To balance the combustion equation of ethane, we follow these steps:
1. Balance the carbon atoms first:
- We have 2 carbon atoms in [tex]\( C_2H_6 \)[/tex].
- Therefore, we need 2 molecules of [tex]\( CO_2 \)[/tex] on the product side.
Our equation now looks like:
[tex]\[
C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O
\][/tex]
2. Balance the hydrogen atoms:
- We have 6 hydrogen atoms in [tex]\( C_2H_6 \)[/tex].
- Therefore, we need 3 molecules of [tex]\( H_2O \)[/tex] on the product side, as each [tex]\( H_2O \)[/tex] contains 2 hydrogen atoms.
Our equation now looks like:
[tex]\[
C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O
\][/tex]
3. Balance the oxygen atoms:
- On the product side, we have [tex]\(4 \times (O \text{ in } CO_2) + 3 \times (O \text{ in } H_2O) = 4 \times 2 + 3 \times 1 = 8 + 3 = 11 \)[/tex] oxygen atoms.
- On the reactant side, each molecule of [tex]\( O_2 \)[/tex] provides 2 oxygen atoms. To get 11 oxygen atoms, we need [tex]\( \frac{11}{2} = 5.5 \)[/tex] molecules of [tex]\( O_2 \)[/tex] (which is not practical as we need whole molecules).
Thus, we double the number of all reactants and products:
[tex]\[
2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O
\][/tex]
4. Final check to ensure the equation is balanced:
- Carbon: [tex]\(2 \times 2 = 4\)[/tex] atoms on both sides.
- Hydrogen: [tex]\(2 \times 6 = 12\)[/tex] atoms on both sides.
- Oxygen: [tex]\(7 \times 2 = 14\)[/tex] on the reactant side and [tex]\(4 \times 2 + 6 \times 1 = 8 + 6 = 14\)[/tex] on the product side.
The balanced equation is:
[tex]\[
\boxed{2} \, C_2H_6 + \boxed{7} \, O_2 \rightarrow \boxed{4} \, CO_2 + \boxed{6} \, H_2O
\][/tex]
