0
$\begingroup$

Below you can see the chemical reaction I'm working with:

$$\ce{CH_3CH_2OH(l) + 3O_2(g) -> 2CO_2 (g) + 3H_2O(g)}$$

I've got so far: https://i.sstatic.net/3zgNk.png

How can I calculate the volume of the product of carbon dioxide gas (the pressure is 1 bar, and the temperature is 20 deg Celsius).

$\endgroup$

1 Answer 1

1
$\begingroup$
  1. You need to determine the limiting reactant. Use the equation and the number of moles of $\ce{CH_3CH_2OH}$ and $\ce{O_2}$.

    In the reaction, you need 1 mol of $\ce{CH_3CH_2OH}$ and 3 mol of $\ce{O_2}$ to get 2 mol of $\ce{CO_2}$. If you have less than 1 mol of $\ce{CH_3CH_2OH}$ but 3 mol of $\ce{O_2}$, you cannot get 2 mol of $\ce{CO_2}$. In such a case, $\ce{CH_3CH_2OH}$ is the limiting reactant. The amount of $\ce{CO_2}$ will thus be lower than 2 mol.

  2. Using the number of moles of the limiting reactant, you should be able to calculate the number of moles of $\ce{CO_2}$ produced. Use proportions!

  3. The last step will involve using the ideal gas equation, by assuming that $\ce{CO_2}$ behaves like an ideal gas. $$PV=nRT$$

    You have the pressure $P$, the temperature $T$, $n$ (the number of moles of the $\ce{CO_2}$), $R$ is a constant. Find the volume $V$.

$\endgroup$
5
  • $\begingroup$ If you'd like to take the time, i would like to know what the exact answer to my question is, i mean - the result. $\endgroup$ Commented May 15, 2013 at 18:32
  • 3
    $\begingroup$ @FrederikWitte You're not trying to use the site to get your homework done, are you? $\endgroup$
    – Jerry
    Commented May 15, 2013 at 18:39
  • $\begingroup$ Just a side note: Pressure is usually referred to with a small $p$. $\endgroup$ Commented Apr 2, 2014 at 2:29
  • $\begingroup$ @Martin It can be represented by both. $\endgroup$
    – Jerry
    Commented Apr 2, 2014 at 4:57
  • $\begingroup$ @Jerry, you are right. IUPAC goldbook suggests a small $p$. $\endgroup$ Commented Apr 2, 2014 at 8:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.