Problem 109 - Chemistry

Question

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from \(7.45 \mathrm{~g}\) of propane.

Step-by-Step Solution

Answer

The balanced equation for the combustion of propane is \(\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2} \mathrm{O}\). The amount of carbon dioxide that could be produced from 7.45 g of propane at STP is 11.36 l.

Step 1: Write the unbalanced chemical equation

The unbalanced chemical equation for the combustion of propane is: \(\mathrm{C}_{3} \mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\)

Step 2: Balance the chemical equation

To balance the equation, ensure that there is an equal number of each type of atom on both sides of the equation. The balanced equation is: \(\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2} \mathrm{O}\)

Step 3: Convert the mass of propane to moles

Since molar mass of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)) is 44.09 g/mol, the moles of propane can be calculated as: \(7.45 \, \text{g} / 44.09 \, \text{g/mol} = 0.169 \, \text{moles}\)

Step 4: Use the balanced equation to find moles of carbon dioxide

From the balanced equation, 1 mol of propane produces 3 mol of carbon dioxide, hence \(0.169 \, \text{moles} \times 3 = 0.507 \, \text{moles of } \mathrm{CO}_{2}\)

Step 5: Convert moles of carbon dioxide to liters

At Standard Temperature and Pressure (STP), 1 mol of any gas occupies 22.4 l. Therefore, to find the volume of the carbon dioxide gas: \(0.507 \, \text{moles} \times 22.4 \, \text{l/mol} = 11.36 \, \text{l}\)

Key Concepts

Balancing Chemical Equations

Understanding how to balance chemical equations is crucial because it ensures the law of conservation of mass is respected, meaning the total mass of reactants equals the total mass of products. In a balanced equation, the number of atoms for each element and total charge must be the same on both sides.
For example, the combustion of propane requires balancing hydrogen (H), carbon (C), and oxygen (O) atoms. You'll need three CO₂ molecules to account for the carbon atoms from one C₃H₈ molecule, and four H₂O to balance out the eight hydrogen atoms. Oxygen atoms from O₂ adjust last to ensure both sides are balanced, resulting in a 5:3:4 ratio with propane and oxygen.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It is essential for predicting the amounts of substances consumed and produced in a given reaction.
Using the balanced propane combustion reaction, stoichiometry allows us to calculate that 1 mol of C₃H₈ will produce 3 mols of CO₂. This stoichiometric relationship helps determine the volume of CO₂ produced from a specific mass of propane, as seen in the exercise. Converting grams to moles, you apply the mole ratio from the balanced equation next, thus connecting mass with the molar volume of gases.

Molar Volume of a Gas

The molar volume is the volume occupied by one mole of any gas at a specified condition, typically at standard temperature and pressure (STP). The molar volume is an essential concept when calculating the volume of gas produced or consumed in a reaction.
At STP, molar volume is 22.4 liters per mole. In the case of our exercise, after finding the moles of CO₂, we multiply by the molar volume to convert moles to liters, thus obtaining the actual volume of gas produced.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) refer to the standard condition for measuring the volume of gases, which is 0 degrees Celsius and 1 atmosphere of pressure. This standardization allows chemists to compare volumes of gases from different reactions easily because all gases occupy 22.4 liters per mole at STP.
In our exercise, we use STP to simplify the conversion from moles of CO₂ to volume. Without the concept of STP, comparing the volumes of gases would be much more complex, as gas volumes vary significantly with changes in temperature and pressure.