Gaseous iodine pentafluoride, IF 5 , can be prepared by the reaction of solid iodine and gaseous fluorine: I 2 (s) + 5 F 2 (g) → 2 IF 5 (g) A 5.00-L flask containing 10.0 g of I 2 is charged with 10.0 g of F 2 , and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 °C. (b) What is the mole fraction of IF 5 in the flask?

Welcome back everyone by oxidizing sodium fluoride with manganese oxide. Flooring gas can be created. Were given the following equation where two moles of sodium fluoride solid react with two moles of sulfuric acid and one mole of manganese oxide solid To produce two moles or sorry, one mole of sodium Sulfate solid, one mole of manganese sulfate solid and two moles of liquid water and one mole of flooring gas. Assume that the produced gas has a volume of .631 leaders at 26°C, a pressure of 700 and 62 millimeters of mercury and is saturated with water vapor at a partial pressure of 27.9 millimeters of mercury. What is the mole fraction of flooring gas in the gas? So what we want to recall is Dalton's law, which we can apply in terms of this described reaction, where our flooring gas is collected through the bubbling of our water vapor. And so we can say that the total pressure is equal to the pressure of our flooring gas added to our pressure of our water vapor. And this should equal 762 mm of mercury according to our prompt. So to solve for our pressure of our flooring gas, we would isolate it and say that that is equal to the difference between our total pressure and our pressure of water. Now, according to the prompt, we are given our pressure of water as 27.9 mm of mercury. So we're just going to plug in our knowns and what we're going to have is that our pressure of our flooring gas is equal to the total pressure given in the prompt as 762 millimeters of mercury subtracted from. And sorry, this should be a subtraction sign here. So this is subtracted from our pressure of our water vapor given as 27.9 millimeters of mercury. So this is going to give us our pressure of our flooring gas equal to a value of 734.1 millimeters of mercury. So now to get our mole fraction of our flooring gas, we're going to recall that we take the pressure of our component. So being our flooring gas divided by the total pressure Plus our pressure of our or rather it would just be our total pressure since it would include everything in our system. And so plugging in what we know we just saw for our pressure of our flooring gas as 734.1 mm of mercury. Just slide this over and this is divided by our total pressure which was given to us in the prompt as 762 mm of Mercury. So what this will simplify too is a value of .963 and this would be our final answer as the mole fraction of flooring gas in our gas produced. So what's highlighted in yellow is our final answer. I hope everything I explained was clear. If you have any questions please leave them down below, and I will see everyone in the next practice video.

Ch.10 - Gases

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Brown 14th Edition

Ch.10 - Gases

Problem 126b

Chapter 10, Problem 126b

Gaseous iodine pentafluoride, IF₅, can be prepared by the reaction of solid iodine and gaseous fluorine: I₂(s) + 5 F₂(g) → 2 IF₅(g) A 5.00-L flask containing 10.0 g of I₂ is charged with 10.0 g of F₂, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 °C. (b) What is the mole fraction of IF₅ in the flask?

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First, calculate the number of moles of I2 and F2 using their molar masses. The molar mass of I2 is approximately 253.8 g/mol, and the molar mass of F2 is approximately 38.0 g/mol. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).

Determine the limiting reagent by comparing the mole ratio of I2 and F2 to the stoichiometry of the reaction. The balanced equation shows that 1 mole of I2 reacts with 5 moles of F2. Calculate the moles of each reactant and compare to find which one is the limiting reagent.

Calculate the moles of IF5 produced using the stoichiometry of the reaction. Since the reaction produces 2 moles of IF5 for every 1 mole of I2 consumed, use the moles of the limiting reagent to find the moles of IF5 formed.

Calculate the total moles of gas in the flask after the reaction. This includes the moles of IF5 produced and any excess moles of the non-limiting reagent that remain unreacted.

Finally, calculate the mole fraction of IF5 in the flask. The mole fraction is given by the formula: \( \text{mole fraction of IF5} = \frac{\text{moles of IF5}}{\text{total moles of gas}} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced equation. It allows us to determine the amount of product formed or reactants consumed by using mole ratios derived from the coefficients in the balanced equation. In this case, the stoichiometric coefficients indicate that 1 mole of I2 reacts with 5 moles of F2 to produce 2 moles of IF5.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture, defined as the ratio of the number of moles of that component to the total number of moles of all components in the mixture. It is a dimensionless quantity and is useful for understanding the composition of gaseous mixtures, such as the one formed in this reaction. The mole fraction of IF5 can be calculated after determining the moles of all species present.

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law is essential for understanding the behavior of gases under various conditions. In this scenario, it can be used to find the total number of moles of gas in the flask after the reaction, which is necessary for calculating the mole fraction of IF5.

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Related Practice

Textbook Question

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of -164 °C. One possible strategy is to oxidize the methane to methanol, CH₃OH, which has a boiling point of 65 °C and can therefore be shipped more readily. Suppose that 3.03×10⁸ m³ of methane at atmospheric pressure and 25 °C is oxidized to methanol. (a) What volume of methanol is formed if the density of CH₃OH is 0.791 g/mL?

Textbook Question

Gaseous iodine pentafluoride, IF₅, can be prepared by the reaction of solid iodine and gaseous fluorine: I₂(s) + 5 F₂(g) → 2 IF₅(g) A 5.00-L flask containing 10.0 g of I₂ is charged with 10.0 g of F₂, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 °C. (a) What is the partial pressure of IF5 in the flask?

Textbook Question

Gaseous iodine pentafluoride, IF₅, can be prepared by the reaction of solid iodine and gaseous fluorine: I₂(s) + 5 F₂(g) → 2 IF₅(g) A 5.00-L flask containing 10.0 g of I₂ is charged with 10.0 g of F₂, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 °C. (d) What is the total mass of reactants and products in the flask?

Textbook Question

A 6.53-g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces 1.72 L of carbon dioxide gas at 28 °C and 99.06 kPa pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture.