Problem 63 - General Chemistry: Principles and Modern Applications

Question

How many moles of \(\mathrm{NO}(\mathrm{g})\) can be produced in the reaction of \(3.00 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g})\) and \(4.00 \mathrm{mol} \mathrm{O}_{2}(\mathrm{g}) ?\) $$ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

Step-by-Step Solution

Answer

The maximum possible amount of \(\mathrm{NO}\) that can be produced is 3.00 moles.

Step 1: Analyzing the Chemical Reaction

Look at the balanced chemical equation: \(4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \longrightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2}\mathrm{O} \). This tells us the molar relationship between reactants and products. The coefficients represent the number of moles. In this case for every 4 moles of \(\mathrm{NH}_{3}\) and 5 moles of \(\mathrm{O}_{2}\), 4 moles of \(\mathrm{NO}\) and 6 moles of \(\mathrm{H}_{2}\mathrm{O}\) are produced.

Step 2: Identify the Limiting Reagent

A reaction goes to completion when one of the reactants gets completely consumed, which is referred to as the limiting reagent. We have: 1. For 4 moles of \(\mathrm{NH}_{3}\), we need \(5/4 \times 4 = 5\) moles of \(\mathrm{O}_{2}\). But we have 4.00 moles of \(\mathrm{O}_{2}\), 2. For 5 moles of \(\mathrm{O}_{2}\), we need \(4/5 \times 5 = 4\) moles of \(\mathrm{NH}_{3}\). We have 3.00 moles of \(\mathrm{NH}_{3}\).So, \(\mathrm{NH}_{3}\) is the limiting reagent.

Step 3: Calculate the Amount of Product

The amount of product formed is proportional to the amount of the limiting reagent. Since 4 moles of \(\mathrm{NH}_{3}\) produces 4 moles of \(\mathrm{NO}\), 3.00 moles of \(\mathrm{NH}_{3}\) (the limiting reagent) will produce \(4/4 \times 3 = 3.00\) moles of \(\mathrm{NO}\). Therefore, the maximum amount of \(\mathrm{NO}\) that can be produced is 3.00 moles.

Key Concepts

Limiting Reagent

In a chemical reaction, the limiting reagent is the reactant that is consumed first, stopping the reaction from continuing further. This concept is crucial as it determines the amount of product formed.
Let's consider our initial amounts: 3.00 moles of \(\) \mathrm{NH}_{3} \(\) and 4.00 moles of \(\) \mathrm{O}_{2} \(\).To determine the limiting reagent, compare the ratio of the amounts available with the balanced equation requirements. When 4 moles of \(\) \mathrm{NH}_{3}\(\) react with 5 moles of \(\) \mathrm{O}_{2} \(\), the reaction is balanced. However, with only 3.00 moles of \(\) \mathrm{NH}_{3} \(\), calculate required \(\) \mathrm{O}_{2} \(\)as \(\) \(\frac{5}{4} imes 3=3.75\). Since we have 4.00 moles of \(\) \mathrm{O}_{2} \(\),\(\) \mathrm{NH}_{3} \(\) limits the reaction.
This means \(\) \mathrm{NH}_{3} \(\)is consumed first.Such a calculation helps efficiently using resources and preventing wastage.Why is determining the limiting reagent important?- It dictates the maximum yield of product.- Ensures you understand often complex real-world reactions.- Offers insights to optimize industrial processes.

Chemical Reaction Balancing

Balancing chemical equations is the process of equating the number of atoms for each element on both sides of the equation. This ensures the law of conservation of mass is upheld, meaning matter cannot be created or destroyed.
In our example, the balanced equation is:\[4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2}\mathrm{O}\]This tells us two important things:

• The coefficients (the numbers before molecules) represent the moles needed for a full reaction.
• For every 4 moles of \(\) \mathrm{NH}_{3}\(\), 5 moles of \(\) \mathrm{O}_{2} \(\) are required.The result is 4 moles of \(\) \mathrm{NO} \(\) and 6 moles of \(\) \mathrm{H}_{2}\mathrm{O} \(\).

Chemical balancing is crucial for predicting reaction outcomes.
It provides:- A clear picture of reactant consumption.- The exact products formed.- Insight into mole-mole relationships needed in quantitative chemistry.

Mole Concept

The mole concept is fundamental in chemistry as it relates the mass of a substance to the number of particles it contains. This helps chemists understand large numbers of atoms or molecules in practical terms, using the unit 'mole'.
The mole acts as a bridge between the macroscopic and atomic world. In our example, the calculations are based on a balanced equation with given moles of reactants:

• Every 4 moles of \mathrm{NH}_{3} lead to 4 moles of \mathrm{NO}.
• We determine the moles of \mathrm{NO} simply from the moles of our limiting reagent, \mathrm{NH}_{3}.

A mole is equivalent to \(6.022 \times 10^{23}\) particles, known as Avogadro's number. This relationship allows conversion between grams, moles, and number of atoms or molecules.The benefits of using the mole concept include:- Simplifying stoichiometric calculations.- Allowing easy transition between microscopic and macroscopic chemistry.- Providing a universal scale for understanding chemical reactions.