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The total pressure of a mixture of oxygen and hydrogen is 1.65 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.190 atm when measured at the same values of \(T\) and \(V\) as the original mixture. What was the composition of the original mixture in mole percent?

Short Answer

Expert verified
The original mixture was composed of \(88.5\%\) oxygen and \(11.5\%\) hydrogen by mole. This was determined by calculating the partial pressures of oxygen and hydrogen, finding their mole fractions, and converting these into mole percents.

Step by step solution

01

Calculate partial pressure of oxygen

Initially, we have the total pressure of the mixture as 1.65 atm. After igniting the mixture and removing the water, the remaining pure hydrogen has a pressure of 0.190 atm. Therefore, the partial pressure of oxygen in the original mixture can be found by subtracting the final pressure of hydrogen from the total pressure: \(P_{O_2} = P_{total} - P_{H_2}\) \(P_{O_2} = 1.65\, \mathrm{atm} - 0.190\, \mathrm{atm} = 1.46\, \mathrm{atm}\)
02

Calculate mole fractions

Now that we have the partial pressures of both hydrogen and oxygen, we can find their mole fractions in the mixture. Using the ideal gas law, we have: \(P_{O_2} = x_{O_2}P_{total}\) \(x_{O_2} = \frac{P_{O_2}}{P_{total}}\) \(x_{O_2} = \frac{1.46\, \mathrm{atm}}{1.65\, \mathrm{atm}} = 0.885\) Similarly, for hydrogen: \(P_{H_2} = x_{H_2}P_{total}\) \(x_{H_2} = \frac{P_{H_2}}{P_{total}}\) \(x_{H_2} = \frac{0.190\, \mathrm{atm}}{1.65\, \mathrm{atm}} = 0.115\)
03

Convert mole fractions to mole percents

To convert the mole fractions to mole percents, we simply multiply each by 100: Mole percent of oxygen = \(0.885 \times 100 = 88.5\% \) Mole percent of hydrogen = \(0.115 \times 100 = 11.5\% \) Thus, the original mixture was composed of \(88.5\%\) oxygen and \(11.5\%\) hydrogen by mole.

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Key Concepts

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

Understanding Partial Pressure
If you've worked with gas mixtures, you've encountered the term 'partial pressure.' It's a measure of the contribution each gas in a mixture makes to the total pressure. Think of it as if each gas were alone in the container; its pressure would be 'partial pressure.' For a given mixture held at constant temperature and volume, the total pressure is the sum of the partial pressures of the individual gases. This comes from Dalton's Law of Partial Pressures.

To find a specific gas's partial pressure, like we did for oxygen in our exercise, you subtract the pressure of the other gases from the total pressure. Once the water produced from the reaction was removed, only hydrogen remained, with a partial pressure that was measured. Subtracting this pressure from the initial total pressure gave us the oxygen's partial pressure before the reaction.
The Ideal Gas Law in Action
The ideal gas law is an essential tool in understanding gases. It's usually written as PV=nRT, tying together pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T). For gas mixtures, 'n' can be considered as the sum of moles of all components. But what does it do?

This law allows us to relate the physical properties of a gas under a set of conditions, and in the context of our problem, it helps us correlate partial pressures with mole fractions. A really neat aspect of the ideal gas law is its ability to adapt to situations involving mixtures of gases. When we look at a single component in a mix like oxygen or hydrogen, the ideal gas law can still be applied to find its partial pressure or mole fraction, just like we did in the exercise.
Mole Fraction Explained
Mole fraction may sound complicated, but it's just a way to express the concentration of a component in a mixture. Represented by the letter 'x,' it's the ratio of the number of moles of one component to the total number of moles in the mixture. You can think of it like a piece of pie; the mole fraction tells you how big your slice is compared to the whole pie.

Calculating mole fraction is straightforward. In our exercise, we divided the partial pressure of each gas by the total pressure to get their respective mole fractions. The beauty of mole fraction is that it doesn't depend on pressure or temperature; it's all about the ratio. Converting these fractions into percent form just means multiplying by 100, giving us a percentage that's easy to visualize and understand. Now, when you think of gas compositions, you can picture the mole percent as the 'percentage of the pie' that each gas takes up!

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Most popular questions from this chapter

Approximately how many oxygen molecules arrive each second at the mitochondrion of an active person with a mass of \(84 \mathrm{kg} ?\) The following data are available: Oxygen consumption is about \(40 . \mathrm{mL}\) of \(\mathrm{O}_{2}\) per minute per kilogram of body weight, measured at \(T=300 . \mathrm{K}\) and \(P=1.00 \mathrm{atm} .\) In an adult there are about \(1.6 \times 10^{10}\) cells per kg body mass. Each cell contains about 800 . mitochondria.

A compressed cylinder of gas contains \(2.74 \times 10^{3} \mathrm{g}\) of \(\mathrm{N}_{2}\) gas at a pressure of \(3.75 \times 10^{7} \mathrm{Pa}\) and a temperature of \(18.7^{\circ} \mathrm{C} .\) What volume of gas has been released into the atmosphere if the final pressure in the cylinder is \(1.80 \times 10^{5}\) Pa? Assume ideal behavior and that the gas temperature is unchanged.

Liquid \(\mathrm{N}_{2}\) has a density of \(875.4 \mathrm{kg} \mathrm{m}^{-3}\) at its normal boiling point. What volume does a balloon occupy at \(298 \mathrm{K}\) and a pressure of 1.00 atm if \(3.10 \times 10^{-3} \mathrm{L}\) of liquid \(\mathrm{N}_{2}\) is injected into it? Assume that there is no pressure difference between the inside and outside of the balloon.

Use the ideal gas and van der Waals equations to calculate the pressure when \(2.25 \mathrm{mol} \mathrm{H}_{2}\) are confined to a volume of \(1.65 \mathrm{L}\) at \(298 \mathrm{K}\). Is the gas in the repulsive or attractive region of the molecule-molecule potential?

One liter of fully oxygenated blood can carry 0.18 liters of \(\mathrm{O}_{2}\) measured at \(T=298 \mathrm{K}\) and \(P=1.00 \mathrm{atm}\) Calculate the number of moles of \(\mathrm{O}_{2}\) carried per liter of blood. Hemoglobin, the oxygen transport protein in blood has four oxygen binding sites. How many hemoglobin molecules are required to transport the \(\mathrm{O}_{2}\) in \(1.0 \mathrm{L}\) of fully oxygenated blood?

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