Question

Assume that a car burns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\left(d=0.703 \mathrm{~g} / \mathrm{cm}^{3}\right)\). (a) Write the balanced equation for burning octane in air. forming \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) (b) The car has a fuel efficiency of 32 miles per gallon of octane; determine the volume of \(\mathrm{CO}_{2}\) at \(25{ }^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\) that is generated when the car goes on a 10 -mile trip.

Short answer

The volume of CO2 generated is 1424.13 L.

Step-by-step solution

Write the combustion reaction

The combustion of octane can be represented by the chemical equation: \(C_8H_{18} + O_2 \rightarrow CO_2 + H_2O\). To balance this equation, we ensure that the number of atoms of each element is the same on both sides of the equation.

Balance the combustion equation

Balance the equation by adjusting coefficients: \(2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O\). This shows that burning 2 moles of octane requires 25 moles of oxygen and produces 16 moles of carbon dioxide and 18 moles of water.

Calculate moles of octane used

First, convert the 10-mile trip into gallons used: \(\text{Fuel efficiency} = 32 \text{ miles per gallon}\). For a 10-mile trip: \[\text{Gallons used} = \frac{10}{32} = 0.3125 \text{ gallons}\]. Next, convert gallons to liters knowing 1 gallon = 3.78541 liters: \[0.3125 \text{ gallons} \times 3.78541 \frac{\text{liters}}{\text{gallon}} = 1.1836 \text{ liters}\]. Then, use the density \(0.703 \text{ g/cm}^3\) to find grams: \[1.1836 \text{ L} \times 1000 \frac{\text{cm}^3}{\text{L}} \times 0.703 \frac{\text{g}}{\text{cm}^3} = 831.95 \text{ g}\]. Finally, convert grams to moles: \(\text{Molar mass of} \ C_8H_{18} = 114.22 \text{ g/mol}\): \(831.95 \text{ g} \div 114.22 \text{ g/mol} = 7.28 \text{ moles}\).

Calculate moles of CO2 produced

From the balanced equation, 2 moles of octane produce 16 moles of CO2. For 7.28 moles of octane, we calculate moles of \(CO_2\): \[\frac{16 \text{ moles of } CO_2}{2 \text{ moles of } C_8H_{18}} = 8 \text{ moles of } CO_2 \text{ per mole of } C_8H_{18}\]. Thus, \(7.28 \times 8 = 58.24 \text{ moles of } CO_2\).

Calculate volume of CO2 produced

Using the Ideal Gas Law \(PV = nRT\), where \(P = 1.0 \text{ atm}\), \(V = ?\), \(n = 58.24 \text{ moles}\), \(R = 0.0821 \frac{\text{L atm}}{\text{mol K}}\), \(T = 298 \text{ K (25}"):\) solving for \(V:\) \[V = \frac{nRT}{P} = \frac{58.24 \times 0.0821 \times 298}{1} = 1424.13 \text{ L}\].

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that involves quantitative relationships between the reactants and products in a chemical reaction. It's like a recipe for a chemical reaction. By knowing the amounts of reactants, we can predict how much product will be formed. This is crucial for efficient resource use and environmental considerations. In the combustion reaction of octane, stoichiometry helps us determine the exact amount of oxygen needed to completely react with octane to form carbon dioxide and water. In essence, stoichiometry allows us to use the balanced equation to convert between masses, moles, and molecules, ensuring precise calculations in chemical processes.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that describes the relationship between pressure, volume, temperature, and the number of moles of a gas. It's represented by the formula: \[PV = nRT\] Where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the number of moles of gas.
• \(R\) is the ideal gas constant \(0.0821 \; \frac{L \cdot atm}{mol \cdot K}\).
• \(T\) is the temperature in Kelvin.

The Ideal Gas Law allows us to calculate one variable if the others are known. In the case of the combustion reaction of octane, we use this law to determine the volume of carbon dioxide produced when octane is burned. By knowing the moles of carbon dioxide (from stoichiometry) and the conditions of the reaction (temperature and pressure), we can find out the volume of the gas produced, which is important for understanding emissions.

Balancing Chemical Equations

Balancing chemical equations is the process of ensuring that the number of atoms for each element is equal on both sides of the equation. This is based on the conservation of mass principle, which states that mass cannot be created or destroyed in a chemical reaction. For the combustion of octane, the unbalanced equation was:\[C_8H_{18} + O_2 \rightarrow CO_2 + H_2O\]By balancing it, we get:\[2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O\]Balancing involves careful counting and adjusting coefficients to align all atoms. This step is essential for stoichiometry as it gives the correct ratios of chemicals involved, ensuring precise and meaningful calculations and predictions in any chemical reaction.