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.

The final pressure is 2.0 atm

Explanation:

Step 1: Data given

Number of moles of methanol = 4 moles

Number of moles of oxygen = 3 moles

Volume and temperature are fixed

Original pressure = 1.0 atm

Step 2: The balanced equation

2 CH3OH + 3 O2 → 2 CO2 + 4 H2O  

For 2 moles of CH3OH consumed, we need 3 moles of O2 to produce 2 moles of CO2 and 4 moles of H2O

This means the mol ratio methanol to oxygen = 2:3

Step 3 : Calculate the fina lpressure

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

Number of moles O2 = 3 mol

This means the total number of moles for the reactants = 3 moles

The total number of moles  for the products = 2 + 4 = 6 moles of gas

Since V and T are fixed (and R is constant)

⇒ we can write the gas the gas law as following:

P2/P1 = n2/n1   OR P2 = (n2*P1)/n1

P2 = (6 mol * 1.0 atm ) / 3.0 mol

P2 = 2.0 atm

The final pressure is 2.0 atm

i do not get it   at all