Problem 61 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

(i) A sample of \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) is strongly heated in air. The residue is \(\mathrm{Mn}_{3} \mathrm{O}_{4}\)(ii) The residue is dissolved in \(100 \mathrm{~mL}\) of \(0.1 \mathrm{~N} \mathrm{FeSO}_{4}\) containing dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (iii) The solution reacts completely with \(50 \mathrm{~mL}\) of \(\mathrm{KMnO}_{4}\) solution. (iv) \(25 \mathrm{~mL}\) of the \(\mathrm{KMnO}_{4}\) solution used in step (iii) requires \(30 \mathrm{~mL}\) of \(0.1\) N \(\mathrm{FeSO}_{4}\) solution for complete reaction. Find the amount of \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) present in the sample. [1980]

Step-by-Step Solution

Answer

The sample contains 0.802 g of \( \mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O} \).

Step 1: Identify the Reaction Process

First, understand the chemical reaction process: When \( \mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O} \) is heated, it decomposes to form \( \mathrm{Mn}_{3} \mathrm{O}_{4} \). The residue is then treated with \( \mathrm{FeSO}_{4} \) and reacts with \( \mathrm{KMnO}_{4} \). The task is to determine the initial amount of \( \mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O} \) based on the stoichiometry and volume data of \( \mathrm{KMnO}_{4} \) and \( \mathrm{FeSO}_{4} \).

Step 2: Calculate Normality of KMnO4 Solution

In step (iv), 25 mL of the \( \mathrm{KMnO}_{4} \) solution is titrated with 30 mL of 0.1 N \( \mathrm{FeSO}_{4} \). Using the equivalence point equation, \( \mathrm{VN} = \mathrm{VN} \), we find the normality of \( \mathrm{KMnO}_{4} \): \[25 \times N_{\mathrm{KMnO}_{4}} = 30 \times 0.1\]Solving for \( N_{\mathrm{KMnO}_{4}} \), we have:\[N_{\mathrm{KMnO}_{4}} = \frac{30 \times 0.1}{25} = 0.12 \text{ N}\]

Step 3: Use Normality to Find Quantity of KMnO4 Used

Using the determined normality from step (ii), calculate the quantity of \( \mathrm{KMnO}_{4} \) that reacted in step (iii), where 50 mL of it was used:\[\text{Equivalents of } \mathrm{KMnO}_{4} = 50 \times 0.12 \times 10^{-3} = 0.006 \text{ equivalents}\]

Step 4: Calculate the Equivalents of Mn3O4

Since \( \mathrm{FeSO}_{4} \) reduces \( \mathrm{Mn}_{3} \mathrm{O}_{4} \) back to \( \mathrm{Mn}^{2+} \), we use the fact that each mole of \( \mathrm{FeSO}_{4} \) is equivalent to one electron transfer. As the stoichiometry requires five moles of electrons per mole of \( \mathrm{Mn}_{3} \mathrm{O}_{4} \), so:\[\text{Equivalents of } \mathrm{Mn}_{3} \mathrm{O}_{4} = \text{Equivalents used in reaction} = 0.006 \]

Step 5: Determine Amount of MnSO4·4H2O

Each formula unit of \( \mathrm{Mn}_{3} \mathrm{O}_{4} \) forms from 3 formula units of \( \mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O} \). Given the equivalents of \( \mathrm{Mn}_{3} \mathrm{O}_{4} \) from step (iv), calculate the mass of the hydrate:\[ \text{Equivalents of } \mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O} = \frac{0.006}{5} \times 3 = 0.0036 \]Using the molar mass \( M = 223.1 \text{ g/mol} \):\[0.0036 \times 223.1 = 0.80196 \text{ g}\]Thus, the initial mass of \( \mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O} \) in the sample is 0.802 g.

Key Concepts

Chemical Reactions

In the exercise provided, a chemical reaction occurs when a sample of \( \mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O} \) is heated. This heating process leads to a decomposition reaction. During this decomposition, the hydrated manganese sulfate loses water molecules and transforms into manganese(II, III) oxide, \( \mathrm{Mn}_{3} \mathrm{O}_{4} \). This type of reaction changes the chemical structure and properties of the original compound.

Understanding chemical reactions involves recognizing initial reactants, observing the changes that take place, and identifying the final products. Reactions can be affected by various factors such as temperature, concentration, and presence of other substances. In this instance, the initial substance is a compound that undergoes thermal decomposition, resulting in the residue \( \mathrm{Mn}_{3} \mathrm{O}_{4} \).

Chemical reactions are fundamental in stoichiometry as they allow for the calculation of how much product is formed from given reactants, and vice-versa. This entails using balanced equations to ascertain the relationship between reactants consumed and products formed.

Titration Calculations

Titration is a popular technique used in chemistry to determine the concentration of a solution. In this problem, we encounter a back titration approach. Initially, we know that \( \mathrm{KMnO}_{4} \) reacts with \( \mathrm{FeSO}_{4} \), and its concentration was first determined by titrating with \( \mathrm{FeSO}_{4} \).

The equation \( \mathrm{VN} = \mathrm{VN} \) from step 2 comes from equating the volume and normality of two solutions at the equivalence point during titration. This equation helps calculate the unknown normality of \( \mathrm{KMnO}_{4} \) solution. Once the normality of \( \mathrm{KMnO}_{4} \) was found, it was used in the second titration setup to figure out how much \( \mathrm{KMnO}_{4} \) was needed to react with the residue.

In titration calculations, it is crucial to keep the concepts of normality and equivalence points clear. Here, normality refers to the number of equivalents of a solute per liter of solution, thus crucial for determining concentrations in titrations. By using standard solutions and knowing how to use the formula \( \mathrm{VN} = \mathrm{VN} \), students can solve for unknown concentrations effectively.

Chemical Equivalents

Understanding chemical equivalents is key in stoichiometry and is applied throughout this solution. An equivalent in chemistry is a measure indicating the reactive capacity of a chemical species. In step 4, equivalents are used to establish a relationship between \( \mathrm{Mn}_{3} \mathrm{O}_{4} \) and the titrant \( \mathrm{KMnO}_{4} \).

Equivalents allow for conversion between moles of different substances based on how they react with each other. For example, with \( \mathrm{Mn}_{3} \mathrm{O}_{4} \), you know that the substance requires several moles of electrons for its conversion when reacting in the solution. This particular relationship defines how many moles of ions, elements, or compounds participate in a given reaction.

• An equivalent refers to the amount that will react with or supply one mole of hydrogen ions in a reaction.
• In redox reactions like the one in the exercise, equivalents account for the electron transfer.

Thus, calculating equivalents allows you to balance reactions and compute how much of a reactant is consumed or a product is formed, giving insight into the stoichiometry of the reaction.