Question

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is a minor component of natural gas and is used in domestic cooking and heating. (a) Balance the following equation representing the combustion of propane in air: $$ \mathrm{C}_{3} \mathrm{H}_{8}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$ (b) How many grams of carbon dioxide can be produced by burning 3.65 mol of propane? Assume that oxygen is the excess reactant in this reaction.

Short answer

482.506 grams of carbon dioxide are produced.

Step-by-step solution

Understand the Reaction

The combustion of propane involves reacting it with oxygen to produce carbon dioxide and water. The unbalanced chemical equation is:\[ \mathrm{C}_{3} \mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O} \] Our goal is to balance this equation and use it in calculations.

Balance the Chemical Equation

To balance the equation, ensure that the number of atoms of each element is equal on both sides. - Start with carbon: Propane has 3 carbon atoms, so we need 3 \( \mathrm{CO}_{2} \): \[ \mathrm{C}_{3} \mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O} \] - Balance hydrogen next: Propane has 8 hydrogen atoms, so we need 4 \( \mathrm{H}_{2} \mathrm{O} \): \[ \mathrm{C}_{3} \mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2} \mathrm{O} \] - Balance oxygen lastly: We have \( 3 \times 2 + 4 \times 1 = 10 \) oxygen atoms on the right. Use 5 \( \mathrm{O}_{2} \) on the left: \[ \mathrm{C}_{3} \mathrm{H}_{8} + 5 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2} \mathrm{O} \] Thus, the balanced equation is: \[ \mathrm{C}_{3} \mathrm{H}_{8} + 5 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2} \mathrm{O} \]

Calculate Moles of Carbon Dioxide

According to the balanced equation, burning 1 mole of \( \mathrm{C}_{3} \mathrm{H}_{8} \) produces 3 moles of \( \mathrm{CO}_{2} \). Therefore, burning 3.65 moles of \( \mathrm{C}_{3} \mathrm{H}_{8} \) produces:\[ 3.65 \text{ moles of } \mathrm{C}_{3} \mathrm{H}_{8} \times \frac{3 \text{ moles of } \mathrm{CO}_{2}}{1 \text{ mole of } \mathrm{C}_{3} \mathrm{H}_{8}} = 10.95 \text{ moles of } \mathrm{CO}_{2} \]

Convert Moles of Carbon Dioxide to Grams

The molar mass of \( \mathrm{CO}_{2} \) is approximately 44.01 g/mol. Thus, the mass of carbon dioxide produced is:\[ 10.95 \text{ moles } \times 44.01 \text{ g/mol} = 482.506 \text{ grams} \] Therefore, burning 3.65 moles of propane produces 482.506 grams of \( \mathrm{CO}_{2} \).

Key concepts

Combustion Reaction

In chemistry, a combustion reaction is an important type of chemical reaction where a substance combines with oxygen to produce heat and light. This process is commonly referred to as burning. When propane, a hydrocarbon with the formula \(\mathrm{C}_{3}\mathrm{H}_{8}\), combusts, it reacts with oxygen \(\mathrm{O}_{2}\) from the air. The products of this reaction are carbon dioxide \(\mathrm{CO}_{2}\) and water \(\mathrm{H}_{2}\mathrm{O}\).
To represent this occurrence, the chemical equation must first reflect each participating element accurately, illustrating the transformation of reactants into products. The reaction invariably releases energy, making it exothermic. This energy release is fundamental to many everyday activities like heating and cooking.

Stoichiometry Calculations

Stoichiometry is an essential concept in chemistry, dealing with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction. In the context of our propane combustion reaction, stoichiometry involves using the balanced chemical equation to determine the exact proportions of reactants and products.
To execute stoichiometry calculations, one must begin by balancing the chemical equation. This ensures the law of conservation of mass is upheld, where the same number of each type of atom exists before and after the reaction. For the combustion of \(\mathrm{C}_{3}\mathrm{H}_{8}\), the balanced equation is:
\[ \mathrm{C}_{3}\mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O} \]

• From this, you can derive mole ratios essential for further calculations.
• For instance, 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide.
• These relationships are vital in determining unknown quantities like how much CO_2 is produced from a specific amount of propane.

Molar Mass

Molar mass is a crucial concept in chemistry that relates a substance's mass to its amount in moles. It essentially allows you to convert between the mass of a substance and the number of moles, using a unit called grams per mole (g/mol). Each element in the periodic table is measured in atomic mass units, and the molar mass is the sum of these weights for a compound.
For carbon dioxide \(\mathrm{CO}_{2}\), the molar mass calculation involves summing the atomic masses of its constituent elements: carbon (C) and oxygen (O).

• Carbon’s atomic mass is approximately 12.01 g/mol, while each oxygen atom is approximately 16.00 g/mol.
• Since there are two oxygen atoms in \(\mathrm{CO}_{2}\), you add: \(12.01\) + \(2 \times 16.00 = 44.01\) g/mol.

This value indicates the grams of \(\mathrm{CO}_{2}\) per mole, and is utilized in calculations to determine the mass of \(\mathrm{CO}_{2}\) produced during the combustion of propane, as detailed in our exercise.

Oxidation Process

The oxidation process is an integral part of combustion reactions. It involves the transfer of electrons between substances, often resulting in the release of energy. In the context of propane combustion, oxidation refers to the reaction of propane with oxygen to form new products, carbon dioxide, and water.
During this process, propane is oxidized. This means it loses electrons, and oxygen is reduced as it gains electrons. This transfer is characteristic of oxidation-reduction (redox) reactions.
In our reaction:

• The hydrocarbon \(\mathrm{C}_{3}\mathrm{H}_{8}\) is broken down, and oxygen molecules bond with carbon and hydrogen.
• Propane's carbon atoms form bonds with oxygen, producing \(\mathrm{CO}_{2}\).
• Similarly, hydrogen atoms in propane are oxidized to form water \(\mathrm{H}_{2}\mathrm{O}\).

This transformation underscores the fundamental principles of oxidation in chemical reactions, particularly how reactants are converted into new, more stable compounds through electron exchange.