Problem 111 - 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

(a) 1: 1

Step 1: Understand the Reactions Involved

First, let's look at the chemical reactions involved. When aluminum reacts with dilute sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \)), the reaction is \( 2\mathrm{Al} + 3\mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + 3\mathrm{H}_2 \). Similarly, when aluminum reacts with sodium hydroxide (\( \mathrm{NaOH} \)), the reaction is \( 2\mathrm{Al} + 2\mathrm{NaOH} + 6\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{NaAl(OH)}_4 + 3\mathrm{H}_2 \). In both cases, the product is hydrogen gas.

Step 2: Calculate Moles of Aluminium

To calculate the number of moles of aluminum, use the molar mass of aluminum. Given the mass of aluminum is 2 grams and the molar mass of aluminum is approximately 27 g/mol, the number of moles is calculated using the formula: \( \text{moles of Al} = \frac{2}{27} \approx 0.074 \text{ moles} \).

Step 3: Determine Moles of Hydrogen Gas from Reactions

In both reactions, 2 moles of aluminum produce 3 moles of \( \mathrm{H}_2 \). This means that \( \frac{3}{2} \) moles of \( \mathrm{H}_2 \) are produced per mole of Al reacting. Thus, for the reactions of 0.074 moles of Al with \( \mathrm{H}_2\mathrm{SO}_4 \) and \( \mathrm{NaOH} \), \( \text{moles of } \mathrm{H}_2 = 0.074 \times \frac{3}{2} \approx 0.111 \text{ moles} \).

Step 4: Considering the Ratio of Reaction Stoichiometry

Since the stoichiometry of hydrogen gas production is identical for both reactions (both produce \( 3\mathrm{H}_2 \) for \( 2\mathrm{Al} \)), the amount of \( \mathrm{H}_2 \) produced in each reaction will be the same. Thus, the volumes of hydrogen gas produced in both reactions will be equal.

Step 5: Final Answer

Therefore, the ratio of the volumes of hydrogen evolved when aluminum is treated with excess \( \mathrm{H}_2 \mathrm{SO}_4 \) and excess \( \mathrm{NaOH} \) is \( 1:1 \).

Key Concepts

Chemical Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand the quantitative relationships between reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate the proportions of elements and compounds involved.

• In a chemical equation, coefficients indicate the number of moles of each substance. This allows for the determination of how much reactant is needed to produce a certain amount of product.
• For example, in the reactions involving aluminum, both interactions with sulfuric acid and sodium hydroxide produce hydrogen gas. The stoichiometry reveals that 2 moles of aluminum yield 3 moles of hydrogen gas in both instances.
• This consistency in stoichiometry means that for a given amount of aluminum, the same moles of hydrogen (\(\frac{3}{2} \text{ moles}\) per mole of aluminum) are produced, regardless of whether sulfuric acid or sodium hydroxide is used.

Aluminium Reactions

Aluminium is a reactive metal that reacts with both acids and bases to release hydrogen gas. These interactions illustrate aluminum's amphoteric properties, meaning it can react with both types of substances.

• When aluminum reacts with sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4}\)), it forms aluminum sulfate and hydrogen gas. The balanced equation is \(2\mathrm{Al} + 3\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} + 3\mathrm{H}_{2}\).
• Similarly, the reaction with sodium hydroxide (\(\mathrm{NaOH}\)) and water results in sodium aluminate and hydrogen. This is given by \( 2\mathrm{Al} + 2\mathrm{NaOH} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{NaAl(OH)}_{4} + 3\mathrm{H}_{2}\).
• Although the reaction conditions (acidic versus basic) differ, the stoichiometry remains unchanged in terms of hydrogen gas production.

Sodium Hydroxide Reactions

Sodium hydroxide, or lye, is a highly caustic base known for its ability to react with metals, like aluminum, leading to the evolution of hydrogen gas.

• The reaction of sodium hydroxide with aluminum is unique because it involves aluminum dissolving in the alkaline medium to form sodium aluminate, \(\mathrm{NaAl(OH)}_{4}\), a complex ion. This is indicative of aluminum's amphoteric properties.
• Water is also crucial in this reaction, supplying additional hydrogen atoms needed to form hydrogen gas. Without water, sodium hydroxide would not react with aluminum as effectively.
• In homes or industries, this reaction is often utilized for cleaning and etching aluminum surfaces, as it thoroughly dissolves any oxide layer, offering a clean and reactive metal surface.

Sulfuric Acid Reactions

Sulfuric acid is a strong acid frequently employed in chemical reactions for its ability to donate protons. Its reaction with metals, like aluminum, chiefly results in the release of hydrogen gas.

• Aluminum's reaction with sulfuric acid produces aluminum sulfate and hydrogen gas, as demonstrated in the equation: \( 2\mathrm{Al} + 3\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} + 3\mathrm{H}_{2}\).
• This type of acid-metal interaction is not uncommon, with hydrogen gas evolution being a telltale history of such reactions, showcasing aluminum's ability to act as a reducing agent.
• Industrially, this reaction can be harnessed to produce large amounts of hydrogen, which is valuable for applications in areas like fuel and chemical synthesis.