Question

A gaseous mixture of propane and butane of volume 3 litre on complete combustion produces 11.0 litre \(\mathrm{CO}_{2}\) under standard conditions of temperature and pressure. The ratio of volume of butane to propane is: (a) \(1: 2\) (b) \(2: 1\) (c) \(3: 2\) (d) \(3: 1\)

Short answer

The ratio of the volume of butane to propane is 2:1.

Step-by-step solution

- Write the Balanced Chemical Equations for Combustion

We shall write the balanced chemical equations for the combustion of butane (C4H10) and propane (C3H8): For Butane: C4H10 + 13/2 O2 -> 4 CO2 + 5 H2OFor Propane: C3H8 + 5 O2 -> 3 CO2 + 4 H2OThese equations show that burning 1 mole of butane produces 4 moles of CO2, and burning 1 mole of propane produces 3 moles of CO2. Under standard conditions, moles translate directly to volume.

- Define the Let Statements

Let the volume of butane be 'x' litres and the volume of propane be 'y' litres. We are given the total volume of the mixture is 3 litres. Therefore, we can write the equation x + y = 3.

- Write the Volume Ratios

Using the balanced equations, we can relate the volume of gas to the volume of CO2 produced: Volume ratio for butane to CO2 is 1:4, and the volume ratio for propane to CO2 is 1:3.

- Establish the Relation with CO2 Volume

From the combustion equations, 4 litres of CO2 are produced from 1 litre of butane and 3 litres of CO2 from 1 litre of propane. Therefore, we can write the volume of CO2 produced as 4x + 3y.

- Set up the Equation with Given CO2 Volume

We know that 11 litres of CO2 are produced. Therefore, we can write the following equation: 4x + 3y = 11.

- Solve the Equations for x and y

Now we have two equations, x + y = 3 and 4x + 3y = 11. Solving these simultaneous linear equations will give us the volumes of butane (x) and propane (y).

- Solve for x (Volume of Butane)

Multiply the first equation by 3 to facilitate the elimination method: 3(x + y) = 3(3) => 3x + 3y = 9. Now we subtract this new equation from the CO2 equation: (4x + 3y) - (3x + 3y) = 11 - 9 which simplifies to x = 2.

- Solve for y (Volume of Propane)

Substitute the value of x into the first equation: x + y = 3 => 2 + y = 3 which simplifies to y = 1.

- Calculate the Ratio of Butane to Propane

With the volumes of butane (x) and propane (y) now known, calculate their ratio: x:y = 2:1.

Key concepts

Stoichiometry

At the core of solving chemical equation problems is stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction.

Understanding stoichiometry is essential for determining the amounts of substances consumed and produced in a reaction. In the case of our combustion exercise, we utilize stoichiometry to relate the volumes of butane and propane to the volume of carbon dioxide produced.

When balancing the chemical equations for the combustion of butane and propane, we noted that 1 mole of butane produces 4 moles of CO2, and 1 mole of propane produces 3 moles of CO2. Once we establish these mole ratios, they can be applied to the volumes of gases involved because, under standard conditions, volumes of gases are directly proportional to the number of moles if the temperature and pressure are consistent (as per Avogadro's law).

Thus, we effectively set up mathematical equations that reflect these stoichiometric relationships, allowing us to solve for the unknown volumes of butane and propane using the given volume of CO2 (11 liters).

Gas Laws

The gas laws are a series of equations that describe how gases behave under varying conditions of temperature, pressure, and volume.

In this exercise, we are dealing with gases under 'standard conditions' which implies a standard temperature and pressure (STP). At STP, one mole of any gas occupies 22.4 liters. This important principle allows us to deduce that the ratio between the volumes of gases in a reaction at STP is the same as the ratio between the number of moles of gases.

Avogadro's Law

Specifically, Avogadro's law states that equal volumes of ideal gases, at the same temperature and pressure, contain the same number of molecules. Therefore, we can use the volume of CO2 produced to infer the volume of the reactant gases when given the balanced chemical equation.

This exercise provides a practical example of these gas laws in action. By using the balanced equations and the concept of molar volume at STP, we can set up ratios that help us solve for the unknown volumes of butane and propane.

Mole Concept

The mole concept is a fundamental chemical principle that provides a bridge between the atomic world and the macroscopic world we live in.

A mole is simply a specific number of particles; specifically, Avogadro's number of particles, which is approximately particles. It allows chemists to count atoms, molecules, and ions in a manner that can be correlated with mass and volume, especially in dealing with gases.

Application in Combustion

When we say 1 mole of butane yields 4 moles of CO2, we are using the mole concept to relate discrete chemical entities (molecules) to measurable quantities (liters of gas).

In our example, the total volume of the gaseous mixture (butane and propane) is 3 liters, and it produces 11 liters of CO2. The mole concept enables us to calculate the volume of individual gases (butane and propane) in the mixture by relating it to the volume of CO2 produced.

Through the systematic use of the mole concept in the step-by-step solution, we are able to discern the ratio of volumes between butane and propane in the gaseous mixture, ultimately determining this ratio to be 2:1.