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Each step in the following process has a yield of 60.0%. CH₂ +4Cl₂ ⇒ CCI₂ + 4HCI CCI + 2 HF →→ CCI₂F₂ + 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. moles HCI: 30 Incorrect mol

Each step in the following process has a yield of 60.0%. CH₂ +4Cl₂ ⇒ CCI₂ + 4HCI CCI + 2 HF →→ CCI₂F₂ + 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. moles HCI: 30 Incorrect mol

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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### Reaction Yield and HCl Production In a chemical process where each step has a yield of 60.0%, consider the following reactions: 1. \( \text{CH}_4 + 4 \text{Cl}_2 \rightarrow \text{CCl}_4 + 4 \text{HCl} \) 2. \( \text{CCl}_4 + 2 \text{HF} \rightarrow \text{CCl}_2\text{F}_2 + 2 \text{HCl} \) The \(\text{CCl}_4\) produced in the first reaction is used as a reactant in the second reaction. The problem is to determine the total amount of HCl produced when 5.00 moles of \(\text{CH}_4\) react, assuming \(\text{Cl}_2\) and \(\text{HF}\) are present in excess. #### Given Information: - Yield for each step is 60.0%. - Initial moles of \(\text{CH}_4\): 5.00 moles. #### Incorrect Attempt: - Answer submitted: 30 moles HCl. Note: The calculation of the correct amount of HCl produced involves considering the yield percentage and the stoichiometry of the reactions.
Transcribed Image Text:### Reaction Yield and HCl Production In a chemical process where each step has a yield of 60.0%, consider the following reactions: 1. \( \text{CH}_4 + 4 \text{Cl}_2 \rightarrow \text{CCl}_4 + 4 \text{HCl} \) 2. \( \text{CCl}_4 + 2 \text{HF} \rightarrow \text{CCl}_2\text{F}_2 + 2 \text{HCl} \) The \(\text{CCl}_4\) produced in the first reaction is used as a reactant in the second reaction. The problem is to determine the total amount of HCl produced when 5.00 moles of \(\text{CH}_4\) react, assuming \(\text{Cl}_2\) and \(\text{HF}\) are present in excess. #### Given Information: - Yield for each step is 60.0%. - Initial moles of \(\text{CH}_4\): 5.00 moles. #### Incorrect Attempt: - Answer submitted: 30 moles HCl. Note: The calculation of the correct amount of HCl produced involves considering the yield percentage and the stoichiometry of the reactions.
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