Problem 64 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

One gram of an alloy of aluminium and magnesium when treated with excess of dil. HCl forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen, collected over mercury at \(0^{\circ} \mathrm{C}\) has a volume of \(1.20\) litres at \(0.92\) atm. pressure. Calculate the composition of the alloy. \([\mathrm{H}=1, \mathrm{Mg}=24, \mathrm{Al}=27]\)

Step-by-Step Solution

Answer

Approximately 43.2% magnesium and 56.8% aluminum.

Step 1: Start with Given Information - Hydrogen Reaction

One gram of alloy when treated with excess dilute HCl produces hydrogen gas. We are given that the volume of hydrogen gas ( V ) is 1.20 liters at 0.92 atm and 0°C.

Step 2: Use Ideal Gas Law to Find Hydrogen Moles

The ideal gas law is \(PV = nRT\). At 0°C (273 K) and in atm and liters, \(R = 0.0821\). Substituting values, \(n = \frac{PV}{RT} = \frac{0.92 \times 1.20}{0.0821 \times 273}\). Calculate \(n\) to find moles of hydrogen.

Step 3: Calculate Moles of Hydrogen

Substituting the numbers into the equation \(n = \frac{0.92 \times 1.20}{0.0821 \times 273}\), we find \(n \approx 0.0489\) moles of hydrogen gas.

Step 4: Relate Moles of Hydrogen to Metal Reaction

Each mole of hydrogen gas corresponds to the reaction of one mole of metal equivalent either from magnesium (producing 1 mole H₂ per 2 moles Mg) or aluminum (producing 1 mole H₂ per 2 moles Al).

Step 5: Write Reaction Equations for Magnesium and Aluminium

\(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\) and \(2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\). Each mole of magnesium produces 1 mole of hydrogen, and each mole of aluminum produces 1.5 moles of hydrogen.

Step 6: Use Stoichiometry to Relate Masses

Let \(x\) grams be the mass of magnesium. Then \((1-x)\) grams is the mass of aluminum. Using molecular weights: \(\frac{x}{24} + \frac{(1-x)}{27} = 0.0489 \, \text{moles of hydrocarbons}\).

Step 7: Solve for Magnesium Mass \(x\)

Substituting into the stoichiometric equation, we have \(\frac{x}{24} + \frac{(1-x)}{27} = 0.0489\). Solving gives \(x = 0.432 \, \text{grams Mg}\).

Step 8: Calculate the Composition of the Alloy

The mass of magnesium is \(0.432\) grams, and therefore, the mass of aluminum is \(0.568\) grams \((1-0.432)\). Convert these to percentages: \( \text{Mg: } \frac{0.432}{1} \times 100 = 43.2\%\) and \(\text{Al: } \frac{0.568}{1} \times 100 = 56.8\%\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a crucial tool in understanding how gases behave under different conditions. The equation is written as \( PV = nRT \), where:

• \( P \) represents the pressure of the gas in atmospheres (atm),
• \( V \) is the volume in liters,
• \( n \) is the number of moles of gas,
• \( R \) is the ideal gas constant \( (0.0821 \, \text{L atm/mol K}) \), and
• \( T \) is the temperature in Kelvin.

In this context, we start by noting the conditions given: a hydrogen gas volume of 1.20 liters at 0.92 atm pressure and at a temperature of \(0^{\circ} \mathrm{C} \), equivalent to 273 K. Using the ideal gas law, we can solve for \( n \), the moles of hydrogen, by rearranging the equation to \( n = \frac{PV}{RT} \). With these values, we can calculate the amount of hydrogen gas produced by the chemical reaction between the alloy and hydrochloric acid.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It helps us predict the amount of product that can be formed from a given amount of reactants. In this exercise, it is used to determine the masses of magnesium (Mg) and aluminum (Al) in the alloy.
The reactions of these metals with hydrochloric acid \( (\text{HCl}) \) are expressed as:

• \( \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \)
• \( 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \)

Each mole of magnesium produces 1 mole of hydrogen, while 1 mole of aluminum produces 1.5 moles of hydrogen. By setting up a stoichiometric equation such as \( \frac{x}{24} + \frac{(1-x)}{27} = 0.0489 \), where \( x \) represents the mass of magnesium, and solving it, we find how each metal contributes to the production of hydrogen gas, allowing us to determine their proportions in the alloy.

Chemical Reactions with HCl

When metals react with hydrochloric acid, they typically form metal chlorides and release hydrogen gas. This principle forms the basis of our alloy composition calculation, where the metals involved are aluminum and magnesium.
For aluminum, the reaction can be represented as:

• \( 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \)

This means that for every 2 moles of aluminum, 3 moles of hydrogen gas are produced. Similarly, the reaction for magnesium is:

• \( \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \)

Here, 1 mole of magnesium yields 1 mole of hydrogen gas.
These equations enable us to understand how the evolution of hydrogen gas relates to the composition of the starting alloy. By calculating the moles of hydrogen gas produced and using stoichiometry, the exact mass of each metal in the alloy can be determined, which ultimately allows us to compute the alloy's chemical composition as percentages.