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A 15.0-L rigid container was charged with 0.500 atm of kryp- ton gas and 1.50 atm of chlorine gas at 350.°C. The krypton and chlorine react to form krypton tetrachloride. What mass of krypton tetrachloride can be produced assuming 100% yield?

A 15.0-L rigid container was charged with 0.500 atm of kryp- ton gas and 1.50 atm of chlorine gas at 350.°C. The krypton and chlorine react to form krypton tetrachloride. What mass of krypton tetrachloride can be produced assuming 100% yield?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question 71 Has got me stumped. I figured that I would combine the pressures and use that as the overall pressure and then calculate the total moles and make a ratio between the total moles of reactant to the product but it did not work because the answer is has got me stumped. I figured that I would combine the pressures and use that as the overall pressure and then calculate the total moles and make a ratio between the total moles of reactant to the product but it did not work because the answer is 33.2

Certainly! Here is the transcription of the handwritten notes as if they were to appear on an educational website: --- ### Chemistry Calculation **Problem #71 - Summary** The reaction of Krypton (Kr) with Iodine (I) is being considered: \[ \text{Kr(g) + I}_{2(g)}\rightarrow \text{KrI}_{4(g)} \] The Ideal Gas Law equation is used: \[ PV = nRT \] **Given/Calculated Values:** - **V₁** = 15 L (does not change) - Convert Volume: - 1 cm = 0.85 dm (1 L = 1.5 dm³) - Conversion factors: 0.85 dm and 1.5 dm - **Temperature (T)** = 350 K - Convert Temperature (not provided conversion) - Resulting T = 623 K (does not change) **Moles of Reactant:** \[ P_TV = n_TRT \] Given the calculation: \[ \frac{(2.0)(15)}{1.08206(623)} = n_T = 0.5868 \text{ moles reactant} \] - **Convert to moles reactant**: \[ \frac{0.5868 \text{ moles reactant}}{1 \text{ mol KrI}_4} = \frac{83.91 g}{\text{rel KrI}_4} = 166.4 \text{ g} \] **Note:** An annotation at the bottom indicates the realization of a conceptual oversight: - "Did not work. Why? We did just 1 get." --- This transcription translates the handwritten calculations into typed format, typical for educational settings. If any diagrams or graphs were present in the image, there would be a detailed explanation here, but there are none in this case.
Transcribed Image Text:Certainly! Here is the transcription of the handwritten notes as if they were to appear on an educational website: --- ### Chemistry Calculation **Problem #71 - Summary** The reaction of Krypton (Kr) with Iodine (I) is being considered: \[ \text{Kr(g) + I}_{2(g)}\rightarrow \text{KrI}_{4(g)} \] The Ideal Gas Law equation is used: \[ PV = nRT \] **Given/Calculated Values:** - **V₁** = 15 L (does not change) - Convert Volume: - 1 cm = 0.85 dm (1 L = 1.5 dm³) - Conversion factors: 0.85 dm and 1.5 dm - **Temperature (T)** = 350 K - Convert Temperature (not provided conversion) - Resulting T = 623 K (does not change) **Moles of Reactant:** \[ P_TV = n_TRT \] Given the calculation: \[ \frac{(2.0)(15)}{1.08206(623)} = n_T = 0.5868 \text{ moles reactant} \] - **Convert to moles reactant**: \[ \frac{0.5868 \text{ moles reactant}}{1 \text{ mol KrI}_4} = \frac{83.91 g}{\text{rel KrI}_4} = 166.4 \text{ g} \] **Note:** An annotation at the bottom indicates the realization of a conceptual oversight: - "Did not work. Why? We did just 1 get." --- This transcription translates the handwritten calculations into typed format, typical for educational settings. If any diagrams or graphs were present in the image, there would be a detailed explanation here, but there are none in this case.
The text appears to be from a chemistry textbook. Here's the transcription and explanation for an educational website: --- **Problem 77** A 150.0-L rigid container was charged with 0.500 atm of krypton gas and 1.50 atm of chlorine gas at 350°C. The krypton and chlorine react to form krypton tetrachloride. What mass of krypton tetrachloride can be produced assuming 100% yield? **Problem 78** What volume of O₂(g) at 350°C and a pressure of 5.25 atm is needed to completely convert 5.00 g of sulfur to sulfur trioxide? The reactions pertinent to the conversions are outlined below: S(s) + O₂(g) → SO₂(g) 2SO₂(g) + O₂(g) → 2SO₃(g) --- **Explanation** In Problem 77, the chemical reaction involves the combination of krypton gas and chlorine gas to form krypton tetrachloride. Using the ideal gas law and stoichiometry, you can determine the mass of krypton tetrachloride formed under given conditions. In Problem 78, the task is to find out the volume of oxygen gas needed to completely oxidize sulfur to sulfur trioxide at specified temperature and pressure. The chemical equations provided show a stepwise oxidation of sulfur, first to sulfur dioxide and then to sulfur trioxide. For accurate calculations, one would: 1. Determine the moles of reactants using the ideal gas law for gases. 2. Use stoichiometry and balanced chemical equations to find the relationships between reactants and products. 3. Calculate the desired quantities, such as mass or volume, based on reactions’ stoichiometry. This practice encourages an understanding of gas laws, stoichiometry, and chemical reaction principles in chemistry.
Transcribed Image Text:The text appears to be from a chemistry textbook. Here's the transcription and explanation for an educational website: --- **Problem 77** A 150.0-L rigid container was charged with 0.500 atm of krypton gas and 1.50 atm of chlorine gas at 350°C. The krypton and chlorine react to form krypton tetrachloride. What mass of krypton tetrachloride can be produced assuming 100% yield? **Problem 78** What volume of O₂(g) at 350°C and a pressure of 5.25 atm is needed to completely convert 5.00 g of sulfur to sulfur trioxide? The reactions pertinent to the conversions are outlined below: S(s) + O₂(g) → SO₂(g) 2SO₂(g) + O₂(g) → 2SO₃(g) --- **Explanation** In Problem 77, the chemical reaction involves the combination of krypton gas and chlorine gas to form krypton tetrachloride. Using the ideal gas law and stoichiometry, you can determine the mass of krypton tetrachloride formed under given conditions. In Problem 78, the task is to find out the volume of oxygen gas needed to completely oxidize sulfur to sulfur trioxide at specified temperature and pressure. The chemical equations provided show a stepwise oxidation of sulfur, first to sulfur dioxide and then to sulfur trioxide. For accurate calculations, one would: 1. Determine the moles of reactants using the ideal gas law for gases. 2. Use stoichiometry and balanced chemical equations to find the relationships between reactants and products. 3. Calculate the desired quantities, such as mass or volume, based on reactions’ stoichiometry. This practice encourages an understanding of gas laws, stoichiometry, and chemical reaction principles in chemistry.
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