{1 mole}$ of $\text{CaC}{{\text{O}}_{\text{3}}}$ . So, $\text{0}\text{.03 moles}$ of $\text{CaO}$ should be prepared by decomposition of $\text{0}\text{.03 moles}$ of $\text{CaC}{{\text{O}}_{\text{3}}}$ . The $\text{CaC}{{\text{O}}_{\text{3}}}$ is formed from $\text{CaC}{{\text{l}}_{\text{2}}}$ . So, the moles of $\text{CaC}{{\text{l}}_{\text{2}}}$ will also be $\text{0}\text{.03 moles}$. This implies that mass of $\text{CaC}{{\text{l}}_{\text{2}}}$ present in mixture must be $\text{0}\text{.03 }\!\!\times\!\!\text{ 111 g = 3}\text{.33 g}$. As $\text{111 g}$ is the molecular mass of $\text{CaC}{{\text{l}}_{\text{2}}}$. So, the percentage of $\text{CaC}{{\text{l}}_{\text{2}}}$ in the mixture should be expressed using the formula $\dfrac{\text{Mass of substance}}{\text{Total mass of compound}}\text{ }\!\!\times\!\!\text{ 100}$.
Percentage of $Cacl_2$ = $\dfrac{3.33 }{ 5.55}$$\times$ 100 = 60 %
Putting values in the above equation, we get $\text{60 }\!\!%\!\!\text{ }$.
So, the percentage of $\text{CaC}{{\text{l}}_{\text{2}}}$ present in the mixture is $\text{60 }\!\!%\!\!\text{ }$. Hence Option \[a\] is the correct option.

Hence, the correct option is a.

Note: Weight of sodium chloride can be evaluated by subtracting mass of calcium chloride from the total weight of the compound and from there we can easily calculate its moles and percentage composition in the compound.