To balance the chemical equation for the combustion of ethane ([tex]$C_2H_6$[/tex]), we need to ensure the same number of atoms of each element on both the reactant and product sides of the equation.
The unbalanced equation is:
[tex]\[
\square \, C_2H_6 + \square \, O_2 \rightarrow \square \, CO_2 + \square \, H_2O
\][/tex]
We start by balancing the carbon (C) atoms:
- On the reactant side, there are 2 carbon atoms in [tex]$C_2H_6$[/tex].
- To balance, we need 2 molecules of [tex]$C_2H_6$[/tex], so the coefficient for [tex]$CO_2$[/tex] should be 4 because each [tex]$CO_2$[/tex] molecule has 1 carbon atom.
Next, we balance the hydrogen (H) atoms:
- On the reactant side, there are 6 hydrogen atoms in [tex]$C_2H_6$[/tex].
- To balance, we need 3 molecules of [tex]$H_2O$[/tex], as each [tex]$H_2O$[/tex] molecule has 2 hydrogen atoms (3 x 2 = 6).
Finally, we balance the oxygen (O) atoms:
- On the product side, there are [tex]$4 \times 2 + 6 \times 1 = 8 + 6 = 14$[/tex] oxygen atoms.
- Thus, we need 7 molecules of [tex]$O_2$[/tex] on the reactant side because each [tex]$O_2$[/tex] molecule has 2 oxygen atoms (7 x 2 = 14).
Now, the balanced chemical equation is:
[tex]\[
2 \, C_2H_6 + 7 \, O_2 \rightarrow 4 \, CO_2 + 6 \, H_2O
\][/tex]
So, the coefficients are:
- [tex]$C_2H_6$[/tex]: 2
- [tex]$O_2$[/tex]: 7
- [tex]$CO_2$[/tex]: 4
- [tex]$H_2O$[/tex]: 6
Filling in the blanks, we get:
[tex]\[
\boxed{2} \, C_2H_6 + \boxed{7} \, O_2 \rightarrow \boxed{4} \, CO_2 + \boxed{6} \, H_2O
\][/tex]
