Problem 1 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The ammonia \(\left(\mathrm{NH}_{3}\right)\) released on quantitative reaction of \(0.6 \mathrm{~g}\) urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) with sodium hydroxide \((\mathrm{NaOH})\) can be neutralized by: [Main Jan. 07, 2020 (II)] (a) \(200 \mathrm{~mL}\) of \(0.4 \mathrm{~N} \mathrm{HCl}\) (b) \(200 \mathrm{~mL}\) of \(0.2 \mathrm{~N} \mathrm{HCl}\) (c) \(100 \mathrm{~mL}\) of \(0.2 \mathrm{~N} \mathrm{HCl}\) (d) \(100 \mathrm{~mL}\) of \(0.1 \mathrm{~N} \mathrm{HCl}\)

Step-by-Step Solution

Answer

(a) 200 mL of 0.4 N HCl can neutralize the ammonia.

Step 1: Urea Reaction Equation

The reaction of urea with sodium hydroxide can be represented as:\[ \mathrm{NH}_2\mathrm{CONH}_2 (aq) + 2\mathrm{NaOH} (aq) \rightarrow 2\mathrm{NH}_3 (g) + \mathrm{Na_2CO_3} (aq) \]This equation shows that 1 mole of urea produces 2 moles of ammonia.

Step 2: Calculate Moles of Urea

Urea has a molar mass of 60 g/mol. Therefore, the moles of urea in 0.6 g is calculated as:\[ \text{Moles of urea} = \frac{0.6 \, \text{g}}{60 \, \text{g/mol}} = 0.01 \, \text{moles} \]

Step 3: Calculate Moles of Ammonia

From the reaction equation, each mole of urea produces 2 moles of ammonia. Therefore,\[ \text{Moles of NH}_3 = 0.01 \times 2 = 0.02 \, \text{moles} \]

Step 4: Stoichiometry of Neutralization Reaction

The neutralization reaction of ammonia with hydrochloric acid is:\[ \mathrm{NH}_3 (g) + \mathrm{HCl} (aq) \rightarrow \mathrm{NH}_4Cl (aq) \]This equation shows a 1:1 mole ratio between ammonia and hydrochloric acid.

Step 5: HCl Required for Neutralization

We need 0.02 moles of HCl to neutralize 0.02 moles of ammonia. Using normality, 1 N HCl provides 1 mole of HCl per liter. Thus, for 0.4 N HCl,\[ \text{Volume} = \frac{0.02}{0.4} \text{ L} = 0.05 \text{ L} = 50 \text{ mL} \]

Step 6: Check the Provided Options

We need 50 mL of 0.4 N HCl. The closest option available is (a) 200 mL of 0.4 N HCl, which provides more than enough HCl and will neutralize all the ammonia released.

Key Concepts

Stoichiometry

In chemical reactions, stoichiometry is all about measuring the right amounts of reactants and products. It is based on the idea that substances react in precise, predictable ratios. These ratios are usually derived from balanced chemical equations. Knowing stoichiometry helps chemists figure out how much of each substance to use or expect in a reaction.

For example, in the urea-sodium hydroxide reaction, the balanced equation tells us that 1 mole of urea produces 2 moles of ammonia. This knowledge is crucial because it helps calculate the quantities needed or produced in a reaction.

• The balanced equation gives a clear ratio of reactants to products.
• It allows for the calculation of expected yields in a reaction.
• Stoichiometry helps in optimizing the reactant amounts to reduce waste.

Neutralization Reaction

Neutralization reactions are a type of chemical reaction where an acid and a base react to form salt and water. They are crucial in various applications, including industrial processes and biological systems. In the case of the reaction between ammonia and hydrochloric acid, the reaction is simple yet significant.

The reaction is represented by:

• \( \mathrm{NH}_3 (g) + \mathrm{HCl} (aq) \rightarrow \mathrm{NH}_4Cl (aq) \)

This equation shows that one mole of ammonia reacts with one mole of hydrochloric acid in a 1:1 ratio to produce ammonium chloride.

• Neutralization is essential for making sure that the pH is balanced in chemical processes.
• This reaction type has practical uses in antacids and chemical manufacturing.
• In this exercise, the goal was to find the suitable amount of HCl to completely neutralize the ammonia produced.

Molar Mass Calculation

Molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. It is usually expressed in grams per mole (g/mol) and is used to convert between the mass of a compound and the moles of a compound.

For the urea in our problem, its molar mass was calculated to be 60 g/mol. By knowing the molar mass, we can easily convert the given mass of urea to moles:

• The formula used is: \( \text{Moles} = \frac{\text{mass in grams}}{\text{molar mass}} \).
• Understanding molar mass is critical for stoichiometry calculations.
• It helps in determining the proportions of ingredients in chemical reactions.

In simpler terms, it is one of the basic building blocks for understanding how much of a substance is present and how it will behave in a reaction. For students, mastering molar mass calculations is essential for success in chemistry studies.