Question

Propane C3H8 is a hydrocarbon tht is commonly used as a fuel

Answer 1

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)

$25g C3H8*\frac{1mol C3H8}{44gC3H8} *\frac{5molO2}{1molC3H8}* \frac{32gO2}{1mol O2}*\frac{1L air}{0.275g}$

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) $Hof C3H8= -(-285.8\frac{KJ}{mol}*4mol +-393.5*3mol$

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:

$25g\frac{2219.2KJ}{44g C3H8}$ = 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