Question

A sample of a mixture of $CaC{l_2}$ and $N{a_2}C{O_3}$ weighing 4.22g was treated to precipitate all the Ca as $CaC{O_3}$. This $CaC{O_3}$ is heated and quantitatively converted into 0.959g of $CaO$. Calculate the percentage of $CaC{l_2}$ in the mixture. (Atomic mass of Ca=40, O=16, C=12 and Cl=35.5g/mole)
A. 55.28%
B. 37.3%
C. 45.00%
D. 49.01%

Answer 1

Hint: We will find out the reactions of each step given in the question such as the precipitation and then the product was heated to give the resultant product. Then we will find out the weights of the compounds and then the percentage of $CaC{l_2}$ as per asked in the question. Refer to the solution below.

Complete step-by-step answer:
The molecular masses of the compounds are given in the question.
Molecular mass of $CaC{l_2}$ is 111g
Molecular mass of $N{a_2}C{O_3}$ is 106g
Molecular mass of $CaC{O_3}$ is 100g
Molecular mass of $CaO$ is 56g
Weight of $CaC{l_2}$ + $N{a_2}C{O_3}$= 4.22g
The first equation is-
$ \Rightarrow CaC{l_2} + N{a_2}C{O_3}\xrightarrow{{precipitation}}CaC{O_3}$
According to this equation, we can say that-
111g of $CaC{l_2}$ gives 100g of $CaC{O_3}$. (given in the question)
Equation for the decomposition of $CaC{O_3}$ is-
$ \Rightarrow CaC{O_3}\xrightarrow{\Delta }CaO + C{O_2}$
Now, according to this equation, it is clear that-
100g of $CaC{O_3}$ gives 56g of $CaO$. (given in the question)
So, the weight $CaC{O_3}$ required to form 0.959g $CaO$ = Molecular weight of $CaC{O_3}$/Molecular weight of $CaO$ multiplied by 0.959.
$
   \Rightarrow \dfrac{{100}}{{56}} \times 0.959 \\
    \\
   \Rightarrow 1.7125g \\
$
$CaC{O_3}$ formed will be 1.7125g
We already know that 111g of $CaC{l_2}$ gives 100g of $CaC{O_3}$.
Then, the weight of $CaC{l_2}$ required to form 1.7125g of $CaC{O_3}$ = Molecular weight of $CaC{l_2}$/Molecular weight of $CaC{O_3}$ multiplied by 1.7125.
$
   \Rightarrow \dfrac{{111}}{{100}} \times 1.7125 \\
    \\
   \Rightarrow 1.9009g \\
$
$CaC{l_2}$ formed will be 1.9009g
Therefore, percentage of $CaC{l_2}$ in the mixture = weight of $CaC{l_2}$/weight of $CaC{l_2}$+ $N{a_2}C{O_3}$ multiplied by 100.
$
   \Rightarrow \dfrac{{1.901}}{{4.22}} \times 100 \\
    \\
   \Rightarrow 45.04 \simeq 45\% \\
$
Hence, option C is the correct option.

Note: Precipitation is a mechanism that determines the formation of a material. The material shaped is labeled the "precipitate" while the reaction is in a liquid solution. It is called the precipitant that makes the solid form. The precipitate stays in equilibrium without adequate gravity (settling) to pull together stable particles.