Problem 33
Question
How many moles of acidified \(\mathrm{FeSO}_{4}\) can be completely oxidized by one mole of \(\mathrm{KMnO}_{4}\) ? (a) 20 (b) 10 (c) 5 (d) \(0.5\)
Step-by-Step Solution
Verified Answer
(c) 5
1Step 1: Understanding the Reaction
In an acidified medium, the balanced redox reaction between \( ext{KMnO}_4\) and \( ext{FeSO}_4\) is written as: \[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^2+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+} \] This tells us that one mole of \( ext{KMnO}_4\) reacts with five moles of \( ext{FeSO}_4\).
2Step 2: Stoichiometry Calculation
According to the balanced equation, the stoichiometric ratio between \( ext{KMnO}_4\) and \( ext{FeSO}_4\) is 1:5. Therefore, one mole of \( ext{KMnO}_4\) can oxidize five moles of \( ext{FeSO}_4\).
Key Concepts
KMnO4FeSO4Stoichiometry
KMnO4
Potassium permanganate, represented by the chemical formula \( \text{KMnO}_4 \), is a strong oxidizing agent. An oxidizing agent is a substance that has the ability to accept electrons from another substance, thereby oxidizing the other substance. This process often involves the oxidizing agent undergoing reduction. In the context of the redox reaction occurring in an acidified solution, \( \text{KMnO}_4 \) accepts electrons from the ferrous ion \( \text{Fe}^{2+} \) in \( \text{FeSO}_4 \). The manganese in \( \text{KMnO}_4 \) starts in the +7 oxidation state and is reduced to the +2 state, forming \( \text{Mn}^{2+} \). This reduction is accompanied by the formation of water molecules and occurs with the consumption of hydrogen ions, \( \text{H}^+ \), emphasizing the acidified medium's role in facilitating the reaction.Important aspects of \( \text{KMnO}_4 \):
- Color and State: It is a purple crystalline solid.
- Solubility: Dissolves in water to give a purple/pink solution.
- Applications: Widely used in redox titrations and as a disinfectant and antiseptic.
FeSO4
Ferrous sulfate, denoted as \( \text{FeSO}_4 \), is an iron compound in which iron is in the +2 oxidation state. It functions as a reducing agent within chemical reactions. Reducing agents donate electrons to other substances, causing their own oxidation.In redox reactions, \( \text{FeSO}_4 \) is oxidized by strong oxidants like \( \text{KMnO}_4 \). During this reaction, the ferrous ion \( \text{Fe}^{2+} \) is oxidized to the ferric ion \( \text{Fe}^{3+} \), while manganese ions in \( \text{KMnO}_4 \) are reduced. This interplay of electron exchange highlights the dynamic common in redox reactions.Key points about \( \text{FeSO}_4 \):
- Color and Structure: Typically appears as pale green crystals or blue-green heptahydrate crystals.
- Uses: Employed in water treatment, as a nutritional supplement, and in reducing chromate in cement.
- Nature: Acts as a reducing agent making it pivotal in redox reactions.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. In the context of a balanced redox reaction featuring \( \text{KMnO}_4 \) and \( \text{FeSO}_4 \), stoichiometry helps determine the quantities of substances involved.The balanced reaction:\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}\]Here, it's illustrated that one mole of \( \text{MnO}_4^- \) ions reacts with five moles of \( \text{Fe}^{2+} \) ions. Therefore, in a stoichiometric ratio of 1:5, one mole of \( \text{KMnO}_4 \) suffices to oxidize five moles of \( \text{FeSO}_4 \).Stoichiometry emphasizes:
- Mole Ratio: Critical to determining the amounts of reactants and products involved.
- Balance: Ensures mass and charge conservation across the reaction.
- Applications: Used in calculating the necessary quantities of reactants for reactions or expected product yields.
Other exercises in this chapter
Problem 28
The molecular weight of \(\mathrm{O}_{2}\) and \(\mathrm{SO}_{2}\) are 32 and 64 respectively. At \(15^{\circ} \mathrm{C}\) and \(150 \mathrm{~mm} \mathrm{Hg}\)
View solution Problem 31
At STP the density of a gas \((\mathrm{mol} . \mathrm{wt}=45)\) in \(\mathrm{g} / \mathrm{L}\) is (a) \(11.2\) (b) 1000 (c) 2 (d) 224
View solution Problem 34
A compound possess \(8 \%\) sulphur by mass. The least molecular mass is (a) 200 (b) 400 (c) 155 (d) 355
View solution Problem 35
The vapour density of ozone is (a) 24 (b) 16 (c) 48 (d) 72
View solution