Problem 133 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(1)\) when \(1 \mathrm{~mol}\) of ammonia and \(1 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) are made to react to completion then (a) \(1.0 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is produced (b) \(1.0 \mathrm{~mol}\) of NO will be produced (c) all the ammonia will be consumed (d) all the oxygen will be consumed

Step-by-Step Solution

Answer

The correct option is (d) all the oxygen will be consumed.

Step 1: Write down the balanced equation

The given chemical reaction is already balanced: \[4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(l)\] This means 4 moles of ammonia react with 5 moles of oxygen to produce 4 moles of nitric oxide and 6 moles of water.

Step 2: Determine limiting reactant

To determine the limiting reactant, compare the mole ratio of the reactants. From the balanced equation, 4 moles of NH\(_3\) require 5 moles of O\(_2\). Thus, 1 mole of NH\(_3\) would need \(1.25\) moles of O\(_2\) to react completely. Since only 1 mole of O\(_2\) is available, O\(_2\) is the limiting reactant.

Step 3: Calculate moles of products formed from limiting reactant

Because O\(_2\) is the limiting reactant, we calculate the products based on its 1 mole. From this, using stoichiometry: - 5 moles O\(_2\) produce 6 moles H\(_2\)O, so 1 mole O\(_2\) produces \(\frac{6}{5}\) moles of H\(_2\)O.- 5 moles O\(_2\) produce 4 moles NO, so 1 mole O\(_2\) produces \(\frac{4}{5}\) moles of NO.

Step 4: Evaluate each option based on calculations

Let's evaluate the options:- (a) 1.0 mol of H\(_2\)O is produced: We calculated \(\frac{6}{5} = 1.2\), not zero or 1.0 mol, so \(1.0\) mol of H\(_2\)O is incorrect.- (b) 1.0 mol of NO will be produced: We calculated \(\frac{4}{5} = 0.8\), not 1.0 mol, so \(1.0\) mol of NO is incorrect.- (c) All ammonia will be consumed: Only 0.8 mol of NO can be produced and 0.8 mol of NH\(_3\) will react (from the mole ratio of NH\(_3\) to NO being \(1:1\)). Not all of NH\(_3\) is consumed as 1 mole was available, so incorrect.- (d) All oxygen will be consumed: Since O\(_2\) is the limiting reactant, all 1 mole of it will be consumed, making (d) correct.

Key Concepts

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between the substances involved in a reaction.
It simply helps to measure the exact amounts of reactants and products in a chemical reaction. To understand this better, think of a recipe. Much like how a recipe details the correct ratio of ingredients to make a dish, stoichiometry provides the proportions of various reactants and products in a chemical equation.
Understanding stoichiometry is crucial because it allows chemists to predict how much of each chemical is needed and how much product will be formed. For example, in the equation given:

• 4 moles of NH extsubscript{3} (ammonia) react with 5 moles of O extsubscript{2} (oxygen).
• This produces 4 moles of NO (nitric oxide) and 6 moles of H extsubscript{2}O (water).

Knowing stoichiometry helps determine not only these ratios but also aligns with finding the limiting reactant, which is the reactant that will be completely used up first, limiting the formation of products.

Balanced Chemical Equation

A balanced chemical equation represents a chemical reaction with the same number of atoms for each element on both sides of the equation.
Balanced equations are crucial because they adhere to the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction. In our exercise's equation:\[4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(l)\]

• The number of nitrogen (N) atoms on both sides is 4.
• The number of hydrogen (H) atoms is 12.
• And the number of oxygen (O) atoms is 10.

By ensuring the equation is balanced, it becomes easier to accurately employ stoichiometry. This ensures correct predictions of how much reactant is needed and how much product is formed, as in the limiting reactant scenario we calculated.

Mole Ratio

The mole ratio is derived from the coefficients of the reactants and products in a balanced chemical equation.
It is used to convert between moles of one substance and moles of another. This concept is imperative because it connects the reactants to the possible products.
Essentially, it dictates the proportions in which substances react and form.In the current balanced reaction:\[4 \text{NH}_3 : 5 \text{O}_2 : 4 \text{NO} : 6 \text{H}_2\text{O}\]It tells us:

• 4 moles of NH extsubscript{3} are needed to react with 5 moles of O extsubscript{2},
• To produce 4 moles of NO and 6 moles of water.

To identify the limiting reactant, we use these mole ratios. For one mole of NH extsubscript{3}, it needs 1.25 moles of O extsubscript{2} but only has 1, making O extsubscript{2} the limiting factor.
Thus, this mole ratio determines the output: 0.8 moles of NO and 1.2 moles of H extsubscript{2}O.