In a 2-step process to obtain HCl, if we begin with 2.00 moles of CHâ‚„ and the yield of each step is 70.0%, the amount of HCl produced is 7.56 mol.
Let's consider the first step of the process.
CH₄ + 4 Cl₂ ⟶ CCl₄ + 4 HCl
The molar ratio of CHâ‚„ to HCl is 1:4. Considering the percent yield is 70.0%, the moles of HCl obtained from 2.00 moles of CHâ‚„ are:
[tex]2.00 molCH_4 \times \frac{4molHCl}{1molCH_4} \times 70.0\% = 5.60 mol HCl[/tex]
The molar ratio of CHâ‚„ to CClâ‚„ is 1:1. Considering the percent yield is 70.0%, the moles of CClâ‚„ obtained from 2.00 moles of CHâ‚„ are:
[tex]2.00 molCH_4 \times \frac{1molCCl_4}{1molCH_4} \times 70.0\% = 1.40 mol CCl_4[/tex]
1.40 moles of CClâ‚„ react in the second step.
CCl₄ + 2 HF ⟶ CCl₂F₂ + 2 HCl
The molar ratio of CClâ‚„ to HCl is 1:2. Considering the percent yield is 70.0%, the moles of HCl obtained from 1.40 moles of CClâ‚„ are:
[tex]1.40 molCCl_4 \times \frac{2molHCl}{1molCCl_4} \times 70.0\% = 1.96 mol HCl[/tex]
5.60 moles of HCl were obtained in the first step and 1.96 moles in the second step. The total amount of HCl obtained is:
[tex]5.60 mol + 1.96 mol = 7.56 mol[/tex]
In a 2-step process to obtain HCl, if we begin with 2.00 moles of CHâ‚„ and the yield of each step is 70.0%, the amount of HCl produced is 7.56 mol.
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