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Question: 10g of \[\mathrm{MnO}_{2}\] on reaction with HCl forms 2.24 L of \[\mathrm{Cl}_{2}\](g) at NTP, the ...

10g of MnO2\mathrm{MnO}_{2} on reaction with HCl forms 2.24 L of Cl2\mathrm{Cl}_{2}(g) at NTP, the percentage impurity of MnO2\mathrm{MnO}_{2} is:
A) 87%
B) 25%
C) 33.3%
D) 13%

Explanation

Solution

Initially, from the balanced chemical equation we will find out the number of moles in STP. After a complete reaction with the chlorine present , the rest manganese from the given 10g will give us the impurity.

Complete step by step answer:
Stoichiometry is the quantitative relationships between the reactants and products formed. The term has been derived from the greek words- “stoicheion” meaning element and “metron” meaning measurement. Thus it is important for us to have a balanced chemical equation.
The Law of conservation of mass dictates that the quantity of each element does not change in a chemical reaction. Thus, each side of the chemical equation must represent the same quantity of any particular element. Likewise, the charge is conserved in a chemical reaction. Therefore, the same charge must be present on both sides of the balanced equation.
Now, the balanced chemical reaction involved in the question is:
MnO2\mathrm{MnO}_{2} +4HCl→MnCl2\mathrm{MnCl}_{2} + Cl2\mathrm{Cl}_{2} +$$2 \mathrm{H}_{2} \mathrm{O}

Here, as per the given equation, 1 mole of $$\mathrm{Cl}_{2}$$ is formed from 1 mole of $$\mathrm{MnO}_{2}$$ Therefore, Number of moles of $$\mathrm{Cl}_{2}$$ is n= 2.24L/22.4L= 0.1 mol Therefore, Mass of $$\mathrm{Cl}_{2}$$ is W= 87g/mol × 0.1 mol = 8.7 g Impurity in $$\mathrm{MnO}_{2}$$ = 10g−8.7g = 1.3g Now, %age impurity of $$\mathrm{MnO}_{2}$$ = (1.3g/10g) X 100 = 13% **So, the correct answer is “Option D”.** **Note:** Always calculate these stoichiometric problems after balancing the chemical equation. In case it is not balanced, we should balance first and then proceed with the calculations.