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Ch.10 - Gases: Their Properties & Behavior

Chapter 10, Problem 90c

Natural gas is a mixture of hydrocarbons, primarily methane (CH4) and ethane (C2H6). A typical mixture might have Xmethane = 0.915 and Xethane = 0.085. Let's assume that we have a 15.50 g sample of natural gas in a volume of 15.00 L at a temperature of 20.00 °C. (c) What is the partial pressure of each component in the sample in atmospheres?

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Hi everyone. This problem reads acid rain is primarily caused by sulfur dioxide and nitrogen dioxide released into the air. The mole ratios of the gasses are usually the following. In a 11.68 g sample of the mixture with the volume of 17.83 liters and a temperature of 23 degrees Celsius. What is the partial pressure of each component? So the question that we want to answer is is the partial pressure of each component, and those components are sulfur dioxide and nitrogen dioxide. Okay, so partial pressure is equal to the mole fraction times the total pressure. So what that means is the partial pressure of sulfur dioxide is going to equal the mole fraction of sulfur dioxide where X represents mole fraction times the total pressure. Okay. And the partial pressure of nitrogen dioxide is going to equal the mole fraction of nitrogen dioxide times the total pressure. Okay, so we know what the mole fraction is for both. Because those values are given in the problem, we're given the mole fraction of sulfur dioxide and the mole fraction of nitrogen dioxide. However, we don't know what the total pressure is. So in order for us to solve for the total pressure, we need the ideal gas law which is P. V. Equals N. R. T. Where P is pressure. V. Is volume. N is moles are is gas constant and T is temperature. So because we want to solve for the total pressure, we need to isolate this variable by dividing both sides by volume. Okay, so we get pressure is equal to N. R. T over V. So let's go ahead and calculate our total pressure. Okay? So the first thing that we need to do is calculate the total number of moles. So in the problem we know that we have 11.68 g sample. Okay? So we need to go from grams of gas to moles of gas. All right. And in order for us to do this, we need the molar mass of the mixture. Okay? So first we need to calculate the average molar mass of the mixture. The way that we're going to do that is we're going to So let's go ahead. And right here number one is we need to calculate the average molar mass of the mixture. This is the only way we're going to be able to convert from grams of gas to moles of mass because we need that molar mass of the mixture. Okay, so our average molar mass of the mixture is going to equal our mole fraction of sulfur dioxide times the molar mass of sulfur dioxide plus the mole fraction of nitrogen dioxide times the molar mass of nitrogen dioxide. So let's go ahead and Do this calculation. Okay, so we know that the mole fraction of sulfur dioxide is 0.700. That was given in the problem. And the molar mass of Sulfur dioxide using the periodic table is 64.07 g per mole Plus the mole fraction of nitrogen dioxide is given in the problem, it's 0.300. And the molar mass of nitrogen dioxide is 46.01 g per mole. Okay, so the average molar mass of the mixture is equal to 58.652. Okay, So this is graham. So that is the conversion that we're going to use. So in one mole of our mixture There is .652g of gas. All right. So our grams of gas cancel and we're left with moles. And that's what we want to plug into our ideal gas law. We're solving for moles of gas. So once we do this calculation, we get 0.1991 moles of gas. All right. So let's go ahead and plug that in here. So we have 0.1. Let me do it in black. So it can match. So we have 0. moles of gas R is r gas constant, which we should have this value memorized at 0.8206 Leaders times atmosphere over mole times kelvin. And temperature Our temperature needs to be in Calvin because our gas constant are the temperatures in the unit of Kelvin. So, we need to convert 23°C to Kelvin. And the way that we do that is we have 23°C plus 273 gives us Kelvin. So this is the value we're going to use for the temperature So Times 296 Kelvin. And this is all over the volume. The volume that is given is 17 point 83 leaders. Okay, so let's go ahead and move this down so that we can write our total pressure. So pressure is equal to 0. atmospheres. Okay, So now that we know that we can go back up here to solve for the partial pressures of each component. Okay, so we said that the partial pressure of sulfur dioxide is equal to the mole fraction of sulfur dioxide times the total pressure. Okay, so we're going to go ahead and plug in the mole fraction of sulfur dioxide is 0.700 Times the total pressure we just calculated is 0. atmospheres. So the partial pressure of sulfur dioxide is 0. atmospheres. And for nitrogen dioxide it's equal to the mole fraction of nitrogen dioxide times the total pressure. So the mole fraction of nitrogen dioxide is 0. times the total pressure. 0.2712 atmospheres is equal to zero 0814 Atmospheres. Okay, so this is going to be our final answer. Four. The partial pressure of sulfur dioxide and the partial pressure of nitrogen dioxide. Okay, that is it. For this problem. I hope this was helpful
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Chlorine gas was first prepared in 1774 by the oxidation of NaCl with MnO2: 2 NaCl(s) + 2 H2SO4(l) + MnO2(s) → Na2SO4(s) + MnSO4(s) + 2 H2O(g) + Cl2(g) Assume that the gas produced is saturated with water vapor at a partial pressure of 28.7 mm Hg and that it has a volume of 0.597 L at 27 °C and 755 mm Hg pressure. (b) How many grams of NaCl were used in the experiment, assuming complete reaction?

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Textbook Question

Natural gas is a mixture of hydrocarbons, primarily methane 1CH42 and ethane 1C2H62. A typical mixture might have Xmethane = 0.915 and Xethane = 0.085. Let's assume that we have a 15.50 g sample of natural gas in a volume of 15.00 L at a temperature of 20.00 °C. (a) How many total moles of gas are in the sample?

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Textbook Question

Natural gas is a mixture of hydrocarbons, primarily methane (CH4) and ethane (C2H6). A typical mixture might have Xmethane = 0.915 and Xethane = 0.085. Let's assume that we have a 15.50 g sample of natural gas in a volume of 15.00 L at a temperature of 20.00 °C. (b) What is the pressure of the sample in atmospheres?

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Textbook Question

Gaseous compound Q contains only xenon and oxygen. When 0.100 g of Q is placed in a 50.0-mL steel vessel at 0 °C the pressure is 0.229 atm. (a) What is the molar mass of Q, and what is a likely formula?

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Gaseous compound Q contains only xenon and oxygen. When 0.100 g of Q is placed in a 50.0-mL steel vessel at 0 °C the pressure is 0.229 atm. (b) When the vessel and its contents are warmed to 100 °C, Q decomposes into its constituent elements. What is the total pressure, and what are the partial pressures of xenon and oxygen in the container?

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