Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

\(10 \mathrm{~mL}\) of propane are mixed with \(70 \mathrm{~mL}\) of oxygen and exploded. What would be the volume of residual gases after explosion and after exposure of the residual gases to alkali? All volume measurements are made at the same temperature and pressure. (a) \(55 \mathrm{~mL}, 25 \mathrm{~mL}\) (b) \(40 \mathrm{~mL}, 15 \mathrm{~mL}\) (c) \(50 \mathrm{~mL}, 20 \mathrm{~mL}\) (d) \(35 \mathrm{~mL}, 40 \mathrm{~mL}\)

Step-by-Step Solution

Answer

The residual gases after explosion is 50 mL, and post-alkali exposure is 20 mL. Option (c): \(50 \mathrm{~mL}, 20 \mathrm{~mL}\).

Step 1: Write the Balanced Chemical Equation

The combustion of propane, \( \text{C}_3\text{H}_8 \), with oxygen \( \text{O}_2 \) can be expressed as: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]. This represents the complete combustion of propane at constant temperature and pressure.

Step 2: Determine the Initial Reactant Volumes

We have \( 10 \text{ mL} \) of propane \( \text{C}_3\text{H}_8 \) and \( 70 \text{ mL} \) of oxygen \( \text{O}_2 \). According to the balanced equation, 1 mole of propane needs 5 moles of oxygen. So, \( 10 \text{ mL} \) of propane needs \( 50 \text{ mL} \) of oxygen.

Step 3: Identify Limiting Reactant and React Excess

Since there are \( 70 \text{ mL} \) of \( \text{O}_2 \) and only \( 50 \text{ mL} \) is required, oxygen is in excess. Thus, after the reaction, \( 70 - 50 = 20 \text{ mL} \) of oxygen remains unreacted.

Step 4: Calculate Volume of Initial Residual Gases

Complete reaction of propane forms \( 3 \text{CO}_2 \) molecules, equating to \( 3 \times 10 = 30 \text{ mL} \) of \( \text{CO}_2 \). Hence, residual gases are \( 30 \text{ mL} \text{ CO}_2 + 20 \text{ mL} \text{ unreacted } \text{O}_2 = 50 \text{ mL} \).

Step 5: Calculate Volume After Exposure to Alkali

Alkali absorbs \( \text{CO}_2 \). When \( 30 \text{ mL} \) of \( \text{CO}_2 \) is absorbed, only the unreacted oxygen remains: \( 20 \text{ mL} \text{ unreacted } \text{O}_2 \).

Key Concepts

Limiting Reactant

In chemistry, the concept of the limiting reactant is crucial to understanding chemical reactions. The limiting reactant is the substance that is completely consumed first in a chemical reaction and thus determines the maximum amount of product that can be formed. In our exercise involving the combustion of propane and oxygen, we have different volumes: 10 mL of propane and 70 mL of oxygen.
When we look at the balanced chemical equation: \[\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\]the stoichiometry tells us that 1 mole of propane reacts with 5 moles of oxygen. For the given volumes, 10 mL of propane would ideally require 50 mL of oxygen to react completely.
Let's apply the concept:

• Since we have 70 mL of oxygen available, but only 50 mL is needed, propane is the limiting reactant.
• Once the propane is used up, the reaction stops even though there is still some oxygen left, indicating that the oxygen is in excess.
• The leftover or excess reactant can be calculated by subtracting the required amount from the initial amount, resulting in 20 mL of excess oxygen.

Understanding the limiting reactant helps you determine not just how much product will form, but also what will remain after the reaction ends.

Balanced Chemical Equation

A balanced chemical equation is fundamental for understanding the proportions of reactants and products involved in a chemical reaction. To achieve balance, the number of atoms of each element should be equal on both sides of the equation. This reflects the law of conservation of mass, which states that matter can neither be created nor destroyed in a chemical reaction.
Considering our specific exercise, the balanced equation for the combustion of propane is crucial: \[\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\]This equation tells us several important things:

• 1 molecule of propane reacts with 5 molecules of oxygen.
• The reaction produces 3 molecules of carbon dioxide and 4 molecules of water for each molecule of propane consumed.

Thus, balancing the equation helps you predict the amount of each product produced and the amount of reactants consumed. It serves as a roadmap to calculate the volumes involved in the reaction, ensuring that all calculations regarding limiting reactants and excess are accurate. Remember, the balance of the equation is the backbone of chemical calculations.

Gas Laws

Gas laws are a set of theoretical principles that describe how gases behave under different conditions. They are particularly useful when dealing with reactions involving gases, like the combustion of propane in our problem. The exercise assumes all volume measurements are made at the same temperature and pressure, aligning with the gas laws that describe relationships between volume, temperature, and pressure.
Here are some important gas laws to consider:

• Boyle's Law: States that the pressure of a gas is inversely proportional to its volume when temperature is constant. Hence, if the volume doubles, the pressure halves, provided temperature does not change.
• Charles's Law: States that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is constant.
• Avogadro's Law: This law is important for reactions involving gases. It states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. This is why volume measurements can be related directly to moles of gases in reactions.

For our specific exercise, the volumes of gases directly relate to the moles of gases through Avogadro's Law, simplifying our calculations. By using these gas laws, you can predict how gases will respond to changes in conditions, which is crucial when analyzing the results of combustion reactions.