Let 4 moles of methanol (liquid) combust in 3 moles of gaseous oxygen to form gaseous carbon dioxide and water vapor. suppose this occurs in a chamber of fixed volume and fixed temperature. if the original pressure is 1.0 atm, what is the final pressure in the chamber. express your answer in atm. enter a numerical value, do not enter units. assume liquids take up negligible volume.

The final pressure is 2.0 atm

Explanation:

Step 1: Data given

Number of moles methanol = 4 mol

Number of moles oxygen = 3 mol

original pressure is 1.0 atm

Step 2: The balanced equation

2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l)

Step 3: Calculate final pressure

For 2 moles of methanol consumed, we need 3 moles O2 to produce 2 moles CO2 and 4 moles H2O

We started with 3 moles of O2 gas.  

Since methanol is not a gas, it doesn't count for the pressure.

V and T are fixed

This means the final pressure can be given by:

P2/P1 = n2/n1  

 with n2 = number moles of products

with n1 = number of moles of reactants

P2 = (6.0 mol*1.0 atm) / (3.0 mol )

P2 = Final pressure = 2.0 atm

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .