[johnnyjohn993]

2LiOH + CO₂  Li₂CO₃ + H₂O
The pressure of carbon dioxide inside the cabin of a submarine having a volume of 2.4x10⁵ L is
7.9x10^-3 atm at 312K. A solution of LiOH of negligible volume is indtroduced into the cabin. Eventually the pressure of CO₂ falls to 1.2x10^-4 atm. How many grams of Li₂CO₃ are formed by this process?

My attempt :
 PV=nRT

n= PV/RT
n=           (7.9x10^-3 atm)(2.4x10⁵ L)
    __________________________________________
       (0.08206 atm L/K mol ) (312 K)


n= 74.05 mol CO₂

NOw i dont know what to do next PLease help. I dont know how to get the value of moles of CO₂ at 1.2x10^-4 atm.

[sjb]

2LiOH + CO₂  Li₂CO₃ + H₂O
The pressure of carbon dioxide inside the cabin of a submarine having a volume of 2.4x10⁵ L is
7.9x10^-3 atm at 312K. A solution of LiOH of negligible volume is indtroduced into the cabin. Eventually the pressure of CO₂ falls to 1.2x10^-4 atm. How many grams of Li₂CO₃ are formed by this process?

My attempt :
 PV=nRT

n= PV/RT
n=           (7.9x10^-3 atm)(2.4x10⁵ L)
    __________________________________________
       (0.08206 atm L/K mol ) (312 K)


n= 74.05 mol CO₂

NOw i dont know what to do next PLease help. I dont know how to get the value of moles of CO₂ at 1.2x10^-4 atm.

The same way you calculated the original number of moles.