The combustion reaction is as expressed,
CxHy + O2 --> CO2 + H2O
The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.
Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.
moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14
The empirical formula for the hydrocarbon is therefore, CH₂.
Answer:
A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:
C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol
The value given for ΔH∘rxnΔHrxn∘ means that:
a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.
b. the reaction is endothermic.
c. the enthalpy of formation of propane is 2202 kJ/mol.
d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.
e. None of these.
