Problem 159 - A Complete Resource Book in Chemistry for JEE Main

Question

\(14.96 \mathrm{~g}\) of a mixture containing \(\mathrm{n}\)-hexane and ethyl alcohol are reacted with sodium metal to give \(200 \mathrm{ml}\) of \(\mathrm{H}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(760 \mathrm{~mm}\) pressure. The percentage of ethyl alcohol in the mixture is \([\mathrm{R}=0.082 \mathrm{~L}\) atm \(\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right]\)

Step-by-Step Solution

Answer

The percentage of ethyl alcohol in the mixture is approximately 4.92%.

Step 1: Understanding the Chemical Reaction

Sodium metal reacts with ethyl alcohol as follows: \( 2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2 \). This means that 2 moles of ethyl alcohol produce 1 mole of hydrogen gas.

Step 2: Use Ideal Gas Law to Find Moles of \(\text{H}_2\)

Apply the ideal gas law equation: \( PV = nRT \). Given \( P = 1 \text{ atm} \), \( V = 0.2 \text{ L} \), \( T = 27 + 273 = 300 \text{ K} \). Solve for \( n \) (moles of \( \text{H}_2 \)): \[ n = \frac{PV}{RT} = \frac{1 \times 0.2}{0.082 \times 300} \]. After calculating,\[ n \approx 0.008 \text{ moles of } \text{H}_2 \].

Step 3: Relate Moles of \(\text{H}_2\) to Ethyl Alcohol

Since 2 moles of ethyl alcohol produce 1 mole of hydrogen, \[ 0.008 \text{ moles of } \text{H}_2 \text{ correspond to } 2 \times 0.008 = 0.016 \text{ moles of ethyl alcohol.} \]

Step 4: Calculate Mass of Ethyl Alcohol

The molar mass of ethyl alcohol, \( \text{C}_2\text{H}_5\text{OH} \), is \( 46 \text{ g/mol} \). Thus, the mass of ethyl alcohol is: \[ 0.016 \times 46 = 0.736 \text{ g}. \]

Step 5: Find Percentage of Ethyl Alcohol in Mixture

Calculate the percentage of ethyl alcohol in the mixture:\[ \text{Percentage} = \left( \frac{\text{mass of ethyl alcohol}}{\text{total mass of mixture}} \right) \times 100 = \left( \frac{0.736}{14.96} \right) \times 100. \]After calculating,\[ \text{Percentage of Ethyl Alcohol} \approx 4.92\% \].

Key Concepts

Chemical Reaction with Sodium

When ethyl alcohol (ethanol) is introduced to sodium metal, an interesting chemical reaction occurs. These elements react to produce hydrogen gas (\(\text{H}_2\)) and sodium ethoxide. The equation for this chemical reaction is:

• \(2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2\)

This indicates that for every 2 moles of ethyl alcohol, 1 mole of hydrogen gas is produced.

The reaction highlights how the ethyl alcohol's hydroxyl group reacts with sodium to liberate hydrogen gas. This process also forms sodium ethoxide, contributing to the production of \(\text{H}_2\) and changes the nature of the substances involved in the mixture. Understanding this reaction is crucial when looking to determine chemical compositions and stoichiometry in a mixture.

Mole Calculation

Moles are a convenient unit in chemistry used to express chemical amounts. Calculating moles involves applying a balance between volume, temperature, pressure, and gas constant, known as the ideal gas law, represented by:

• \(PV = nRT\)

In this expression, \(P\) denotes pressure, \(V\) is volume, \(n\) stands for moles, \(R\) is the gas constant, and \(T\) signifies temperature in Kelvin.

Using the ideal gas law, you can determine the number of moles of hydrogen gas (\(\text{H}_2\)) produced from the reaction of ethyl alcohol with sodium. In this particular calculation:

• Volume \(V = 0.2 \text{ L}\)
• Pressure \(P = 1 \text{ atm}\)
• Temperature \(T = 300 \text{ K}\)
• Gas constant \(R = 0.082 \text{ L atm mol}^{-1} \text{ K}^{-1}\)

Substituting these values into the equation calculates the number of moles of hydrogen gas. Through these calculations, the amount can be used to relate how much ethyl alcohol is reacting based on stoichiometric ratios given by the balanced equation for the reaction.

Percentage Composition

The percentage composition of a mixture informs us what fraction of the whole is made up of a specific substance. In this exercise, we are focusing on the percentage of ethyl alcohol in the mixture. To compute this, you need to know both the mass of ethyl alcohol and the total mass of the mixture.

First, calculate the mass of ethyl alcohol by multiplying the moles of ethyl alcohol (from the reaction and mole calculation) by its molar mass:

• Molar Mass of Ethyl Alcohol \( = 46 \text{ g/mol}\)

Then, using the formula:

• \(\text{Percentage} = \left( \frac{\text{Mass of Ethyl Alcohol}}{\text{Total Mass of Mixture}} \right) \times 100\)

By substituting the values, you determine what portion of the mixture is ethyl alcohol. This calculation reveals the purity of the sample in terms of the weight percentage of ethyl alcohol and helps in understanding the overall composition of the mixture.