Problem 83 - A Complete Resource Book in Chemistry for JEE Main

Question

\(2 \mathrm{~g}\) of aluminium is treated separately with excess of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and excess of \(\mathrm{NaOH}\). The ratio of the volumes of hydrogen evolved is (a) \(1: 1\) (b) \(1: 2\) (c) \(2: 1\) (d) \(2: 3\)

Step-by-Step Solution

Answer

The ratio of the volumes of hydrogen evolved is (a) 1: 1.

Step 1: Identify the Reactions

Aluminum reacts with dilute sulfuric acid, \[2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2\]and with sodium hydroxide,\[2Al + 2NaOH + 6H_2O \rightarrow 2NaAl(OH)_4 + 3H_2\]In both reactions, hydrogen gas \(H_2\) is produced.

Step 2: Calculate Moles of Aluminium

The molar mass of aluminum is approximately \(27 \, \text{g/mol}\). Thus, the moles of aluminum are calculated by dividing the mass given: \(\frac{2 \, g}{27 \, g/mol} \approx 0.074 \, mol\).

Step 3: Calculate Moles of Hydrogen from H2SO4 Reaction

In the reaction with sulfuric acid, \(3 \text{ moles of } \text{H}_2\) are produced for every \(2 \text{ moles of } \text{Al}\). Using the ratio, \(0.074 \, mol \, Al \times \frac{3 \text{ moles } H_2}{2 \text{ moles } Al} = 0.111 \, mol \, H_2\).

Step 4: Calculate Moles of Hydrogen from NaOH Reaction

In the reaction with sodium hydroxide, \(3 \text{ moles of } \text{H}_2\) are produced for every \(2 \text{ moles of } \text{Al}\), the same as the sulfuric acid reaction, giving us \(0.111 \, mol \, H_2\) as well.

Step 5: Determine Volume Ratio of Hydrogen Evolved

Since both reactions give the same number of moles of \(H_2\), the volume ratio of hydrogen evolved is the same, as volume is directly proportional to the number of moles. Hence, the ratio is \(1:1\).

Key Concepts

Aluminum and Sulfuric Acid Reaction

When aluminum reacts with sulfuric acid, it can produce impressive results. In this chemical reaction, aluminum (Al) meets up with sulfuric acid (H_2SO_4) to form a new compound and release hydrogen gas (H_2). This reaction can be represented by the balanced chemical equation: \[2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2\]

• Aluminum, being a reactive metal, donates electrons and gets oxidized.
• Sulfuric acid plays the role of the acid in this reaction, providing protons (H^+) that are reduced to hydrogen gas.
• This forms aluminum sulfate (Al_2(SO_4)_3), a compound made of aluminum, sulfur, and oxygen.
• The reaction is robust, especially with dilute sulfuric acid, leading to hydrogen gas generation that is visible as bubbles.

This type of reaction not only explains basic principles of oxidation and reduction but also provides a real-life example of how metals can transform during a chemical process.

Aluminum and Sodium Hydroxide Reaction

Much like its reaction with sulfuric acid, aluminum can interact with sodium hydroxide (NaOH). This time, the reaction also produces hydrogen gas. The balanced chemical equation for this activity reads:\[2Al + 2NaOH + 6H_2O \rightarrow 2NaAl(OH)_4 + 3H_2\]

• Aluminum donates electrons to the hydroxide ions forming sodium aluminate (NaAl(OH)_4).
• Water acts as an additional reactant, facilitating the entire reaction process and aiding in hydrogen production.
• The release of hydrogen gas occurs as a result of aluminum's readiness to provide electrons, allowing the liberation of hydrogen from the hydroxide molecules.

With both reactions, hydrogen gas generation is a common thread, showcasing aluminum’s remarkable reactivity. This aluminum-sodium hydroxide interaction is a favorite demonstration in chemistry due to its rapid and noticeable production of gas.

Hydrogen Gas Evolution

In both the sulfuric acid and sodium hydroxide reactions, the focus is on the hydrogen gas that bubbles out. Both reactions produce 3 moles of hydrogen gas from 2 moles of aluminum.

• Hydrogen gas ( H_2 ) appears as bubbles and can be observed as an indication of the reaction's progression.
• This visibility makes it an exciting experiment for learners to see firsthand in a controlled environment.
• The ratio of hydrogen evolution in both reactions remains the same, confirming the predictable outcomes of chemical equations.

Thus, hydrogen gas evolution is not just the measure of an effective chemical process but also a visual treat, showing the consistency and reliability of chemical reactions as per stoichiometry.

Molar Mass Calculation

Understanding how to calculate molar mass is crucial in stoichiometry and chemical reactions. To find the molar mass of aluminum, we use its atomic information:

• Aluminum has an atomic mass of about 27 ext{ g/mol}.
• Calculating moles is done by dividing the given mass by the molar mass: \(\frac{2 ext{ g}}{27 ext{ g/mol}} \approx 0.074 ext{ mol}\).
• This calculation is pivotal as it sets the stage for determining how much product forms, in this case, hydrogen gas.

Mastering molar mass calculation enables one to understand and predict the outcomes of reactions, ensuring accurate assessments and successful completion of chemical activities.