Question

Pentane (C5H12), when it burns, generates heat according to the following:

​C5H12 + 8 O2 → 5 CO2 + 6 H2O + 3509 kJ of heat energy

a) The molar mass of pentane is 72.2 g / mole. The density of pentane is 0.626 g / mL.

Determine the number of moles of pentane in 1,250-mL of pentane (4 points)

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b) How much heat energy in kJ is generated by the combustion of the stated 1,250-mL of pentane? ​[4 points]

c) How many grams and moles of oxygen will be needed to totally consume the stated 1,250-mL of pentane? [4 points]

d) The oxygen source for this combustion is a tank with a pressure of 12.50 atm and a temperature of 300.0 K. All of the oxygen in the tank was used up. What is the volume in the tank? [6 points]

Note: The ideal gas constant has the usual value previously discussed in class.

Answer 1

C5H12 + 8 O2 → 5 CO2 + 6 H2O + 3509 kJ of heat energy

a) The molar mass of pentane is 72.2 g / mole.

The density of pentane is d = 0.626 g / mL.

Volume of pentane, V = 1,250-mL

Mass of pentane , m = volume x density

                               = 1250 mL x 0.626 g/mL

                                = 782.5 g

So number of moles , n = mass/molar mass

                                  = 782.5g/(72.2g/mole))

                                  = 10.84 moles

(b) 1 mole of pentane produces 3509 kJ

10.84 moles of pentane produces 10.84 mol x 3509 kJ /mol = 38030.4 kJ

(c) 1 mole of pentane reacts with 8 moles of O2

10.84 moles of pentane reacts with 8x10.84 moles = 86.72 moles of O2

So mass of O2 reacted , m = Number of moles x molar mass

                                        = 86.72 mol x 32 g/mol

                                        = 2775 g of oxygen

(d) We know that ideal gas equation is PV = nRT

Where

T = Temperature = 300 K

P = pressure = 12.50 atm

n = No . of moles = 86.72 mole

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

V= Volume of the gas = ?

Plug the values we get V = (nRT)/P = 170.9 L