The answer is #3.5cents#.

The balanced equation for propane's combustion is

#C_3H_(8(g)) + 5O_(2(g)) -> 3CO_(2(g)) + 4H_2O_((l))#

In order to determine the heat given off by this reaction, we need to find the value of the enthalpy of combustion, #DeltaH#; since this is an exothermic reaction, we would expect this value to be negative.

Since it was not given to you, we'll calculate it by using the standard enthalpies of formation (#DeltaH_f^@# - these values are usually given). So,

#H_2O#: #DeltaH_f^@ = -258.8 (kJ)/(mol)#;

#CO_2#: #DeltaH_f^@ = -393.5 (kJ)/(mol)#;

#O_2#: #DeltaH_f^@ = 0 (kJ)/(mol)#;

#C_3H_8#: #DeltaH_f^@ = -103.9 (kJ)/(mol)#.

#DeltaH = (4 mol es) * (-258.8 (kJ)/(mol)) + (3 mol es) * (-393.5 (kJ)/(mol)) - (1 mol e) * (-103.9 (kJ)/(mol)) = -2112 kJ = -2112 * 10^(-3)MJ = 2.1 MJ#

The heat given off is #q = - DeltaH = 2.1 MJ#

This is the heat generated when **1 mole** of propane is used; since we need to generate more heat, #4.8 MJ# to be exact, more propane needs to be used. The ratio of the two value for heat will give us the ratio for moles:

#(4.8MJ)/(2.1MJ) = 2.3 ->#we need 2.3 times more moles of everything to produce #4.8MJ#, which means that the number of propane moles is

#n_(C_3H_8) = 2.3 * 1 mol e= 2.3# moles

We know that propane's molar mass is #44.0 g/(mol)#, which means that the mass of propane needed is

#m_(C_3H_9) = 2.3 mol es * 44.0 g/(mol) = 101g#

Since we know its density, we can find its volume

#V = m/(rho) = (101g)/(2.01g/(cm^3)) = 50.3 cm^3 = 50.3 * 10^(-3)L= 0.0503L#

Therefore, the cost of generating that much heat is

#co st = 0.0503L * (0.69 $)/L = 0.035 $ = 3.5 cents#