Question #8b019
1 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
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
#"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
#"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
#"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"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,