Hello,

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 P_{𝑓𝑖𝑛𝑎𝑙} : P_{i𝑛𝑖𝑡𝑖𝑎𝑙}.

First, I said that P_{𝑓𝑖𝑛𝑎𝑙} : P_{i𝑛𝑖𝑡𝑖𝑎𝑙} = (n_{𝑓𝑖𝑛𝑎𝑙}/V_{𝑓𝑖𝑛𝑎𝑙}) : (n_{i𝑛𝑖𝑡𝑖𝑎𝑙}/V_{i𝑛𝑖𝑡𝑖𝑎𝑙}), 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, n_{NH3} = (5 * P)/(R * T), n_{Cl2} = (5 * P)/(R * T), etc. For V, 5.0 L of Cl2 is given, while V_{NH3} = 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,

James