To write the combustion reaction equation for $C_2H_6$, we react it with oxygen ($O_2$) to produce carbon dioxide ($CO_2$) and water ($H_2O$). Step 1: Write the unbalanced equation. $$ C_2H_6 + O_2 \rightarrow CO_2 + H_2O $$ Step 2: Balance the carbon atoms. There are 2 carbon atoms on the left ($C_2H_6$), so we need 2 molecules of $CO_2$ on the right. $$ C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O $$ Step 3: Balance the hydrogen atoms. There are 6 hydrogen atoms on the left ($C_2H_6$), so we need 3 molecules of $H_2O$ on the right ($3 \times 2 = 6$). $$ C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O $$ Step 4: Balance the oxygen atoms. Count the oxygen atoms on the right side: From $2CO_2$: $2 \times 2 = 4$ oxygen atoms. From $3H_2O$: $3 \times 1 = 3$ oxygen atoms. Total oxygen atoms on the right = $4 + 3 = 7$. Since oxygen on the left is in the form of $O_2$, we need $\frac{7}{2}$ molecules of $O_2$ to get 7 oxygen atoms. $$ C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O $$ Step 5: Clear the fraction by multiplying the entire equation by 2. $$ 2(C_2H_6) + 2\left(\frac{7}{2}O_2\right) \rightarrow 2(2CO_2) + 2(3H_2O) $$ $$ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O $$ Step 6: Verify the balance. Left side: Carbon: $2 \times 2 = 4$ Hydrogen: $2 \times 6 = 12$ Oxygen: $7 \times 2 = 14$ Right side: Carbon: $4 \times 1 = 4$ Hydrogen: $6 \times 2 = 12$ Oxygen: $(4 \times 2) + (6 \times 1) = 8 + 6 = 14$ The equation is balanced. The final balanced combustion reaction equation for $C_2H_6$ is: $$ \boxed{2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O} $$