Question

10g of $Mn{O_2}$ on reaction with $HCl$ forms $2.24L$ of $C{l_2}\left( g \right)$ at $NTP$, the percentage impurity of $Mn{O_2}$ is $Mn{O_2} + 4HCl \to MnC{l_2} + Cl{ _2} + 2{H_2}O$
A.$87\% $
B.$25\% $
C.$33.3\% $
D.$13\% $

Answer 1

Hint: We can calculate the percentage impurity using the mass of impurity and mass of $Mn{O_2}$ multiplied by 100. We have to calculate the moles of $C{l_2}\left( g \right)$ using the volume of $C{l_2}\left( g \right)$ and molar volume. Then, we have to calculate the mass of $Mn{O_2}$ using the moles of $Mn{O_2}$ and molar mass of $Mn{O_2}$. From the calculated mass of $Mn{O_2}$, we have to calculate the mass of impurity in $Mn{O_2}$ using calculated mass of $Mn{O_2}$ and total mass of $Mn{O_2}$. From the mass of impurity in $Mn{O_2}$, we can calculate the percentage impurity.

Complete answer:
Given data contains
Mass of $Mn{O_2}$ is $10g$.
Volume of $C{l_2}\left( g \right)$ is $2.24L$.
The given equation is,
$Mn{O_2} + 4HCl \to MnC{l_2} + Cl{ _2} + 2{H_2}O$
From the given equation, we can see that one mole of $Mn{O_2}$ forms one mole of $C{l_2}$.
Let us now calculate the number of moles of chlorine gas. We can calculate the number of moles of chlorine gas using the molar volume and given volume of chlorine at $NTP$. We know that the value of molar volume is $22.4L$.
We can write the formula to calculate the number of moles of chlorine gas as,
Number of moles of chlorine gas$ = \dfrac{{{\text{Given volume}}}}{{{\text{Molar volume}}}}$
Let us now substitute the values of given volume and molar volume in the expression to calculate the number of moles of chlorine gas.
Number of moles of chlorine gas$ = \dfrac{{{\text{Given volume}}}}{{{\text{Molar volume}}}}$
Number of moles of chlorine gas=$\dfrac{{2.24L}}{{22.4mol/L}}$
Number of moles of chlorine gas=$0.1mol$
The number of moles of chlorine gas is $0.1mol$.
We know that one mole of one mole of $Mn{O_2}$ forms one mole of $C{l_2}$.
So, $0.1mol$ of $Mn{O_2}$ forms $0.1mol$ of $C{l_2}$.
From the moles of $Mn{O_2}$, we can calculate the mass of $Mn{O_2}$ using the molar mass of $Mn{O_2}$.
We have to know that the molar mass of $Mn{O_2}$ is $87g/mol$.
Mass of $Mn{O_2}$=$0.1\not{{mol}} \times 87\dfrac{g}{{\not{{mol}}}}$
Mass of $Mn{O_2}$=$8.7g$
The mass of $Mn{O_2}$ is $8.7g$.
We can now calculate the impurity of $Mn{O_2}$ using the difference of total mass of $Mn{O_2}$ and the calculated mass of $Mn{O_2}$.
The formula to calculate the impurity of $Mn{O_2}$ is written as,
Impurity in $Mn{O_2}$$ = {\text{Total mass of }}Mn{O_2} - {\text{Calculated mass of }}Mn{O_2}$
Let us substitute the values of total mass of $Mn{O_2}$ and calculated mass of $Mn{O_2}$.
Impurity in $Mn{O_2}$=$10g - 8.7g$
Impurity in $Mn{O_2}$=$1.3g$
We have calculated the impurity in $Mn{O_2}$ as $1.3g$.
With all the calculated values, let us now calculate the percentage of impurity.
We can calculate the percentage of impurity using the formula of mass percentage.
We know that the formula to calculate the mass percentage is,
Mass percentage$ = \dfrac{{{\text{Grams of impurity of Mn}}{{\text{O}}_2}}}{{{\text{Total mass of Mn}}{{\text{O}}_2}}} \times 100\% $
Let us substitute the values of impurity in $Mn{O_2}$ and total mass in $Mn{O_2}$,
Mass percentage$ = \dfrac{{{\text{Grams of impurity of Mn}}{{\text{O}}_2}}}{{{\text{Total mass of Mn}}{{\text{O}}_2}}} \times 100\% $
Mass percentage$ = \dfrac{{1.3g}}{{10g}} \times 100\% $
Mass percentage$ = 13\% $
The percentage of impurity is $13\% $.
Therefore, the option (D) is correct.

Note:
An alternate way to solve this problem is,
The given equation is,
$Mn{O_2} + 4HCl \to MnC{l_2} + Cl{ _2} + 2{H_2}O$
Given,
Mass of $Mn{O_2}$ is 10g.
Volume of $C{l_2}\left( g \right)$ is $2.24L$.
We have to know that the molar mass of $Mn{O_2}$ is $87g/mol$.
$87g$ of $Mn{O_2}$ gives $22.4L$ of $C{l_2}\left( g \right)$.
Therefore, $2.24L$ of $C{l_2}\left( g \right)$ gives $8.7g$ of $Mn{O_2}$.
So, 10g of $Mn{O_2}$ is taken.
The percentage of impurity is taken as, $10g - 8.7g = 1.3g$
Mass percentage$ = \dfrac{{{\text{Grams of impurity of Mn}}{{\text{O}}_2}}}{{{\text{Total mass of Mn}}{{\text{O}}_2}}} \times 100\% $
Mass percentage$ = \dfrac{{1.3g}}{{10g}} \times 100\% $
Mass percentage$ = 13\% $
The percentage of impurity is $13\% $. Therefore, the option (D) is correct.