Question

Question #8b019

Answer 1

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.