Problem 99 - Chemistry: The Central Science

Question

Assume that a single cylinder of an automobile engine has a volume of \(524 \mathrm{~cm}^{3}\). (a) If the cylinder is full of air at \(74^{\circ} \mathrm{C}\) and \(0.980 \mathrm{~atm}\), how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) (b) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ?

Step-by-Step Solution

Answer

(a) The number of moles of O2 present in the cylinder is approximately 0.00398 mol. (b) The mass of octane (C8H18) that can be combusted by this quantity of O2 is approximately 0.0363 g.

Step 1: Understand the problem and list given information

We are given: - The volume of the cylinder: V = 524 cm³ - The temperature of the air in the cylinder: T = 74°C = 347 K (converting to Kelvin by adding 273) - The pressure in the cylinder: P = 0.980 atm - The mole fraction of O2 in the air: X_O2 = 0.2095

Step 2: Use the Ideal Gas Law to find the number of moles of air

Using the Ideal Gas Law, PV = nRT, we can calculate the number of moles of air in the cylinder. First, convert the volume from cm³ to Liters. 1 cm³ = 0.001 L V = 524 cm³ × (0.001 L/cm³) = 0.524 L Now use the Ideal Gas Law: \(P = 0.980 atm\) \(V = 0.524 L\) \(R = 0.0821 \frac{L atm}{K mol}\) \(T = 347 K\) Rearranging the Ideal Gas Law for n: \(n = \frac{PV}{RT}\)

Step 3: Calculate the number of moles of air

Now, plug in the given values to find the number of moles of air: \(n = \frac{(0.980 atm)(0.524 L)}{(0.0821 \frac{L atm}{K mol})(347 K)}\) Calculate n: \(n \approx 0.0190 mol\)

Step 4: Find the number of moles of O2 present in the air

To find the number of moles of O2 present in the air, multiply the total number of moles of air by the mole fraction of O2. \(n_{O2} = n_{air} \times X_{O2}\) \(n_{O2} = (0.0190 mol)(0.2095) = 0.00398 mol\)

Step 5: Calculate the grams of C8H18 combusted

For every mole of octane (C8H18) combusted, 12.5 moles of O2 are required. Using the stoichiometric relationship: \(C_8H_{18} + 12.5O_2 → 8CO_2 + 9H_2O\) We can calculate how many moles of octane can be combusted with the given moles of O2: \(n_{C_8H_{18}} = \frac{n_{O2}}{12.5}\) \(n_{C_8H_{18}} = \frac{0.00398 mol}{12.5} = 0.0003184 mol\) Now, convert this to grams of octane. The molar mass of C8H18 is about 114 g/mol. \(mass_{C8H18} = n_{C8H18} \times M_{C8H18}\) \(mass_{C8H18} = (0.0003184 mol)(114 \frac{g}{mol}) \approx 0.0363 g\)

Step 6: Final Answers

(a) The number of moles of O2 present in the cylinder is approximately 0.00398 mol. (b) The mass of octane (C8H18) that can be combusted by this quantity of O2 is approximately 0.0363 g.

Key Concepts

Understanding the Ideal Gas Law

The Ideal Gas Law, represented by the equation PV = nRT, is a fundamental principle in chemistry that describes the behavior of an ideal gas. In this equation, 'P' stands for pressure in atmospheres (atm), 'V' is the volume in liters (L), 'n' represents the number of moles of the gas, 'R' is the ideal gas constant, and 'T' is the temperature in Kelvin (K).

When it comes to solving problems like the one in our exercise, the Ideal Gas Law allows us to calculate the number of moles of a gas when the pressure, volume, and temperature are known. It is critical to convert all the units to the standard ones that fit the gas constant (R), which in this case is liters and atmospheres for volume and pressure, respectively, and Kelvin for temperature. Remember, Kelvin can be obtained by adding 273.15 to the Celsius temperature.

Getting a Grip on Mole Fraction

The mole fraction, denoted by 'X', is the ratio of the number of moles of a component to the total number of moles of all components in a mixture. It is a unitless quantity reflecting the proportion of a substance within a mixture.

In our exercise, we used the mole fraction of oxygen (O2) in air, which was given as 0.2095. This piece of information was crucial to determine how much O2 was present in the cylinder. With the total moles of air calculated using the Ideal Gas Law, we multiplied this figure by the mole fraction to find the moles of O2. The beauty of mole fraction is that it simplifies calculations in mixtures, making it a handy tool for chemists.

Combustion Reactions Explained

Combustion reactions are a type of chemical reaction where a substance, typically a hydrocarbon, reacts with oxygen to produce carbon dioxide, water, and energy in the form of heat or light. In our exercise, we examined the combustion of octane (C8H18), which is a component of gasoline.

The balanced combustion equation is crucial since it gives us the stoichiometry—the quantitative relationship between reactants and products in a chemical reaction. For octane, the ratio of O2 to C8H18 in the reaction is 12.5:1, meaning for every mole of octane burned, 12.5 moles of oxygen are required. With the moles of O2 calculated, we can determine how much octane can be combusted. Understanding these ratios is essential for correctly solving real-world problems involving combustion reactions.