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. Enter only a numerical value, do not enter units. Assume liquids take up negligible volume.

Answer 1

Given:

Moles of methanol = 4.0 mol

Moles of O2 = 3.0 mol

Initial pressure = 1.0 atm

Solution:

We know initial pressure and T and V are fixed.

We find initial and final moles of gas and by using p1/n1 = p2 / n2, we can find final pressure.

Calculation of initial moles of gas

Initially there are 3 moles of O2   (moles of liquid methanol are neglected)

So initial moles in chamber n1= 3 mol

Initial pressure = 1.0 atm

Final moles calculation:

To get final moles after combustion we can use combustion reaction of methanol and O2

From this reaction we get mol ratio of CH₄ to O₂ , 1 : 2

Calculation of limiting reactant:

Here we find moles of O2 required for 4 mol and by using actual moles we find limiting reactant.

nO_2=8.0 mol O2

Actually there are only 3 moles of O₂ so O₂ is limiting reactant.

Lets calculate moles of H₂O and CO₂

nH₂O =3 mol

nCO₂ = 1.5 mol

So toal moles of gas in the chamber = nH₂O + nCO₂ = (3.0 +1.5) mol=4.5 mol

Lets find final pressure

P₂ =1.5 atm

So the final pressure in the chamber = 1.5 atm