Given the following values for the heats of formation, what is the number of moles of ethane (C2H6, MW 30.0) required to produce 1.00x10^3 kJ of heat by combustion to CO2 (g) and H2O (l). C2H6 (g) + 7/2O2(g) → 2CO2 (g) + 3 H2O (l)

C2H6 (g) ΔHf° = -84.7 kJ/mol
CO2 (g) ΔHf° = -393.5 kJ/mol
H2O (l) ΔHf° = -285.8 kJ/mol

a. 0.595
b. 1.56
c. 0.641
d. can't be determined without ΔHfo for O2
e. 1.68

0.641 moles of ethane

Explanation:

Based on the equation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)

We can determine ΔH of reaction using Hess's law. For this equation:

Hess's law: ΔH products - ΔH reactants

ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}

Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol

ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}

ΔH = -1559.7kJ/mol

That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:

1.00x10³kJ * (1mole ethane / 1559.7kJ) =

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.