To balance the combustion reaction of ethane, we follow these steps for each element involved:
[tex]\( C_2H_6 + O_2 \rightarrow CO_2 + H_2O \)[/tex]
1. Carbon (C):
- There are 2 carbon atoms on the left side (in [tex]\(C_2H_6\)[/tex]).
- Therefore, we need 2 carbon atoms on the right side, which means 2 [tex]\(CO_2\)[/tex] molecules.
[tex]\[ C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O \][/tex]
2. Hydrogen (H):
- There are 6 hydrogen atoms on the left side (in [tex]\(C_2H_6\)[/tex]).
- Each water molecule ([tex]\(H_2O\)[/tex]) contains 2 hydrogen atoms. To balance 6 hydrogen atoms, we need 3 [tex]\(H_2O\)[/tex] molecules.
[tex]\[ C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O \][/tex]
3. Oxygen (O):
- We have oxygen atoms in both [tex]\(O_2\)[/tex] and the products ([tex]\(CO_2\)[/tex] and [tex]\(H_2O\)[/tex]).
- On the right side, we have 2 [tex]\(CO_2\)[/tex] molecules and 3 [tex]\(H_2O\)[/tex] molecules:
[tex]\[
2 \times 2 \text{ O atoms in } CO_2 = 4 \text{ O atoms} \\
3 \times 1 \text{ O atom in } H_2O = 3 \text{ O atoms} \\
\text{Total: } 4 + 3 = 7 \text{ O atoms}
\][/tex]
- Since each [tex]\(O_2\)[/tex] molecule contains 2 oxygen atoms, we need [tex]\(\frac{7}{2} = 3.5\)[/tex] molecules of [tex]\(O_2\)[/tex].
- However, we prefer whole numbers, so we multiply all coefficients by 2:
[tex]\[
2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O
\][/tex]
The balanced equation is:
[tex]\[ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \][/tex]
But the simplest coefficients are:
[tex]\[ \boxed{1}C_2H_6 + \boxed{7}O_2 \rightarrow \boxed{2}CO_2 + \boxed{6}H_2O \][/tex]
