Griffiths Schmidt
07/31/2024 · Primary School
Each step in the following process has a yield of \( 70.0 \% \). If \( 8.50 \mathrm{~mol} \mathrm{CH}_{4} \) reacts, what is the total amount of HCl produced? Assume that \( \mathrm{Cl}_{2} \) and HF are present in excess. moles \( \mathrm{HCl}: \)
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Step-by-step Solution
To find the total amount of HCl produced, we need to determine the number of moles of HCl produced in the reaction.
Given:
- Yield of each step: \( 70.0\% \)
- Moles of CH\(_4\): \( 8.50 \, \text{mol} \)
The reaction is:
\[ \text{CH}_4 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{HCl} \]
Since Cl\(_2\) and HF are present in excess, the limiting reagent is CH\(_4\).
Let's calculate the moles of HCl produced.
Calculate the value by following steps:
- step0: Calculate:
\(8.5\times 0.7\)
- step1: Multiply the numbers:
\(5.95\)
The total amount of HCl produced is \( 5.95 \, \text{mol} \).
Quick Answer
The total amount of HCl produced is \( 5.95 \, \text{mol} \).
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