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Hi Mam (This is Mahee)!

Attached is the question I made mistakes on. (The second screenshot is the question) Screenshot 2026-02-22 213715.pngScreenshot 2026-02-22 213706.png 

R    CH3OH(g)+      HCl(g)⇌       CH3Cl(g)+  H2O(g)       

Initial  0.250 atm   0.600 atm             0   0
Change     -x   -x                  +x   +x
Equilibrium 0.250−x   0.600−x                  x   x


   Kp=P_CH3OHP_HCl /P_CH3ClP_H2O

x2 /(0.250−x) (0.600−x) =4.7×103
        x=0.249962 atm
P(HCl,eq)=0.600-x=0.600-0.249962=0.350"atm"

You can do this question without RICE table also.
Since K_pis very large, reaction goes almost to completion (methanol is limiting), so x≈0.250.
Here temperature and volume are same for both gases so pressures are directly proportional to moles.
HCl is in excess left unused.
Finally, methanol is zero, and convert pressure of methanol to HCl pressure used taking them as mole ratio.
0.250 atm of methanol reacts with 0.250 atm of HCl .Therefore,
final partial pressure of HCl left: 0.600- 0.250 = 0.350 atm