/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 98 A tank contains a mixture of 52.... [FREE SOLUTION] | 91Ó°ÊÓ

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A tank contains a mixture of 52.5 g oxygen gas and 65.1 \(\mathrm{g}\) carbon dioxide gas at \(27^{\circ} \mathrm{C}\) . The total pressure in the tank is 9.21 atm. Calculate the partial pressures of each gas in the container.

Short Answer

Expert verified
The partial pressures of oxygen and carbon dioxide in the container are \(4.84 \: \text{atm}\) and \(4.37 \: \text{atm}\), respectively.

Step by step solution

01

Convert mass to moles

First, we need to convert the given mass of each gas to moles using the molar mass of each gas. The molar mass of oxygen (Oâ‚‚) is 32 g/mol, and the molar mass of carbon dioxide (COâ‚‚) is 44 g/mol. For Oxygen: moles of Oâ‚‚ = mass of Oâ‚‚ / molar mass of Oâ‚‚ moles of Oâ‚‚ = 52.5 g / 32 g/mol = 1.64 mol For Carbon dioxide: moles of COâ‚‚ = mass of COâ‚‚ / molar mass of COâ‚‚ moles of COâ‚‚ = 65.1 g / 44 g/mol = 1.48 mol
02

Calculate total moles and mole fractions

Next, we need to find the total moles of both gases combined, and then calculate the mole fraction for each gas. Total moles = moles of Oâ‚‚ + moles of COâ‚‚ Total moles = 1.64 mol + 1.48 mol = 3.12 mol Now, we can find the mole fraction for each gas: Mole fraction of Oâ‚‚ = moles of Oâ‚‚ / total moles = 1.64 mol / 3.12 mol = 0.526 Mole fraction of COâ‚‚ = moles of COâ‚‚ / total moles = 1.48 mol / 3.12 mol = 0.474
03

Calculate partial pressures using Dalton's Law

Now we can use Dalton's Law of Partial Pressures, where the partial pressure of each gas equals the product of the total pressure and the mole fraction of that gas: Partial pressure of Oâ‚‚ = total pressure * mole fraction of Oâ‚‚ Partial pressure of Oâ‚‚ = 9.21 atm * 0.526 = 4.84 atm Partial pressure of COâ‚‚ = total pressure * mole fraction of COâ‚‚ Partial pressure of COâ‚‚ = 9.21 atm * 0.474 = 4.37 atm The partial pressures of oxygen and carbon dioxide in the container are 4.84 atm and 4.37 atm, respectively.

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Key Concepts

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

Partial Pressure
Partial pressure is a crucial concept in understanding gas mixtures. It refers to the pressure that each gas in a mixture would exert if it were alone in a container. In essence, each gas feels like it is the only one present and contributes individually to the overall pressure. This contribution is known as the partial pressure of the gas.

When we know the total pressure of a gas mixture, like in our exercise (9.21 atm), we can find the individual gas contributions using Dalton's Law of Partial Pressures. According to this law, the partial pressure of a gas is the product of its mole fraction and the total pressure.
  • This means each gas' partial pressure is just its fraction of the number of particles multiplied by the total pressure.
  • For instance, if a gas makes up half of the gas mixture by moles, its partial pressure will be half of the total pressure.
Understanding partial pressures helps in various applications, from breathing systems to predicting reactions in chemistry.
Mole Fraction
Mole fraction is an important concept when dealing with mixtures of gases. It shows what proportion or fraction of the total moles (or amount) in the mixture a particular gas makes up. This fraction helps us understand how much of the container's contents is contributed by each component gas.

To calculate mole fraction, we use the formula:
  • Mole fraction of a gas = Moles of the gas / Total moles of all gases in the mixture.
In the exercise, the mole fraction for oxygen was calculated as 0.526, and for carbon dioxide, it was 0.474. These values mean oxygen makes up 52.6% of the gas mixture by moles, and carbon dioxide makes up 47.4%.
  • Knowing mole fractions is crucial because they help in computing partial pressures, as seen with Dalton's Law.
Mole fractions are unitless, making them convenient for calculations and comparisons.
Gas Mixtures
Gas mixtures consist of more than one type of gas molecule within the same container. The properties of the mixture depend on the behavior and amount of its components. Each gas behaves independently and contributes to the total pressure according to its proportion in the mixture.

In terms of chemistry, gas mixtures behave according to the Ideal Gas Law. However, Dalton's Law simplifies our understanding by allowing us to calculate the contribution of each component as if they do not interact. In real-world applications, knowing the composition of a gas mixture is essential for processes such as combustion, respiration, and in industrial chemical reactions.
  • Understanding gas mixtures involves knowing not just the types but the amounts of gases present.
  • In our example, we worked with oxygen and carbon dioxide, both affecting the total pressure.
The ability to analyze and predict the behavior of gas mixtures is a powerful tool in science and engineering.
Moles Calculation
Calculating moles from a given mass is a foundational skill in chemistry that allows us to understand proportions in reactions and mixtures. Moles, a measure of substance amount, connect tangible mass with the microscopic world of molecules.

To calculate moles, use the formula:
  • Moles = Mass of the substance / Molar mass of the substance
In our exercise, the moles of oxygen and carbon dioxide were determined using their respective molar masses (32 g/mol for Oâ‚‚ and 44 g/mol for COâ‚‚). This calculation enabled us to find out how much of each gas was present and to move forward with computing mole fractions and partial pressures.
  • Moles serve as a bridge between mass and Avogadro's number, which is crucial for understanding chemical equations and reactions.
Mastering moles calculation empowers students to tackle more complex topics such as stoichiometry and gas laws.

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Most popular questions from this chapter

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(\mathrm{s})\): $${\text { Compound }} \frac{\text { Hot }}{\text { \(\mathrm{CuO}(\mathrm{s})\) }} \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ The product gas is then passed through a concentrated solution of \(\mathrm{KOH}\) to remove the \(\mathrm{CO}_{2} .\) After passage through the \(\mathrm{KOH}\) solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a 0.253 -g sample of a compound produced 31.8 \(\mathrm{mL} \mathrm{N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and 726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is 23.8 torr.

Write an equation to show how sulfuric acids in acid rain reacts with marble and limestone. (Both marble and limestone are primarily calcium carbonate.)

In the "Methode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density \(=1.0 \mathrm{g} / \mathrm{cm}^{3} )\) is allowed to take place in a bottle with a total volume of 825 \(\mathrm{mL}\) until 12\(\%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) . Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is \(0.79 \mathrm{g} / \mathrm{cm}^{3} . )\)

An organic compound contains \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) . Combustion of 0.1023 \(\mathrm{g}\) of the compound in excess oxygen yielded 0.2766 \(\mathrm{g} \mathrm{CO}_{2}\) and 0.0991 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) A sample of 0.4831 \(\mathrm{g}\) of the compound was analyzed for nitrogen by the Dumas method (see Exercise 137\() .\) At STP, 27.6 \(\mathrm{mL}\) of dry \(\mathrm{N}_{2}\) was obtained. In a third experiment, the density of the compound as a gas was found to be 4.02 \(\mathrm{g} / \mathrm{L}\) at \(127^{\circ} \mathrm{C}\) and 256 torr. What are the empirical and molecular formulas of the compound?

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