Question

Consider the combustion of butane:

C4H10(g) +O2(g)->CO2(g)+H2O(l) (unbalanced)

Let 2.00 g of butane gas and 3.05 atm of oxygen gas completely react (limiting) in a 2.00-L container at 125C.

2.1 Will there be a net increase or decrease in total pressure after the reaction is complete?

2.2 After complete reaction, what will be the mole fraction of H2O gas in the system?

Answer 1

Solution

Given data

Mass of butane = 2.0 g

Volume = 2.00 L

Pressure = 3.05 atm

Temperature = 125 C +273 = 398 K

Lets first write the balanced reaction equation

2C4H10(g) +13O2(g) ----- > 8CO2(g) + 10H2O(g)

Now lets first calculate the total moles of gas in the 2.00 L container at initial pressure and temperature using the ideal gas law formula

PV=nRT

Where , P =pressure , V= volume , n= moles , R = 0.08206 L atm per K mol Temperature

Now lets put the values in the formula

3.05 atm * 2.00 L = n *0.08206 L atm per K mol * 398 K

(3.05 atm * 2.00 L)/( 0.08206 L atm per K mol * 398 K) = n

0.1868 mol = n

So total moles of the gas before reaction = 0.1868 mol

Now lets calculate moles of butane using its mass

Moles of butane = mass / molar mass

                              = 2.00 g / 58.12 g per mol

                              = 0.034412 mol C4H10

Now lets calculate initial moles of O2

Moles of O2 = total moles – moles of butane

                      = 0.1868 mol – 0.034412 mol

                     = 0.15239 mol O2

Now lets find how many moles of the butane are needed to react with 0.15239 mol O2

Using the mole ratio of the butane and O2 from the balanced reaction equation.

Mole ratio of the butane and O2 is 2 : 13

0.15239 mol O2 * 2 mol C4H10 / 13 mol O2 = 0.023445 mol C4H10

So the C4H10 is the excess reagent because moles of C4H10 are more than needed to react with 0.15239 mol O2

So lets find excess moles of C4H10

Excess moles of C4H10 = 0.034412 mol – 0.023445 mol =0.010967 mol C4H10 remain unreacted

Since O2 is the limiting reactant therefore calculate the moles of the COI2 and H2O using the moles of O2

Mole ratio of the O2 and CO2 is 13 : 8

0.15239 mol O2 * 8 mol CO2 / 13 mol O2 = 0.09378 mol CO2

0.15239 mol O2 * 10 mol H2O / 13 mol O2 = 0.11722 mol H2O

Now lets find the total moles present in the container after the reaction

Total moles = 0.010967 mol C4H10 + 0.09378 mol CO2 +0.11722 mol H2O = 0.222 mol

So before reaction we had 0.1868 mol gas in the container

And after reaction we have 0.222 mol gases in the container

2.1)Since moles of gas increases therefore the net increase in pressure will observe after the reaction is complete.

2.2) mole fraction of the H2O after the completion of reaction is as follows.

Mole fraction of the H2O = moles of H2O / total moles of gas

                                                     = 0.11722 mol / 0.222 mol

                                                     = 0.528