Question #8b019

1 Answer
Dec 9, 2016

Answer:

Here's what I got.

Explanation:

Start by identifying the reactants and the products. On the reactants' side, you have

  • ethane, #"C"_ 2"H"_ (6(g))#
  • oxygen gas, #"O"_ (2(g))#

On the products' side, you have

  • carbon dioxide, #"CO"_ (2(g))#
  • water vapor, #"H"_ 2"O"_ ((g))#
  • the ehnthalpy change of combustion, #DeltaH_"comb" = "342 kcal mol"^(-1)#

Notice that heat is added to the products' side because it is being evolved, i.e. it is being given off by the reaction. Also, keep in mind that you get #"342 kcal"# of heat given off when #1# mole of ethane undergoes combustion.

Start by writing the unbalanced chemical equation

#"C" _ 2"H"_ (6(g)) + "O"_ (2(g)) -> "CO"_ (2(g)) + "H"_ 2"O"_ ((g)) + "342 kcal"#

To balance the atoms of carbon, multiply the carbon dioxide by #2#

#"C" _ 2"H"_ (6(g)) + "O"_ (2(g)) -> 2"CO"_ (2(g)) + "H"_ 2"O"_ ((g)) + "342 kcal"#

To balance the atoms of hydrogen, multiply the water by #3#

#"C" _ 2"H"_ (6(g)) + "O"_ (2(g)) -> 2"CO"_ (2(g)) + 3"H"_ 2"O"_ ((g)) + "342 kcal"#

To balance the atoms of oxygen, multiply the oxygen molecule by #7/2#

#"C" _ 2"H"_ (6(g)) + 7/2"O"_ (2(g)) -> 2"CO"_ (2(g)) + 3"H"_ 2"O"_ ((g)) + "342 kcal"#

Now, in order to get rid of the fractional coefficient, multiply all the chemical species and the enthalpy change by #2#

#2"C" _ 2"H"_ (6(g)) + 7"O"_ (2(g)) -> 4"CO"_ (2(g)) + 6"H"_ 2"O"_ ((g)) + "684 kcal"#

Notice that in this version of the chemical equation, #2# moles of ethane undergo combustion, which means that the reaction gives off twice as much heat as it gives off when #1# mole undergoes combustion.