The following two-step process has a 50.0 % yield for each step. CH4 + 4 Cl2 – CC14 + 4 HC1 CCI4 + 2 HF – CC12F2 + 2 HC1 The CCI4 that is formed in the first step of the process is used as the reactant for the second step. Assuming 7.00 mol of CH4 are used in the reaction with excess amounts of both Cl2 and HF , how many total moles of HCl would be formed? HCI mol

Chemistry & Chemical Reactivity
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ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
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Chapter4: Stoichiometry: Quantitative Information About Chemical Reactions
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The following two-step process has a 50.0 % yield for each step.
CH4 + 4 Cl2 – CCl4 + 4 HC1
CC4 + 2 HF – CCI2F2 + 2 HCI
The CC14 that is formed in the first step of the process is used as the reactant for the second step. Assuming 7.00 mol
of CH4 are used in the reaction with excess amounts of both Cl2 and HF, how many total moles of HCl would be
formed?
HCI
mol
Transcribed Image Text:The following two-step process has a 50.0 % yield for each step. CH4 + 4 Cl2 – CCl4 + 4 HC1 CC4 + 2 HF – CCI2F2 + 2 HCI The CC14 that is formed in the first step of the process is used as the reactant for the second step. Assuming 7.00 mol of CH4 are used in the reaction with excess amounts of both Cl2 and HF, how many total moles of HCl would be formed? HCI mol
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