Each step in the following process has a yield of 70.0%.
CH4 + 4Cl₂ CC14 + 4 HCI
CCl4 + 2 HF CC1₂F₂ + 2HCl
The CC formed in the first step is used as a reactant in the second step. 4
If 2.50 mol CH4 reacts, what is the total amount of HCl produced? Assume that Cl₂ and HF are present in excess.
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Answer:
CCl4 + 2HF ---> CCl2F2 + 2HCl
Explanation:
Each step in the process below has a 70.0% yield.
CH4 + 4Cl2 ---> CCl4 + 4HCL
CCl4 + 2HF ---> CCl2F2 + 2HCl
The CCl4 formed in the first step is used as a reactant in the second step. If 6.50 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.
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