Consider the combustion of propane:
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(l) ΔH = –2221 kJ
Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.
The reaction of combustion of propane is
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(l) ΔH = –2221 kJ
The enthalpy of combustion of 1 mole of propane is -2221 kJ.
Number of moles of propane in 5 g of propane = 5/44.1 = 0.113 mol
Therefore, dH for 5 g (0.113 mol) of propane = –2221 x 0.113 kJ = - 251.81 kJ
Consider the combustion of propane: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(l) ΔH...
The combustion of propane, C3H8, occurs via the reaction C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) with heat of formation values given by the following table: Substance ΔH∘f (kJ/mol) C3H8 (g)= -104.7 CO2(g)= −393.5 H2O(g)= −241.8 Calculate the enthalpy for the combustion of 1 mole of propane.
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