Answer:
503.5 kJ
Explanation:
The combustion of reaction of propane, C3H8, is:
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l)
The enthalpy of the reaction can be calculated by the enthalpy of formation of the substances, which is the enthalpy of the reaction that produces the substances by only its constituents. The values can be found at a thermodynamic table. The standard condition is 25°C and 1 atm, so:
H°f, C3H8(g) = -103.85 kJ/mol
H°f, O2(g) = 0
H°f, CO2(g) = -393.51 kJ/mol
H°f, H2O(l) = -285.83 kJ/mol
So, the enthalpy of the reaction is:
ΔH°rxn = ∑n*H°f products - ∑n*H°f reactants, where n is the coefficient of the substance:
ΔH°rxn = [3*(-393.51) + 4*(-285.83)] - (-103.85)
ΔH°rxn = -2220 kJ/ mol of C3H8
The heat produced is the value of the enthalpy multiplied by the number of moles of the fuel. The molar mass of C3H8 is 44.10 g/mol, so, in 10.0 g:
n = 10.0/44.10
n = 0.2268 mol
So, the heat is:
Q = -2220 * 0.2268
Q = -503.5 kJ
The minus signal indicates that the heat is being lost, so 503.5 kJ of heat is produced.
