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9.45 g of liquid hexane (C6H14) is introduced into a 10.0 L vessel containing 13.15 atm of oxygen gas at 21ºC and ignited, yielding carbon dioxide and water. If the vessel is then cooled to –10ºC, what will be the gas pressure inside the vessel?

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mickymike92

The pressure of the gas in the vessel will be 10.92 atm

Gas pressure, volume, number of moles, temperature and Ideal has equation

The Ideal gas equation gives the relationship between gas pressure, volume, number of moles and temperature.

The Ideal gas equation is given below:

  • PV = nRT

where

  • P is pressure
  • V is volume
  • n is number of moles
  • R is molar gas constant
  • T is temperature

From data provided

P = 13.15 atm; V = 10.0 L; T = 21°C = 294 K

R = 0.082?057 L.atm/K.mol

making n subject of formula to find initial number of moles of oxygen gas

n = PV/RT

n = 13.15 x 10/0.082?057 x 294

n = 5.45 moles oxygen at start

Combustion of of hexane

The equation for the combustion of hexane is given below:

2 C6H14+ 19 O2 --> 12 CO2 + 14 H2O

Number of moles of hexane in 9.45 g

Number of moles = mass/molar mass

molar mass of hexane = 86 g

Number of moles of Hexane = 9.45/86

Number of moles of hexane = 0.11 moles

Moles of oxygen left

2 moles of hexane reacts with  19 moles of oxygen

0.11 moles of hexane will react with 0.11 × 19/2 moles of oxygen

 1.045 mole O2 will be used up

Moles of oxygen left = 5.45 - 1.045

Moles of oxygen left = 4.405 moles

Moles of CO2 produced

2 moles of hexane produces 12 molesof CO2

0.11 moles of hexane produces 0.11 × 12/2

0.11 moles of hexane produces 0.66 mole CO2

Total moles of gas = 4.406 + 0.66 = 5.065 mole of gas in a 10 L container at -10°C

Pressure of gas in vessel

From ideal gas equation;

P = nRT/V

P = 5.065 × 0.082× 263/10

P = 10.92 atm

Therefore, the pressure of the gas in the vessel will be 10.92 atm

Learn more about ideal gas equation and gas pressure at: https://brainly.com/question/12873752

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