Propane, C 3 H 8 , is a hydrocarbon that is commonly used as a fuel for cooking.

Question

Propane, C3H8, is a hydrocarbon that is commonly used as a fuel for cooking.

a) Calculate the volume (in liters) of air at 27 degrees Celsius and 0.93 atmosphere that is needed to burn completely 8.42 grams of propane. Assume that air is 21.0 percent O2 by volume.

c) Assuming that all of the heat evolved in burning 8.42 grams of propane is transferred to 8.21 kilograms of water (specific heat = 4.184 J g-1 deg C-1) initially at 25.2 degrees Celsius, calculate the increase in temperature of the water.

Explanation / Answer

Solution :-

Propane, C3H8, is a hydrocarbon that is commonly used as a fuel for cooking.

a) Calculate the volume (in liters) of air at 27 degrees Celsius and 0.93 atmosphere that is needed to burn completely 8.42 grams of propane. Assume that air is 21.0 percent O2 by volume.

Solution :-

Balanced reaction equation

C3H8 + 5O2 ---- > 3CO2 +4H2O

27 C +273 = 300 K

P= 0.93 atm

Now lets calculate the moles of O2 needed to react with 8.42 g propane using the mole ratio of the propane and O2

(8.42 g C3H8 * 1 mol / 44.1 g)*(5 mol O2 / 1 mol C3H8) =0.95465 mol O2

Now using the moles of O2 lets calculate the volume of the O2

PV= nRT

V= nRT/P

   = 0.95465 mol * 0.08206 L atm per mol K* 300 K /0.93 atm

    = 25.3 L O2

Now using the percent of the O2 in the air lets calculate the volume of the Air

25.3 L O2 * 100 % / 21 % = 120.5 L Air

So the volume of the Air needed is 120.5 L

b) When 1 mol of propane is reacted with pure oxygen, O2, -2220 kJ of heat is released. Calculate the heat of formation, Delta Hof, of propane given that the delta Hof of H2O(l) is -285.3 kJ mol-1 and delta Hof of CO2(g) is -393.5 kJ mol-1. (Note: I expect you to calculate the deltaHof of propane from the data in this problem.)

Solution :-

C3H8 + 5O2 ---- > 3CO2 +4H2O

Delta H rxn = sum of delta H product - sum delta H reactant

-2220 kJ   = [(CO2*3)+(H2O*4)] – [C3H8*1]

-2220 kJ = [(-393.5*3)+(-285.3*4)] - [C3H8*1]

-2220 kJ = -2321.7 - [C3H8*1]

-2220 kJ + 2321.7 kJ = - [C3H8*1]

101.7 kJ = - [C3H8*1]

-101.7 kJ = [C3H8*1]

So the delta H of formation of the propane is -101.7 kJ/mol

c) Assuming that all of the heat evolved in burning 8.42 grams of propane is transferred to 8.21 kilograms of water (specific heat = 4.184 J g-1 deg C-1) initially at 25.2 degrees Celsius, calculate the increase in temperature of the water.

Solution :-

Lets calculate the amount of energy produced by 8.42 g propane

8.42 g propane * 2220 kJ / 44.1 g = 423.9 kJ

423.9 kJ * 1000 J / 1 kJ = 4.239*10^5 J

Now using this energy lets calculate the final temperature of the water

1 kg water = 1000 g

So 8.21 kg = 8210 g

q= m*c*delta T

delta T = q / m*c

             = 4.239*10^5 J / 8210 g * 4.184 J per g C

             = 12.34 C

So the final temperature of the water is = initial temperature + delta T

                                                                        = 25.2 C + 12.34 C

                                                                        = 37.5 C

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