Question

Given,10 g sample of a mixture of \[CaC{{l}_{2}}\] and $NaCl$ is treated to precipitate all the calcium as \[CaC{{O}_{3}}\] . This \[CaC{{O}_{3}}\] is heated to convert all the $Ca$ to \[CaO\] and final mass of \[CaO\] is 1.68g the mass % of \[CaC{{l}_{2}}\] in original mixture is?
1. 33.3 %
2. 16.2 %
3. 30 %
4. 11 %
Please answer with a full explanation.

Answer 1

Hint: The final mass of \[CaO\] is 1.68 g. To begin with, first, we have to calculate the mass of \[CaC{{O}_{3}}\] . Using the mass of \[CaC{{O}_{3}}\] we will find the mass of \[CaC{{l}_{2}}\] . To find the mass of \[CaC{{O}_{3}}\] and \[CaC{{l}_{2}}\] you have to know the molecular masses of \[CaO\] , \[CaC{{O}_{3}}\] and \[CaC{{l}_{2}}\] . At the final step find the percent mass of \[CaC{{l}_{2}}\]in 10 gm of the mixture.

Complete answer:
The first reaction is the precipitation of \[CaC{{O}_{3}}\] into \[CaO\] and $C{{O}_{2}}$, which is as follows:
\[CaC{{O}_{3}}\text{ }\xrightarrow{Heat}\text{ CaO + C}{{\text{O}}_{2}}\]
We are given the final mass of \[CaO\] that is 1.68 g. Let the mass of \[CaC{{O}_{3}}\] be X.
The molecular mass of \[CaO\] = 56u
The molecular mass of \[CaC{{O}_{3}}\] = 100u

Compound$\to $	\[CaC{{O}_{3}}\]	\[CaO\]
Given mass	X	1.68
Molecular mass	100	56

Cross multiplying the mass and molecular mass we get,
$x\text{ }\times \text{ 56 = 1}\text{.68 }\times \text{ 100}$
$x\text{ = }\dfrac{1.68\text{ }\times \text{ 100}}{56}$
$x\text{ = 3 gm}$
Thus, the mass of \[CaC{{O}_{3}}\] = 3 gm.
Now, the second equation is the reaction between \[CaC{{l}_{2}}\] and $N{{a}_{2}}C{{O}_{3}}$ to give \[CaC{{O}_{3}}\] and $NaCl$.
\[CaC{{l}_{2}}\text{ + }N{{a}_{2}}C{{O}_{3}}\text{ }\to \text{ }CaC{{O}_{3}}\text{ + 2NaCl}\]
Let the mass of \[CaC{{l}_{2}}\] be Y.
The molecular mass of \[CaC{{l}_{2}}\] = 111u.
The molecular mass of \[CaC{{O}_{3}}\] = 100u

Compound$\to $	\[CaC{{l}_{2}}\]	\[CaC{{O}_{3}}\]
Given mass	Y	3
Molecular mass	111	100

Cross multiplying the mass and molecular mass we get,
$Y\text{ }\times \text{100 = 3 }\times \text{ 111}$
$Y\text{ = }\dfrac{333}{100}$
$Y\text{ = 3}\text{.33 gm}$
Thus, the mass of \[CaC{{l}_{2}}\] = 3.33 gm.
Now, we have to calculate the percent of \[CaC{{l}_{2}}\] in 10 gm of the reaction mixture.
$\% CaC{l_2}$ = $\dfrac{3.33}{10} \times 100$
The correct answer is Option 1 = 33.3 %.

Note:
In the above reaction, \[CaC{{l}_{2}}\] is converted to \[CaC{{O}_{3}}\] and then to \[CaO\]. You can also use one mole of each compound by dividing the given mass by molecular mass. To find the molecular masses the atomic mass of each element has to be used which is multiplied by the number of each element in a compound.