Question #8b019

Dec 9, 2016

Here's what I got.

Explanation:

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

• ethane, ${\text{C"_ 2"H}}_{6 \left(g\right)}$
• oxygen gas, ${\text{O}}_{2 \left(g\right)}$

On the products' side, you have

• carbon dioxide, ${\text{CO}}_{2 \left(g\right)}$
• water vapor, ${\text{H"_ 2"O}}_{\left(g\right)}$
• the ehnthalpy change of combustion, $\Delta {H}_{\text{comb" = "342 kcal mol}}^{- 1}$

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 $\text{342 kcal}$ of heat given off when $1$ mole of ethane undergoes combustion.

Start by writing the unbalanced chemical equation

$\text{C" _ 2"H"_ (6(g)) + "O"_ (2(g)) -> "CO"_ (2(g)) + "H"_ 2"O"_ ((g)) + "342 kcal}$

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

$\text{C" _ 2"H"_ (6(g)) + "O"_ (2(g)) -> 2"CO"_ (2(g)) + "H"_ 2"O"_ ((g)) + "342 kcal}$

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

$\text{C" _ 2"H"_ (6(g)) + "O"_ (2(g)) -> 2"CO"_ (2(g)) + 3"H"_ 2"O"_ ((g)) + "342 kcal}$

To balance the atoms of oxygen, multiply the oxygen molecule by $\frac{7}{2}$

$\text{C" _ 2"H"_ (6(g)) + 7/2"O"_ (2(g)) -> 2"CO"_ (2(g)) + 3"H"_ 2"O"_ ((g)) + "342 kcal}$

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

$2 \text{C" _ 2"H"_ (6(g)) + 7"O"_ (2(g)) -> 4"CO"_ (2(g)) + 6"H"_ 2"O"_ ((g)) + "684 kcal}$

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.