Problem 130 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction, \(4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \longrightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}\), when one mole of ammonia and one mole of oxygen are made to react to completion, then (a) \(1.0 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is produced (b) all the oxygen is consumed (c) \(1.5 \mathrm{~mol}\) of \(\mathrm{NO}\) is formed (d) all the ammonia is consumed

Step-by-Step Solution

Answer

Option (b) is correct; all the oxygen is consumed.

Step 1: Understand the Reaction

The given reaction is \(4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2} \mathrm{O}\). This means 4 moles of \(\mathrm{NH}_3\) react with 5 moles of \(\mathrm{O}_2\) to produce 4 moles of \(\mathrm{NO}\) and 6 moles of \(\mathrm{H}_2\mathrm{O}\).

Step 2: Determine the Limiting Reactant

With 1 mole of \(\mathrm{NH}_3\) and 1 mole of \(\mathrm{O}_2\), the reaction needs to use the stoichiometric ratios to identify the limiting reactant. The ratio from the equation is \(\frac{4}{5}\) or \(0.8\) \(\mathrm{NH}_3\) to \(\mathrm{O}_2\). Since we have 1 mole of each, \(\mathrm{O}_2\) requires more \(\mathrm{NH}_3\) than available, making \(\mathrm{O}_2\) the limiting reactant.

Step 3: Calculate Product Formation from Limiting Reactant

Using \(\mathrm{O}_2\) as the limiting reactant, calculate products. For every 5 moles of \(\mathrm{O}_2\), 6 moles of \(\mathrm{H}_2\mathrm{O}\) are formed.Thus, \(1\, \mathrm{mol} \, \mathrm{O}_2\) produces \(\frac{6}{5}\, \mathrm{mol} \) \(\mathrm{H}_2\mathrm{O}\) or 1.2 moles. For \(\mathrm{NO}\), \(\frac{4}{5} \times 1\, \mathrm{mol} \, \mathrm{O}_2 = 0.8\, \mathrm{mol} \, \mathrm{NO}\).

Step 4: Analyze and Compare the Options

Review each option regarding the findings:(a) says \(1.0\, \mathrm{mol} \, \mathrm{H}_2\mathrm{O}\) is produced, but we calculated 1.2 moles.(b) states all \(\mathrm{O}_2\) is consumed, which is true as it's the limiting reactant.(c) indicates \(1.5\, \mathrm{mol} \, \mathrm{NO}\) is formed, but we found 0.8 moles.(d) states all \(\mathrm{NH}_3\) is consumed, but some will be left unreacted because \(\mathrm{O}_2\) is limiting.

Key Concepts

Stoichiometric Ratios

Stoichiometric ratios are essential for understanding how different substances interact in a chemical reaction. They are derived from the balanced equation of a reaction and reveal the proportional relationship between reactants and products. Given the balanced chemical equation:\[4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2} \mathrm{O}\]the stoichiometric ratios here can be expressed as follows:

• 4 moles of ammonia (\(\mathrm{NH}_3\)) are required for every 5 moles of oxygen (\(\mathrm{O}_2\)).
• This reaction produces 4 moles of nitric oxide (\(\mathrm{NO}\)) and 6 moles of water (\(\mathrm{H}_2\mathrm{O}\)) when equimolar amounts of ammonia and oxygen are reacted.

Simplifying these ratios helps identify which reactant runs out first in a reaction, which is crucial for determining the limiting reactant. Understanding stoichiometric ratios enables chemists to predict how much of each reactant is needed and how much product can form.

Chemical Reactions

Chemical reactions describe how substances interact to form new products, changing chemical compounds, their configurations, and properties in the process. The equation \(4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2} \mathrm{O}\) represents a chemical reaction where ammonia and oxygen react to form nitric oxide and water. This reaction is characterized by the rearrangement of atoms, where the nitrogen and hydrogen in ammonia react with oxygen to form new substances.When studying chemical reactions, it's important to note:

• Conservation of mass, meaning the mass of reactants equals the mass of products.
• The type of reaction, such as synthesis, decomposition, or combustion like this one.
• The energy changes, as some reactions release energy while others absorb it.

Grasping these fundamental aspects helps understand how chemical reactions proceed and how to control them to achieve desired outcomes.

Ammonia and Oxygen Reaction

The reaction between ammonia and oxygen is a classic example used to understand limiting reactants and product formation. In this scenario, the balanced reaction equation is key:\[4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2} \mathrm{O}\]In practical terms, for the reaction to go to completion, ammonia must be present in a sufficient ratio to the available oxygen. The stoichiometric ratio here is critical for determining how much ammonia would completely react with the given 1 mole of oxygen:

• Using the ratio \(\frac{4}{5}\) for \(\mathrm{NH}_3\) to \(\mathrm{O}_2\), it becomes clear that 1 mole of \(\mathrm{O}_2\) requires 0.8 mol of \(\mathrm{NH}_3\).
• Since there is 1 mole of each available reactant, oxygen is the **limiting reactant** because it's insufficient for the available ammonia amount.

This understanding is essential for predicting which reactant will be exhausted first and which will remain unreacted, affecting the reaction's efficiency and yield.

Product Formation

Product formation depends on the amounts and ratios of reactants present. Given the reaction:\[4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2} \mathrm{O}\]If we start with 1 mole of ammonia and 1 mole of oxygen, calculations must consider oxygen as the limiting reactant. Here's what happens:

• From 1 mole of \(\mathrm{O}_2\), 1.2 moles of water can be produced, as calculated from \(\frac{6}{5} \times 1 = 1.2\) for \(\mathrm{H}_2\mathrm{O}\).
• For \(\mathrm{NO}\), the production is 0.8 moles, calculated as \(\frac{4}{5} \times 1 = 0.8\).

Product amounts differ from intuitive guesses if assuming complete reaction of both reactants. Therefore, understanding limiting reactants and related stoichiometric calculations are vital for accurate predictions of product quantities.