CS2(g) + 3C12(g) = $2C12(g) + CC14(g) When 1.14 mol of CS2 and 4.80 mol of Cl2 are placed in a 2.00-L container and allowed to come to equilibrium, the mixture is found to contain 0.650 mol of CC14. How many moles of Cl2 are present at equilibrium? O a. 3.50 mol O b.0.650 mol Oc. 2.85 mol Od.0.490 mol O e. 1.43 mol
CS2(g) + 3C12(g) = $2C12(g) + CC14(g) When 1.14 mol of CS2 and 4.80 mol of Cl2 are placed in a 2.00-L container and allowed to come to equilibrium, the mixture is found to contain 0.650 mol of CC14. How many moles of Cl2 are present at equilibrium? O a. 3.50 mol O b.0.650 mol Oc. 2.85 mol Od.0.490 mol O e. 1.43 mol
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter14: Chemical Equilibirum
Section: Chapter Questions
Problem 14.13QP: A chemist put 1.18 mol of substance A and 2.85 mol of substance B into a 10.0-L flask, which she...
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