Answer:C3H8 +5O2-- >3CO2+4H2O, 330L of air,Hof C3H8= 2323.7 KJ/mol,dT=75.30
Explanation:
First We need to find a balanced equation to depict the combustion of the propane:
C3H8 +5O2-- >3CO2+4H2O
b)
330L of air
25 g C3H8 are equal to 44g and 5 mol of O2 will react with a mol of C3H8, each mol of O2 weigh 32g and for each L of air contains 0.275g of O2
c)
the enthalpy of formation of propane will be the negative of the sum of the enthalpy of formation multiplied by the mols of the H2O and CO2
Hof C3H8= 2323.7 KJ/mol
d) We know that each 44g of C3H8(1mol) transfer 2219.2 KJ then:
= 1260.9KJ
We use the equation of heat transfered
Q=mCpdT
1260.9KJ= 4KJ *4.186KJ/Kg*°C *dT
We clear the equation
Q/mCp=dT
dT=75.30