The following two-step process has a 60.0 % yield for each step. CH4 + 4 Cl2 –→ CCl + 4 HCl CCl4 + 2 HF → CCl,F2 + 2 HC1 The CC14 that is formed in the first step of the process is used as the reactant for the second step. Assuming 6.00 mol of CH4 are used in the reaction with excess amounts of both Cl, and HF, how many total moles of HCl would be formed?

The following two-step process has a 60.0 % yield for each step. CH4 + 4 Cl2 –→ CCl + 4 HCl CCl4 + 2 HF → CCl,F2 + 2 HC1 The CC14 that is formed in the first step of the process is used as the reactant for the second step. Assuming 6.00 mol of CH4 are used in the reaction with excess amounts of both Cl, and HF, how many total moles of HCl would be formed?

World of Chemistry

7th Edition

ISBN:9780618562763

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter9: Chemical Quantities

Section: Chapter Questions

Problem 52A

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