Question

A 16.0 gram sample of propane gas (C³H⁸) is burned according to the equation C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g).

In an enclosed container with a volume of 2.25L at a temperature of 322K.

If the sample of propane burns completely and no oxygen remains in the container, calculate the mole fractions of CO₂ and H₂O.

Calculate the total pressure in the container after the reaction.

Calculate the partial pressure of CO₂ and H₂O in the container after the reaction.

If 12.1 grams of propane is burned in the presence of 10.2 atm of oxygen at the above temperature and volume conditions, determine the theoretical yield of H₂O.

Answer 1

Number of moles of propane , n = mass/molar mass

                                               = 16.0 g / 44 (g/mol)

                                               = 0.36 mol

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

From the balanced equation,

1 mole of propane produces 3 moles of CO₂ & 4 moles of H₂O

0.36 mole of propane produces 3x0.36=1.08 moles of CO₂ & 4x0.36=1.44 moles of H₂O

So total num ber of moles , N = 1.08 + 1.44 = 2.52 mole

So mole fraction of CO₂ is X _CO2 = number of moles of CO₂ / total number of moles

                                                    = 1.08 / 2.52

                                                    = 0.43

mole fraction of H₂ O is X _H2O = number of moles of H₂ O / total number of moles

                                                    = 1.44 / 2.52

                                                    = 0.57

Calculation of pressure of CO₂:

We know that PV = nRT

Where

T = Temperature = 322 K

P = pressure = ? atm

n = No . of moles = 1.08 mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume = 2.25 L

Plug the values we get p = (nRT) / V

                                     = 12.7 atm

Calculation of pressure of H₂O:

We know that PV = nRT

Where

T = Temperature = 322 K

P = pressure = ? atm

n = No . of moles = 1.44 mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume = 2.25 L

Plug the values we get p' = (nRT) / V

                                     = 16.9 atm

So total pressure = p CO2 + p H2O = 12.7 + 16.9 = 29.6 atm

Partial pressure of CO2 = mole fraction of CO2 x total pressure

                                  = 0.43 x 29.6 atm

                                  = 12.7 atm

Simillarly partial pressure of H2O = 16.9 atm

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

From the balanced equation,

Number of moles of O₂ , n = (PV)/(RT)

                                      = (10.2atm x 2.25 L ) / (0.0821Latm/(mol-K) x 322K)

                                      = 0.87 moles

Number of moles of propane , n = mass/molar mass

                                              = 12.1 g / 44(g/mol)

                                              = 0.275 moles

1 mole of propane reacts with 5 mole of oxygen

N mole of propane reacts with 0.87 mole of oxygen

N = (1x0.87) / 5

   = 0.174 mol

So (0.275 - 0.174 ) moles of propane left unreacted

Since all the mass of oxygen completly reacted it is the limiting reactant.

5 moles O₂(g)produces 4 moles of H₂O(g)

0.87 moles O₂(g)produces Y moles of H₂O(g)

Y = (0.87x4) / 5

   = 0.696 moles

So mass of H2O , m = number of moles x molar mass

                               = 0.696 mol x 18 (g/mol)

                               = 12.53 g