Each step in the following process has a yield of 60.0%. CCl₂ + 4HCI → CCL₂F₂ + 2HCl The CCI, formed in the first step is used as a reactant in the second step. If 5.00 mol CH₂ reacts, what is the total amount of HCI produced? Assume that Cl₂ and HF are present in excess. CH4 + 4Cl₂ CCl₂ + 2 HF moles HCl: 12.48 Incorrect mol

Chemistry for Engineering Students
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Author:Lawrence S. Brown, Tom Holme
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Chapter4: Stoichiometry
Section: Chapter Questions
Problem 4.13PAE: 4.13 For the following reactions, determine the value of x. 4C+S84CS2 mol S8 yields x mol CS2...
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Each step in the following process has a yield of 60.0%.
CCl₂ + 4HCI
→ CCL₂F₂ + 2HCl
The CCI, formed in the first step is used as a reactant in the second step.
If 5.00 mol CH₂ reacts, what is the total amount of HCI produced? Assume that Cl₂ and HF are present in excess.
CH4 + 4Cl₂
CCl₂ + 2 HF
moles HCl:
12.48
Incorrect
mol
Transcribed Image Text:Each step in the following process has a yield of 60.0%. CCl₂ + 4HCI → CCL₂F₂ + 2HCl The CCI, formed in the first step is used as a reactant in the second step. If 5.00 mol CH₂ reacts, what is the total amount of HCI produced? Assume that Cl₂ and HF are present in excess. CH4 + 4Cl₂ CCl₂ + 2 HF moles HCl: 12.48 Incorrect mol
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