Problem 67 - A Complete Resource Book in Chemistry for JEE Main

Question

In the redox reaction \(\mathrm{xMnO}+\mathrm{yPbO}_{2}+\mathrm{zHNO}_{3} \longrightarrow \mathrm{HMnO}_{4}\) \(+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) \(x=2, y=5, z=10\) (b) \(x=10, y=2, z=5\) (c) \(x=3, y=7, z=6\) (d) \(x=5, y=2, z=10\)

Step-by-Step Solution

Answer

Choice (d): \(x = 5, y = 2, z = 10\) balances the equation.

Step 1: Identify Reactants and Products

In this exercise, the given redox reaction is: \(\text{xMnO} + \text{yPbO}_{2} + \text{zHNO}_{3} \longrightarrow \mathrm{HMnO}_{4} + \mathrm{Pb(NO}_{3})_{2} + \mathrm{H}_{2} \mathrm{O}\). We need to balance this equation to find the correct stoichiometric coefficients \(x\), \(y\), and \(z\).

Step 2: Balance Manganese (Mn)

To balance Mn, note that each side must have the same number of Mn atoms. In our reaction, Mn is present in \(\text{xMnO}\) on the left and \(\mathrm{HMnO}_{4}\) on the right. Given the choices, the number of Mn atoms on both sides should be equal.

Step 3: Balance Lead (Pb)

Next, ensure that Pb is balanced. \(\mathrm{yPbO}_{2}\) on the left becomes \(\mathrm{Pb(NO}_{3})_{2}\) on the right. Again, choose \(y\) such that the number of Pb atoms is the same on both sides.

Step 4: Balance Oxygen (O)

Oxygen atoms come from multiple compounds (\(\text{xMnO}\), \(\mathrm{yPbO}_{2}\), \(\mathrm{zHNO}_{3}\)) and form new compounds on the product side. Balance the equation by ensuring the total number of oxygen atoms is equal both reactants and products.

Step 5: Balance Hydrogen (H) and Nitrogen (N)

Finally, balance hydrogen and nitrogen. Hydrogen is found in \(\mathrm{zHNO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) on the right. Nitrogen is present in \(\mathrm{zHNO}_{3}\) and affects the \(\mathrm{Pb(NO}_{3})_{2}\) formation. Confirm all elements are balanced.

Step 6: Verify All Coefficients

Check each choice given: (a) \(x = 2, y = 5, z = 10\)(b) \(x = 10, y = 2, z = 5\)(c) \(x = 3, y = 7, z = 6\)(d) \(x = 5, y = 2, z = 10\) Use each to balance the complete reaction and identify the match.

Step 7: Conclusion

Choice (d) \(x = 5, y = 2, z = 10\) successfully balances the equation as it maintains equality across all elements.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry, representing the conservation of mass where the number of atoms of each element is conserved before and after a reaction. Here’s how you approach it:

• List the chemical formulas of reactants and products and write down the number of each type of atom on both sides of the equation.
• Begin by balancing atoms of elements that appear in only one reactant and one product, often starting with metals.
• Move to non-metals, and finally balance hydrogen and oxygen since they often appear in multiple compounds.

Once all atoms are balanced, the coefficients (x, y, z in our exercise) ensure that the mass is conserved. Essentially, these coefficients tell you the ratio of molecules that react with each other.
Balancing these requires a systematic approach and sometimes a bit of trial and error until everything is perfectly aligned.

Stoichiometry

Stoichiometry is the quantitative heart of chemistry. It involves calculating the amounts of reactants and products involved in a chemical reaction. Key points include:

• Understanding the mole concept, which allows chemists to count atoms, ions, or molecules in a given substance based on its chemical quantity.
• Using a balanced chemical equation, stoichiometry provides the relationship between the amounts of reactants needed and products formed.

For example, in a balanced redox reaction like ours, the coefficients determine the exact ratio of reactants and products, ensuring neither excess nor deficit in the reaction. This is essential in applications ranging from industrial processes to laboratory experiments.

Oxidation-Reduction

Redox reactions, short for reduction-oxidation reactions, are reactions where electrons are transferred between reactants, changing their oxidation states.

• Oxidation involves loss of electrons, whereas reduction involves gain of electrons.
• In the context of our reaction, identifying which reactants are oxidized and which are reduced is crucial.
• This involves assigning oxidation numbers to different elements in the reaction and noticing changes as reactants become products.

Determining the oxidizing and reducing agents helps in understanding how electrons shift and reactants transform. Mastery of redox reactions requires recognizing these electron transfers and balancing charge along with atoms in the chemical equation.