Question

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.

Answer 1

Answer:

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