Consider the combustion of propane: C3H3(g)+5O2(g)->3Co2+6H2O(g) A) Balance the
ID: 894109 • Letter: C
Question
Consider the combustion of propane:
C3H3(g)+5O2(g)->3Co2+6H2O(g)
A) Balance the reaction.
B) Divide all coefficients by the coefficient on propane, so that you have the reaction for the combustion of 1 mole of propane.
C) TraingleHrxn for the combustion of one mole of propane is -2219 kJ. What mass of propane would you need to burn to generate 5.0 MJ of heat?
D) If propane costs about $0.64/L and ahs a density of 2.01 g/cm3 how much would it cost to generate 5.0 MJ of heat by burning propane?
Explanation / Answer
C3H3+ 5O2---> 3CO2+ 4H2O
1 mole of propane produces 2219 KJ
for producing 5*1000 KJ propane required =5000/2219 moles of propane=3.253 moles
mass of propane =moles* molecular weight = 3.253*39 =126.86 gms
density of propane = 126.86 gm
density = 2.01 g/cc
volume = 126.86/2.01 cc =63.11 cc=63.11/1000 l =0.06311 liters
cost per l =0.64dollars
0.063411 liters = 0.06311*0.64= 0.04039 dollars