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

The ratio is (a) 1:1.

Step 1: Determine moles of aluminium

First, we need to determine the number of moles of aluminium (Al) used. Given that the molar mass of aluminium is approximately 27 g/mol, we can calculate the moles of Al using the formula: \(\text{moles of Al} = \frac{\text{mass}}{\text{molar mass}}\). For 2 g of Al, \(\text{moles of Al} = \frac{2}{27} \approx 0.0741\).

Step 2: Write the reactions with H2SO4 and NaOH

Aluminium reacts with dilute sulfuric acid \((\text{H}_2\text{SO}_4)\) and sodium hydroxide \((\text{NaOH})\) to produce hydrogen gas \((\text{H}_2)\). The relevant balanced reactions are:- With \(\text{H}_2\text{SO}_4\): \(2\text{Al} + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{H}_2\)- With \(\text{NaOH}\): \(2\text{Al} + 6\text{NaOH} + 6\text{H}_2\text{O} \rightarrow 2\text{Na}_3\text{AlO}_3 + 3\text{H}_2\)

Step 3: Calculate hydrogen volume ratios

From the reactions, both pathways yield hydrogen gas as a product. Calculating the moles of \(\text{H}_2\) gas evolved:- In both reactions, 2 moles of Al produce 3 moles of \(\text{H}_2\), resulting in \(\text{H}_2\) moles from \(0.0741\) moles of Al as follows: \(\frac{3}{2} \times 0.0741 = 0.11115\) moles of \(\text{H}_2\).- Since the reactions both yield the same proportions and are proportional to each other, the ratio of hydrogen evolved in the two cases is \(1:1\).

Step 4: Choose the correct option

Based on the calculations, the volumes of hydrogen produced in both reactions are equal. Therefore, the ratio of volumes of hydrogen evolved from each reaction pathway is \(1:1\). Hence, the correct answer is (a) \(1:1\).

Key Concepts

Understanding the Mole Concept

The mole concept is a fundamental principle in chemistry that helps us understand and quantify chemical substances. It serves as a bridge between the atomic world and the tangible world we can measure. The "mole" is a unit that represents an exact number of particles, often atoms or molecules. In this context, Avogadro's number is crucial, as it equals approximately \(6.022 \times 10^{23} \) particles per mole.
To determine the mole, you use the formula:

• \( ext{moles} = \frac{\text{mass}}{\text{molar mass}} \)

Here, the molar mass is the mass of one mole of a substance in grams. In the example mentioned, we calculated the moles of aluminium (Al) by dividing the given mass of the aluminium (2 grams) by its molar mass, 27 grams per mole, which reflected approximately \( 0.0741 \) moles of Al. This calculation helps us understand how much aluminium we have in terms of atoms involved in the reaction. Each mole relates directly to measurable quantities of material, making the mole concept essential in calculating and predicting the outcomes of chemical reactions.

Chemical Reactions Involving Aluminium

Chemical reactions are processes where substances, known as reactants, change into new substances called products. In our scenario, aluminium reacts with two different substances, namely dilute sulfuric acid (\( \text{H}_2\text{SO}_4 \)) and sodium hydroxide (\( \text{NaOH} \)). These reactions produce hydrogen gas as a byproduct.
Let's break down these reactions:

• With sulfuric acid: \(2\text{Al} + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{H}_2\)
  Three moles of \( \text{H}_2 \) gas are produced for every two moles of aluminium.
• With sodium hydroxide: \(2\text{Al} + 6\text{NaOH} + 6\text{H}_2\text{O} \rightarrow 2\text{Na}_3\text{AlO}_3 + 3\text{H}_2\)
  Again, three moles of \( \text{H}_2 \) gas are produced from two moles of aluminium.

In both cases, the aluminium is oxidized, losing electrons, while hydrogen ions gain electrons, resulting in hydrogen gas. These balanced reactions demonstrate the stoichiometric relationship, where coefficients indicate the amount of each substance participating in the reaction.

Calculating the Volume of Hydrogen Gas

When aluminium reacts and produces hydrogen, it's handy to know its volume, especially in gas form. The volume of a gas can be determined through various methods, considering temperature and pressure. However, at standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters.
This consideration allows us to convert moles of hydrogen gas to its volume:

• If \(0.11115\) moles of \(\text{H}_2\) are evolved as calculated, the volume at STP would be: \[0.11115 \text{ moles} \times 22.4 \text{ L/mol} \approx 2.49 \text{ L}\]

This calculation is crucial in understanding the physical representation of the gas produced. The gas laws help predict and verify the volume of gas yielded by a chemical reaction. If conditions aren't at STP, we'll use ideal gas law adjustments involving pressure (P), volume (V), temperature (T), and molarity (n), following the relation: \[ PV = nRT \] where R is the ideal gas constant. Through these calculations, students gain crucial insights into both the quantitative and qualitative aspects of chemical reactions.