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.

\[ \text{moles HCl:} \quad \underline{\phantom{xxxxxxxx}}\quad \text{mol} \]
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
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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