Question

\(\mathrm{aHNO}_{3}+\mathrm{bH}_{2} \mathrm{~S} \longrightarrow \mathrm{cNO}+\mathrm{d} \mathrm{H}_{2} \mathrm{O}+3 \mathrm{~S} .\) What are the values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} ?\) (a) \(2,3,2,2\) (b) \(2,3,2,4\) (c) \(3,3,2,4\) (d) \(4,3,1,2\)

Short answer

The balanced coefficients are (b) : \((2,3,2,4)\).

Step-by-step solution

Balance the Atoms Efficiently

We start by writing the unbalanced chemical equation: \( \text{aHNO}_{3} + \text{bH}_{2}\text{S} \rightarrow \text{cNO} + \text{dH}_{2}\text{O} + 3\text{S} \). Begin by balancing the atoms that appear in only one reactant and one product. Here, sulfur \((\text{S})\) appears in \(\text{H}_{2}\text{S}\) and on the product side as elemental sulfur. Since there are 3 sulfur atoms needed on the product side, set \( \text{b} = 3 \).

Balance the Nitrogen Atoms

Next, balance nitrogen \((\text{N})\). Since there is 1 nitrogen atom in \(\text{HNO}_{3}\) and \(\text{NO}\), we set \( \text{c} \), the coefficient of \(\text{NO}\), equal to \( \text{a} \), the coefficient of \(\text{HNO}_{3}\).

Balance the Oxygen Atoms

Balance the oxygen \((\text{O})\) atoms. The equation \(\text{aHNO}_{3}\) gives us \(3\text{a}\) oxygen atoms, split between \(\text{NO}\) and \(\text{H}_{2}\text{O}\). Each \(\text{NO}\) has 1 oxygen, taking \(\text{a}\) oxygen atoms; the remaining \(2\text{a}-\text{b}\) must form \(\text{H}_{2}\text{O}\), so we find \( \text{d} \) to equate the remaining oxygens.

Balance the Hydrogen Atoms

Balance hydrogen (\(\text{H}\)) atoms. There are \(\text{a}\) hydrogens from \(\text{HNO}_{3}\) and \(2\text{b}\) hydrogens from \(\text{H}_{2}\text{S}\). These need to equal \(2\text{d}\) hydrogen atoms in \(\text{H}_{2}\text{O}\). Find \( \text{d} \) such that \(\text{a} + 2\text{b} = 2\text{d}\).

Solve for Coefficients and Double-check

With \( \text{b} = 3 \) and knowing the relationships from Steps 2 to 4, test the proposed coefficient sets for (a), (b), (c), (d). The set that satisfies all atom balances is the answer. By checking these values, choice (b) \((2,3,2,4)\) balances the equation correctly.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry, focusing on the quantitative relationships between reactants and products in a chemical reaction. It is the backbone of understanding chemical equations and predicting how much of each substance is needed or produced in a reaction.
When dealing with stoichiometry, you must first balance the chemical equation. This ensures the Law of Conservation of Mass, which states that matter cannot be created or destroyed, is followed. Each element must have the same number of atoms on both sides of the equation.
In the exercise, the balanced equation is \(2\text{HNO}_3 + 3\text{H}_2\text{S} \rightarrow 2\text{NO} + 4\text{H}_2\text{O} + 3\text{S}\). Here, stoichiometry helps us use coefficients to interpret the proportion of reactants consumed and products formed. These coefficients indicate a ratio:

• 2 moles of \(\text{HNO}_3\) react with 3 moles of \(\text{H}_2\text{S}\).
• This produces 2 moles of \(\text{NO}\), 4 moles of \(\text{H}_2\text{O}\), and 3 moles of \(\text{S}\).

Understanding stoichiometry allows chemists to calculate actual amounts needed in reactions and evaluate reactions' efficiencies.

Chemical Reactions

Chemical reactions involve a transformation between the reactants and products, providing a path through which materials are transformed. Understanding the nature of these reactions is crucial for balancing equations and predicting reaction outcomes.
Each type of chemical reaction - like synthesis, decomposition, single replacement, or double replacement - follows its specific pattern. Here, the reaction is a form of a redox reaction, meaning that it involves the transfer of electrons. Specifically, it is a reduction and oxidation occurring between \(\text{HNO}_3\) and \(\text{H}_2\text{S}\), guiding their transformation into new compounds.
This reaction showcases key aspects of chemical reactions:

• Reactants: \(\text{HNO}_3\) and \(\text{H}_2\text{S}\), which undergo a process to form products.
• Products: \(\text{NO}\), \(\text{H}_2\text{O}\), and \(\text{S}\), resulting from breaking bonds and forming new ones.

Mastery of reaction types and mechanisms helps predict what will happen during a reaction and enables accurate representation and balancing of chemical equations.

JEE Chemistry Questions

JEE (Joint Entrance Examination) chemistry questions often test a student's understanding of core chemical principles, including stoichiometry, balancing reactions, and understanding chemical behavior. For students preparing for exams like the JEE, mastering these concepts is crucial.
The given exercise simulates the kind of question one might face in JEE chemistry, where you must balance complex reactions and understand molecular interactions. Success in such questions requires:

• Precision in balancing equations – ensuring all atoms are counted correctly and total charge is balanced.
• Strong grasp of underlying chemical reactions – knowing how different molecules interact and transform.
• Practice with a variety of chemical reaction problems – since JEE exams test breadth and depth of chemistry knowledge.

Approaching these questions with a solid strategy and clear understanding of concepts can drastically increase a student’s proficiency and confidence in tackling chemistry exams.