Instances to produce heat by burning

  • Thread starter Thread starter jennypear
  • Start date Start date
  • Tags Tags
    Heat
AI Thread Summary
To produce 213.0 kJ of heat from the combustion of propane (C3H8), the standard enthalpy of reaction is -2,044 kJ, which corresponds to the reaction of 5 moles of oxygen. The calculation shows that 408.8 kJ of heat is produced per mole of oxygen. Dividing the desired heat output (213 kJ) by the heat produced per mole of oxygen yields approximately 0.520 moles of O2 needed. Multiplying this by the molar mass of oxygen (32 g/mol) results in a requirement of about 16.7 grams of oxygen for the reaction. The calculations confirm the accuracy of this result.
jennypear
Messages
16
Reaction score
0
Propane, C3H8(g), is used in many instances to produce heat by burning:

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)

The standard enthalpy of reaction, , is -2,044 kJ. How many grams of oxygen would be needed to react with excess propane to produce 213.0 kJ of heat?


heat of formation of CO2 and h2o were given previously
so i started by finding the heat of formation of c3h8
hrx=((sumHproduct) - (sum of Hreact))
-2044kJ=[3mol*-394KJ/mol)+(4mol*-286kJ/mol)]-x
x=-282kJ

not sure how to proceed from here?
 
Chemistry news on Phys.org
jennypear said:
Propane, C3H8(g), is used in many instances to produce heat by burning:

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)

The standard enthalpy of reaction, , is -2,044 kJ. How many grams of oxygen would be needed to react with excess propane to produce 213.0 kJ of heat?

I think: the standard enthalpy of reaction is the enthalpy change per mole of c3h8 and per 5 moles of oxygen.

2044/5 = 408.8= per 1 mole of oxygen

so 213/408.8 moles gives 213 kj.

mass = moles * RMM

so mass = 213/408.8 * 32
=16.7 grams.

I hope this is right... (Which I seriously doubt)
 
Just to say: The standard heat of formation (DH°f) for O2(g) is zero.

Yes your calculations look fine :-)
 
Back
Top