To balance the chemical equation for the combustion of ethane ([tex]\(C_2H_6\)[/tex]), we need to ensure that the number of atoms of each element on the reactant side (left side) is equal to the number of atoms of the same element on the product side (right side). The given unbalanced equation is:
[tex]$ \square \, C_2H_6 + \square \, O_2 \rightarrow \square \, CO_2 + \square \, H_2O $[/tex]
### Step-by-Step Solution:
1. Balance Carbon (C) Atoms:
Ethane ([tex]\(C_2H_6\)[/tex]) has 2 carbon atoms. To balance carbon atoms, we need 2 molecules of carbon dioxide ([tex]\(CO_2\)[/tex]), each containing 1 carbon atom:
[tex]$ C_2H_6 + \square O_2 \rightarrow 2 CO_2 + \square H_2O $[/tex]
2. Balance Hydrogen (H) Atoms:
Ethane ([tex]\(C_2H_6\)[/tex]) has 6 hydrogen atoms. To balance hydrogen atoms, we need 3 molecules of water ([tex]\(H_2O\)[/tex]), each containing 2 hydrogen atoms:
[tex]$ C_2H_6 + \square O_2 \rightarrow 2 CO_2 + 3 H_2O $[/tex]
3. Balance Oxygen (O) Atoms:
Now, let's balance the oxygen atoms. On the right side, we have:
- [tex]\(2 \times CO_2\)[/tex], which gives [tex]\(2 \times 2 = 4\)[/tex] oxygen atoms
- [tex]\(3 \times H_2O\)[/tex], which gives [tex]\(3 \times 1 = 3\)[/tex] oxygen atoms
- In total: [tex]\(4 + 3 = 7\)[/tex] oxygen atoms on the product side.
To balance 7 oxygen atoms on the reactant side using [tex]\(O_2\)[/tex] (each [tex]\(O_2\)[/tex] molecule has 2 oxygen atoms), we need [tex]\(\frac{7}{2}\)[/tex] molecules of [tex]\(O_2\)[/tex]:
[tex]$ C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O $[/tex]
4. Simplify the Coefficients (Optional):
To get whole number coefficients, multiply all coefficients by 2:
[tex]$ 2 C_2H_6 + 7 O_2 \rightarrow 4 CO_2 + 6 H_2O $[/tex]
Thus, the balanced chemical equation is:
[tex]$ \boxed{2} \, C_2H_6 + \boxed{7} \, O_2 \rightarrow \boxed{4} \, CO_2 + \boxed{6} \, H_2O $[/tex]
