Each step in the following process has a yield of 60.0%. CCl₂ + 4HCI CC1₂F₂ + 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 CCI + 2 HF moles HCl:
Each step in the following process has a yield of 60.0%. CCl₂ + 4HCI CC1₂F₂ + 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 CCI + 2 HF moles HCl:
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![Each step in the following process has a yield of 60.0%.
\[ \text{CH}_4 + 4 \text{Cl}_2 \rightarrow \text{CCl}_4 + 4 \text{HCl} \]
\[ \text{CCl}_4 + 2 \text{HF} \rightarrow \text{CCl}_2\text{F}_2 + 2 \text{HCl} \]
The CCl\(_4\), formed in the first step, is used as a reactant in the second step.
If 5.00 mol CH\(_4\) reacts, what is the total amount of HCl produced? Assume that Cl\(_2\) and HF are present in excess.
\[ \text{moles HCl:} \quad \underline{\phantom{xxxxxxxx}}\quad \text{mol} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2659b4c-30a6-48d4-90ed-e5c7a02f7b8e%2Fd986d9ce-106a-4eee-8439-fa4a18d6200e%2Fdkjn4mo_processed.png&w=3840&q=75)
Transcribed Image Text:Each step in the following process has a yield of 60.0%.
\[ \text{CH}_4 + 4 \text{Cl}_2 \rightarrow \text{CCl}_4 + 4 \text{HCl} \]
\[ \text{CCl}_4 + 2 \text{HF} \rightarrow \text{CCl}_2\text{F}_2 + 2 \text{HCl} \]
The CCl\(_4\), formed in the first step, is used as a reactant in the second step.
If 5.00 mol CH\(_4\) reacts, what is the total amount of HCl produced? Assume that Cl\(_2\) and HF are present in excess.
\[ \text{moles HCl:} \quad \underline{\phantom{xxxxxxxx}}\quad \text{mol} \]
Expert Solution
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Step 1
Given reaction :
CH4 + 4Cl2 ------> CCl4 + 4HCl ----(step 1)
CCl4 + 2HF -----> CCl2F2 + 2HCl --(step 2)
Adding step 1 and step 2, we get
CH4 + 4Cl2 + 2HF -----> CCl2F2 + 6HCl
This is the required overall chemical equation.
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This answer isn't correct. It would be correct if each step had a 100% yield.
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