Problem 21 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A \(20.0 \mathrm{~mL}\) solution containing \(0.2 \mathrm{~g}\) impure \(\mathrm{H}_{2} \mathrm{O}_{2}\) reacts completely with \(0.316 \mathrm{~g}\) of \(\mathrm{KMnO}_{4}\) in acid solution. The purity of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (in \(\%\) ) is (mol. wt. of \(\mathrm{H}_{2} \mathrm{O}_{2}=34 ;\) mol. wt. of \(\mathrm{KMnO}_{4}=158\) ) [Main Sep. 04, 2020 (I)]

Step-by-Step Solution

Answer

The purity of \(\mathrm{H}_2\mathrm{O}_2\) is 80%.

Step 1: Write the Balanced Chemical Equation

The balanced chemical reaction for the reaction between hydrogen peroxide (\(\mathrm{H}_2\mathrm{O}_2\)) and potassium permanganate (\(\mathrm{KMnO}_4\)) in acidic solution is:\[5\mathrm{H}_2\mathrm{O}_2 + 2\mathrm{KMnO}_4 + 3\mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{MnSO}_4 + 5\mathrm{O}_2 + 8\mathrm{H}_2\mathrm{O} + \mathrm{K}_2\mathrm{SO}_4\]

Key Concepts

Chemical Reactions

Understanding chemical reactions is fundamental in chemistry. A chemical reaction involves the transformation of reactants into products by breaking and forming bonds. In the case of hydrogen peroxide (\(\mathrm{H}_2\mathrm{O}_2\)) reacting with potassium permanganate (\(\mathrm{KMnO}_4\)) in an acidic solution, each element rearranges into new substances.

To do this, we first must write the balanced chemical equation. Balancing equations ensures that there are equal amounts of each element on both sides of the equation, respecting the law of conservation of mass. In our example, the equation is:
\[5\mathrm{H}_2\mathrm{O}_2 + 2\mathrm{KMnO}_4 + 3\mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{MnSO}_4 + 5\mathrm{O}_2 + 8\mathrm{H}_2\mathrm{O} + \mathrm{K}_2\mathrm{SO}_4\]

• There are 5 \(\mathrm{H}_2\mathrm{O}_2\) molecules and 2 \(\mathrm{KMnO}_4\) molecules reacting.
• The process involves \(\mathrm{H}_2\mathrm{SO}_4\), which aids in the reaction as an acidic medium.
• Products formed include water (\(\mathrm{H}_2\mathrm{O}\)), oxygen gas (\(\mathrm{O}_2\)), manganese sulfate (\(\mathrm{MnSO}_4\)), and potassium sulfate (\(\mathrm{K}_2\mathrm{SO}_4\)).

Each molecule or ion plays a role either by supplying or consuming atoms, making the balancing act crucial for understanding the mechanism of the reaction.

Stoichiometry

Stoichiometry bridges the gap between qualitative chemical equations and quantitative chemical analysis. It's all about calculating the relative quantities of reactants and products in a reaction.

Using the balanced equation from our earlier discussion, stoichiometry can tell us how much of each substance we need or will get. Let's break it down:

• First, determine the mole ratio from the balanced equation. In our reaction: 5 moles of \(\mathrm{H}_2\mathrm{O}_2\) react with 2 moles of \(\mathrm{KMnO}_4\).
• The molecular weights, provided as \(34\, \mathrm{g/mol}\) for \(\mathrm{H}_2\mathrm{O}_2\) and \(158\, \mathrm{g/mol}\) for \(\mathrm{KMnO}_4\), are crucial to converting mass into moles.

In practical terms, given \(0.316\, \mathrm{g}\) of \(\mathrm{KMnO}_4\), let’s convert this to moles: \[\text{moles of } \mathrm{KMnO}_4 = \frac{0.316 \text{ g}}{158 \text{ g/mol}}\approx 2.0\times10^{-3}\text{ moles}\]
This tells us the moles of \(\mathrm{H}_2\mathrm{O}_2\) that reacted. Using stoichiometry: \[\text{moles of } \mathrm{H}_2\mathrm{O}_2 = \frac{5}{2} \times 2.0\times10^{-3} = 5.0\times10^{-3}\text{ moles}\]Stoichiometry equips us with the ability to predict and understand quantities in a reaction, providing a way to double-check our results.

Solution Purity Calculation

Determining the purity of a chemical in a solution is essential in experiments, especially when dealing with real-world applications where impurities can affect outcomes. Here, our task is to find the purity percentage of \(\mathrm{H}_2\mathrm{O}_2\) in a sample.

The concept revolves around how much of the substance is pure versus impure. By analyzing the stoichiometry from our previous section, we can find how much pure \(\mathrm{H}_2\mathrm{O}_2\) was present by using the number of moles reacted.

• The mass of pure \(\mathrm{H}_2\mathrm{O}_2\) is determined as follows: \[\text{mass} = \text{moles} \times \mathrm{molar\ weight} = 5.0\times10^{-3}\text{ moles}\times 34\text{ g/mol} = 0.17\text{ g}\]
• Knowing the impure sample was \(0.2\, \mathrm{g}\), we calculate purity: \[\text{purity percentage} = \left( \frac{0.17}{0.2} \right) \times 100\% \approx 85\%\]

This essentially tells us that 85% of the material is \(\mathrm{H}_2\mathrm{O}_2\), meaning this is a high purity context for chemical use. Such calculations are vital in ensuring the reliability and accuracy of any scientific experimentation or production.