October 01, 2020, 03:02:59 AM
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Topic: Gas Law/Stoichiometry, explanation to an answer  (Read 227 times)

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Offline Jamestp0118

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Gas Law/Stoichiometry, explanation to an answer
« on: July 20, 2020, 02:05:15 AM »

This was the question:

Given reaction  2NH3(𝑔)+3Cl2(𝑔)⟶N2(𝑔)+6HCl(𝑔) , you react 5.0 L of  NH3  with 5.0 L of  Cl2  measured at the same conditions in a closed container. Calculate the ratio of pressures in the container  Pğ‘“ğ‘–ğ‘›ğ‘Žğ‘™ : Piğ‘›ğ‘–ğ‘¡ğ‘–ğ‘Žğ‘™.

First, I said that Pğ‘“ğ‘–ğ‘›ğ‘Žğ‘™ : Piğ‘›ğ‘–ğ‘¡ğ‘–ğ‘Žğ‘™ = (nğ‘“ğ‘–ğ‘›ğ‘Žğ‘™/Vğ‘“ğ‘–ğ‘›ğ‘Žğ‘™) : (niğ‘›ğ‘–ğ‘¡ğ‘–ğ‘Žğ‘™/Viğ‘›ğ‘–ğ‘¡ğ‘–ğ‘Žğ‘™), because P = nRT/V but the RT would cancel out since it is the same on both sides.

I wasn't sure how to find n, because P and T weren't given. So I thought for now, nNH3 = (5 * P)/(R * T), nCl2 = (5 * P)/(R * T), etc. For V, 5.0 L of Cl2 is given, while VNH3 = 10/3 L. (2/3 mol of NH3 is needed for every 1 mol of Cl2, and n is directly proportional to V ?). I used the same idea on for the final n's and V's.

But I felt like I was wrong...so the second idea I tried was just (1+6):(2+3), the ratio of mols in the equation. That definitely didn't work haha. I also tried figuring out the partial pressure of each component, but that didn't work either.

I'm most confused about how to figure out the number of mols and the volume on each side. I'm also confused if the initial volume would be 10 (5L NH3 + 5L of Cl2), or 25/3 (10/3 L NH3 + 5L CL2).

The website I'm studying on only gives the answer, which is 4/3, without the explanation. Since I am trying to self-study, I would be very thankful if someone could explain the steps or the information we're looking for at each step to solve the problem!

Thank you,


Offline AWK

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Re: Gas Law/Stoichiometry, explanation to an answer
« Reply #1 on: July 20, 2020, 02:59:36 AM »
Start from the reaction stoichiometry. What happens with ammonia and an excess of HCl.

Offline Borek

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Re: Gas Law/Stoichiometry, explanation to an answer
« Reply #2 on: July 20, 2020, 03:05:50 AM »
Don't worry about volumes, the only information they give is that the initial mixture was equimolar, so there were 2n moles of gas initially (whatever the n was - it will cancel out, so the exact value doesn't matter). Then, as AWK wrote, it is just a matter of stoichiometry. How many moles of gases (in terms of n) were present after the reaction?
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