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Remember to keep a few extra digits while calculating, and only round your final answer to the requested number of significant digits. • Now it's time for the ideal gas law: PCH₂CH₂ V RT CH₂CH₂ BCH₂CH₂ n. 11₂ n PCH₂CH₂ RT m CH₂CH₂ ANSWER (7.3985 atm) (70.0 L) L'atm mol k 0.082057 (773.15 K) (8.6015 atm) (70.0 L) L'atm mol-k 0.082057 = 9.4906... mol You did remember to convert temperature to kelvins, right? Finally, molar masses take us from moles to mass: MCH,CH, (8.1632 mol) 30.0690- % CH₂CH₂= 100 x- =245.46... B 266.24. mol, m (9.4906 mol) (2.015885 = 19.132... g mol Now you can calculate the mass percentage of CH₂CH₂: CH₂CH₂(9.4906 mol) (28.0532- (773.15 K) mol 266.24 245.46 g +266.24 g+19.132 g Don't forget to round your final answer to 2 significant digits. = 8.1632... mol -9.4906... mol 50.16...%

Remember to keep a few extra digits while calculating, and only round your final answer to the requested number of significant digits. • Now it's time for the ideal gas law: PCH₂CH₂ V RT CH₂CH₂ BCH₂CH₂ n. 11₂ n PCH₂CH₂ RT m CH₂CH₂ ANSWER (7.3985 atm) (70.0 L) L'atm mol k 0.082057 (773.15 K) (8.6015 atm) (70.0 L) L'atm mol-k 0.082057 = 9.4906... mol You did remember to convert temperature to kelvins, right? Finally, molar masses take us from moles to mass: MCH,CH, (8.1632 mol) 30.0690- % CH₂CH₂= 100 x- =245.46... B 266.24. mol, m (9.4906 mol) (2.015885 = 19.132... g mol Now you can calculate the mass percentage of CH₂CH₂: CH₂CH₂(9.4906 mol) (28.0532- (773.15 K) mol 266.24 245.46 g +266.24 g+19.132 g Don't forget to round your final answer to 2 significant digits. = 8.1632... mol -9.4906... mol 50.16...%

Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter5: Gases
Section: Chapter Questions
Problem 5.98PAE
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CH CH (9) CH₂CH₂(g) +H₂ (s) From the equation, you can write the pressure equilibrium constant expression equation: PCH₂CH₂ PH₂ PCH₂CH₂ Now for the ICE table. Let's let X stand for the increase in pressure of CH, CH₂: Initial Change Equilibrium =10. x²+10.x-160=0 16.0-x Substitute the expressions on the "equilibrium" line into the pressure equilibrium constant expression equation and solve for X: PCH₂CH₂ 50.% PCH₂CH₂ 16.0 -10. +√(10.² +4(160)) 2 x=8.6015... or x=-18.601... 16.0-x The second value of x is not physical, because it makes at least one of the equilibrium pressures negative. Substituting the first value into the "equilibrium expressions in the ICE table gives you the equilibrium pressures of everything: PCH, CH₂=16.0-x-16.0-8.6015 = 7.3985... atm = x= 8.6015 atm = x= 8.6015 atm CH₂CH₂ CH₂CH₂ -x ANSWER Remember to keep a few extra digits while calculating, and only round your final answer to the requested number of significant digits. Now it's time for the ideal gas law: V PCH₂CH₂ RT PCH₂CH₂ PH₂ 0 0 PCH₂CH₂ RT % CH₂CH₂ = 100 x Use the quadratic formula. x (7.3985 atm) (70.0 L) L-atm mol k 0.082057 = 9.4906... mol 0.082057 (8.6015 atm) (70.0L) L'atm mol-k mol MCH₂CH₂(9.4906 mol) 28.0532- mol (9.4906 mol) 2.01588 H₂CH₂CH₂ You did remember to convert temperature to kelvins, right? Finally, molar masses take us from moles to mass: MCH₂CH₂=(8.1632 mol) 30.0690 (773.15 K) m mol Now you can calculate the mass percentage of CH₂ CH₂: (773.15 K) =245.46... =266.24... 266.24 g 245.46 g +266.24 g+19.132 g Don't forget to round your final answer to 2 significant digits. = 19.132... = 8.1632... mol -9.4906... mol = 50.16...%
Transcribed Image Text:CH CH (9) CH₂CH₂(g) +H₂ (s) From the equation, you can write the pressure equilibrium constant expression equation: PCH₂CH₂ PH₂ PCH₂CH₂ Now for the ICE table. Let's let X stand for the increase in pressure of CH, CH₂: Initial Change Equilibrium =10. x²+10.x-160=0 16.0-x Substitute the expressions on the "equilibrium" line into the pressure equilibrium constant expression equation and solve for X: PCH₂CH₂ 50.% PCH₂CH₂ 16.0 -10. +√(10.² +4(160)) 2 x=8.6015... or x=-18.601... 16.0-x The second value of x is not physical, because it makes at least one of the equilibrium pressures negative. Substituting the first value into the "equilibrium expressions in the ICE table gives you the equilibrium pressures of everything: PCH, CH₂=16.0-x-16.0-8.6015 = 7.3985... atm = x= 8.6015 atm = x= 8.6015 atm CH₂CH₂ CH₂CH₂ -x ANSWER Remember to keep a few extra digits while calculating, and only round your final answer to the requested number of significant digits. Now it's time for the ideal gas law: V PCH₂CH₂ RT PCH₂CH₂ PH₂ 0 0 PCH₂CH₂ RT % CH₂CH₂ = 100 x Use the quadratic formula. x (7.3985 atm) (70.0 L) L-atm mol k 0.082057 = 9.4906... mol 0.082057 (8.6015 atm) (70.0L) L'atm mol-k mol MCH₂CH₂(9.4906 mol) 28.0532- mol (9.4906 mol) 2.01588 H₂CH₂CH₂ You did remember to convert temperature to kelvins, right? Finally, molar masses take us from moles to mass: MCH₂CH₂=(8.1632 mol) 30.0690 (773.15 K) m mol Now you can calculate the mass percentage of CH₂ CH₂: (773.15 K) =245.46... =266.24... 266.24 g 245.46 g +266.24 g+19.132 g Don't forget to round your final answer to 2 significant digits. = 19.132... = 8.1632... mol -9.4906... mol = 50.16...%
Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds 2.0 × 10¹¹ kg per year. The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 3.1 kg of solid sulfur and 10.0 atm of oxygen gas at 900. °C into an evacuated 40.0 L tank. The engineer believes Kp = 0.039 for the reaction at this temperature. Calculate the mass of solid sulfur she expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits. Note for advanced students: the engineer may be mistaken in her belief about the value of K and the consumption of sulfur you calculate may not be what she actually observes. kg ☐ x10 X
Transcribed Image Text:Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds 2.0 × 10¹¹ kg per year. The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 3.1 kg of solid sulfur and 10.0 atm of oxygen gas at 900. °C into an evacuated 40.0 L tank. The engineer believes Kp = 0.039 for the reaction at this temperature. Calculate the mass of solid sulfur she expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits. Note for advanced students: the engineer may be mistaken in her belief about the value of K and the consumption of sulfur you calculate may not be what she actually observes. kg ☐ x10 X
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