*What is the total pressure inside the container after the reaction?*

*A 2.00L canister is charged with C3H8 and O2 to a total pressure of 1.50 atm at 0.0 degrees C The mole fraction of C3H8 in the canister is 0.25. Assuming the canister is rigid enough to withstand the pressure change, what is the total pressure inside the canister after the reaction takes place according to the equation below. The temperature inside the canister after the reaction was complete is 150.0 degrees C. equation is:*

*C3H8(g)+5O2(g)=3 CO2(g)+4H2O(g)*

**Hello MS S,**

2L container at P = 1.5 atm, T = 0C or 273K

0.25 mol fraction implies the propane C_{3}H_{8} has the fraction making up the pressure in the container that is

Partial pressure says the total pressure is the sum of the pressures from the mol fractions:

P_{T} = P_{prop} + P_{ O2}

Propane pressure is:

.25 * 1.5 atm = 0.375 atm C_{3}H_{8 }

Solve for n, number of moles using PV = nRT

n = PV/RT

n = (0.375atm)(2L)/(0.0821_{Latm/molK})(273K)

n = 0.0335 mol C_{3}H_{8}

Oxygen pressure is_{:}

1.5 atm - 0.375 atm = 1.125 atm Oxygen O_{2 }

inside a 2 Liter container:

Solve for n, number of moles using PV = nRT

n = PV/RT

n = (1.125atm)(2L)/(0.0821_{Latm/molK})(273K)

n = 0.1004 mol O_{2}

We need 5 mol_{ }O_{2 }for every mole of Propane, so,

0.0335 mol C_{3}H_{8 }* 5 mol O_{2} / 1 mol C_{3}H_{8} = 0.1675 mol O_{2 }needed

O_{2} is limiting:

0.1004 mol O_{2 }* 1mol C_{3}H_{8 }/ 5 mol O_{2} = 0.0201 mol C_{3}H_{8 }consumed in the reaction

0.0335 - 0.0201 = 0.0134 mol C_{3}H_{8 }remaining

Also,

0.1004 mol O_{2 }* 3 mol CO_{2 }/ 5 mol O_{2} = 0.0602 mol CO_{2 }produced in the reaction

and,

0.1004 mol O_{2 }* 4 mol H_{2}O_{ }/ 5 mol O_{2} = 0.0803 mol CO_{2 }produced in the reaction

Total pressure depends on the total moles of gases occupying the container after the reaction has finished.

n_{T} = n_{prop} + n_{O2} + n_{CO2} + n_{H2O}

n_{T} = 0.0134 mol C_{3}H_{8} + 0 mol O_{2 }+ 0.0602 mol CO_{2 }+ 0.0803 mol CO_{2}

n_{T }= 0.1539 mol

at T = 150C or 423K in 2L container solve for P:

PV = nRT

P = (0.1539 mol)(0.0821_{Latm/molK})(423K)/(2L)

**P = 2.67 atm**