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Chapter 10: Problem 68

(a) What are the mole fractions of each component in a mixture of \(5.08 \mathrm{~g}\) of \(\mathrm{O}_{2}, 7.17 \mathrm{~g}\) of \(\mathrm{N}_{2}\), and \(1.32 \mathrm{~g}\) of \(\mathrm{H}_{2}\) ? (b) What is the partial pressure in \(\mathrm{atm}\) of each component of this mixture if it is held in a 12.40-L vessel at \(15^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) The mole fractions of O鈧�, N鈧�, and H鈧� in the mixture are 0.148, 0.238, and 0.614, respectively. (b) The partial pressures of O鈧�, N鈧�, and H鈧� in the mixture are 0.299 atm, 0.481 atm, and 1.241 atm, respectively.

Step by step solution

01

Calculate the number of moles of each component

To find the number of moles, we need to divide the mass of each component by its molar mass: Molar mass of O鈧� = 32 g/mol Molar mass of N鈧� = 28 g/mol Molar mass of H鈧� = 2 g/mol Number of moles of O鈧� = \( \frac{5.08\mathrm{~g}}{32\mathrm{~g/mol}} = 0.159\mathrm{~mol} \) Number of moles of N鈧� = \( \frac{7.17\mathrm{~g}}{28\mathrm{~g/mol}} = 0.256\mathrm{~mol} \) Number of moles of H鈧� = \( \frac{1.32\mathrm{~g}}{2\mathrm{~g/mol}} = 0.660\mathrm{~mol} \)
02

Calculate the mole fractions

To calculate the mole fractions, we need to divide the number of moles of each component by the total number of moles: Total number of moles = 0.159 + 0.256 + 0.660 = 1.075 mol Mole fraction of O鈧� = \( \frac{0.159\mathrm{~mol}}{1.075\mathrm{~mol}} = 0.148 \) Mole fraction of N鈧� = \( \frac{0.256\mathrm{~mol}}{1.075\mathrm{~mol}} = 0.238 \) Mole fraction of H鈧� = \( \frac{0.660\mathrm{~mol}}{1.075\mathrm{~mol}} = 0.614 \) (a) Therefore, the mole fractions of O鈧�, N鈧�, and H鈧� in the mixture are 0.148, 0.238, and 0.614, respectively.
03

Calculate the total pressure

To find the partial pressures, we must first find the total pressure of the mixture using the ideal gas law: \( PV = nRT \) Here, V = 12.40 L, n (total number of moles) = 1.075 mol, R = 0.0821 L atm / (K mol), and T = (15掳C + 273.15) = 288.15 K. We can calculate the total pressure P as follows: P = \( \frac{nRT}{V} = \frac{(1.075\mathrm{~mol}) (0.0821\mathrm{~L\cdot atm/(K\cdot mol)}) (288.15\mathrm{~K})}{12.40\mathrm{~L}} = 2.021\mathrm{~atm} \)
04

Calculate the partial pressures

To find the partial pressures of each component, we multiply the total pressure with the mole fraction of the respective component: Partial pressure of O鈧� = \( (2.021\mathrm{~atm})(0.148) = 0.299\mathrm{~atm} \) Partial pressure of N鈧� = \( (2.021\mathrm{~atm})(0.238) = 0.481\mathrm{~atm} \) Partial pressure of H鈧� = \( (2.021\mathrm{~atm})(0.614) = 1.241\mathrm{~atm} \) (b) Therefore, the partial pressures of O鈧�, N鈧�, and H鈧� in the mixture are 0.299 atm, 0.481 atm, and 1.241 atm, respectively.

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Key Concepts

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

Partial Pressure
Understanding partial pressure is essential when dealing with gas mixtures. In a mixture of gases, each gas exerts a pressure independent of the others. This pressure, known as the partial pressure, contributes to the total pressure exerted by the gas mixture. The partial pressure of a gas is proportional to its mole fraction in the mixture. For example, in the provided exercise, the partial pressure of each gas component in the mixture is calculated by multiplying the mole fraction by the total pressure of the gas mixture. It's crucial to remember that the total pressure is the sum of all partial pressures in a mixture.

When calculating partial pressures, one must consider the ideal gas law, which relates the pressure, volume, temperature, and moles of a gas. It's a foundational concept in chemistry that helps us to understand how gases behave under different conditions. This law is used in the exercise to first determine the total pressure in the vessel, which is then used to calculate the partial pressure of each component.
Ideal Gas Law
The ideal gas law is usually expressed as PV = nRT, where P is the pressure of the gas, V is the volume it occupies, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin. This law is based on the assumption that gases are composed of a large number of particles that are in constant random motion and that the interactions between these particles are negligible.

The ideal gas law can be used to solve for any one of these variables if the others are known. In our exercise, we used the ideal gas law to calculate the total pressure exerted by a gas mixture in a container. By rearranging the formula to solve for pressure (P), and plugging in the known values for n, R, and T, we determine the total pressure before finding the partial pressures.
Molar Mass
Molar mass is a property that links the mass of a substance to its number of moles. It is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Knowing the molar mass is indispensable for converting between mass and moles of a substance. For instance, the molar mass of O鈧� is 32 g/mol, which tells us that one mole of oxygen gas weighs 32 grams.

In the exercise, the molar mass of each gas component is crucial for calculating the number of moles, which are then used to find the mole fractions and partial pressures. Each gas has a unique molar mass, and using accurate molar mass values ensures that the mole calculations and subsequent steps in our problem solution are precise.

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Most popular questions from this chapter

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}\), \(15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2}\), and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is \(0.980 \mathrm{~atm}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C}\), calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2} ?\) (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section 3.2.)

(a) Calculate the density of \(\mathrm{NO}_{2}\) gas at \(0.970 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a gas if \(2.50 \mathrm{~g}\) occupies \(0.875 \mathrm{~L}\) at 685 torr and \(35^{\circ} \mathrm{C}\).

A piece of solid carbon dioxide with a mass of \(5.50 \mathrm{~g}\) is placed in a 10.0-L vessel that already contains air at 705 torr and \(24{ }^{\circ} \mathrm{C}\). After the carbon dioxide has totally vaporized, what is the partial pressure of carbon dioxide and the total pressure in the container at \(24{ }^{\circ} \mathrm{C}\) ?

A neon sign is made of glass tubing whose inside diameter is \(2.5 \mathrm{~cm}\) and whose length is \(5.5 \mathrm{~m}\). If the sign contains neon at a pressure of \(1.78\) torr at \(35^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h .\)

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(175,000 \mathrm{ft}^{3}\) of helium. If the gas is at \(23^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\), what mass of helium is in the blimp?
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