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Chapter 10: Problem 99

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 \(\mathrm{H}_{2} \mathrm{O}\) ?

Short Answer

Expert verified
(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 by step solution

01

Understand the problem and list given information

We are given: - The volume of the cylinder: V = 524 cm鲁 - The temperature of the air in the cylinder: T = 74掳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
02

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 cm鲁 to Liters. 1 cm鲁 = 0.001 L V = 524 cm鲁 脳 (0.001 L/cm鲁) = 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: \(n = \frac{PV}{RT}\)
03

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: \(n \approx 0.0190 mol\)
04

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. \(n_{O2} = n_{air} \times X_{O2}\) \(n_{O2} = (0.0190 mol)(0.2095) = 0.00398 mol\)
05

Calculate the grams of C8H18 combusted

For every mole of octane (C8H18) combusted, 12.5 moles of O2 are required. Using the stoichiometric relationship: \(C_8H_{18} + 12.5O_2 鈫� 8CO_2 + 9H_2O\) 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. \(mass_{C8H18} = n_{C8H18} \times M_{C8H18}\) \(mass_{C8H18} = (0.0003184 mol)(114 \frac{g}{mol}) \approx 0.0363 g\)
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攖he 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.

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Most popular questions from this chapter

Hydrogen gas is produced when zinc reacts with sulfuric acid: $$ \mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g) $$ If \(159 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(24^{\circ} \mathrm{C}\) and a barometric pressure of 738 torr, how many grams of Zn have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

Newton had an incorrect theory of gases in which he assumed that all gas molecules repel one another and the walls of their container. Thus, the molecules of a gas are statically and uniformly distributed, trying to get as far apart as possible from one another and the vessel walls. This repulsion gives rise to pressure. Explain why Charles's law argues for the kinetic- molecular theory and against Newton's model.

A \(4.00-\mathrm{g}\) sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a 1.00-L vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of 730 torr and a temperature of \(25^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) reacts with the \(\mathrm{CaO}\) and \(\mathrm{BaO}\), forming \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3}\). When the reaction is complete, the pressure of the remaining \(\mathrm{CO}_{2}\) is 150 torr. (a) Calculate the number of moles of \(\mathrm{CO}_{2}\) that have reacted. (b) Calculate the mass percentage of \(\mathrm{CaO}\) in the mixture.

A gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{Kr}\) has a density of \(1.104 \mathrm{~g} / \mathrm{L}\) at 435 torr and \(300 \mathrm{~K}\). What is the mole percent \(\mathrm{O}_{2}\) in the mixture?

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? (a) Because Xe is a noble gas, there is less tendency for the Xe atoms to repel one another, so they pack more densely in the gas state. (b) Xe atoms have a higher mass than \(\mathrm{N}_{2}\) molecules. Because both gases at STP have the same number of molecules per unit volume, the Xe gas must be denser. (c) The Xe atoms are larger than \(\mathrm{N}_{2}\) molecules and thus take up a larger fraction of the space occupied by the gas. (d) Because the Xe atoms are much more massive than the \(\mathrm{N}_{2}\) molecules, they move more slowly and thus exert less upward force on the gas container and make the gas appear denser.
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