$$ C _ { 2 } H _ { 6 } ( g ) + O _ { 2 } ( g ) \rightarrow CO _ { 2 } ( g ) + H _ { 2 } O ( g ) $$

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1. Write the unbalanced equation:  
   $$ C_{2}H_{6} + O_{2} ightarrow CO_{2} + H_{2}O $$

2. Balance the carbon atoms. Ethane, $$ C_{2}H_{6} $$, has 2 carbons, so assign 2 molecules of $$ CO_{2} $$:  
   $$ C_{2}H_{6} + O_{2} ightarrow 2CO_{2} + H_{2}O $$

3. Balance the hydrogen atoms. Ethane has 6 hydrogens and water has 2 hydrogens per molecule, so assign 3 molecules of $$ H_{2}O $$:  
   $$ C_{2}H_{6} + O_{2} ightarrow 2CO_{2} + 3H_{2}O $$

4. Balance the oxygen atoms. Count the oxygens on the product side:  
   - From $$ 2CO_{2} $$: $$2 	imes 2 = 4$$ oxygens  
   - From $$ 3H_{2}O $$: $$3 	imes 1 = 3$$ oxygens  
   Total oxygens needed: $$4 + 3 = 7$$.

Since oxygen comes as $$ O_{2} $$, this means we need $$ \frac{7}{2} $$ molecules of $$ O_{2} $$:  
   $$ C_{2}H_{6} + \frac{7}{2}O_{2} ightarrow 2CO_{2} + 3H_{2}O $$

5. Multiply the entire equation by 2 to eliminate the fractional coefficient:  
   $$ 2C_{2}H_{6} + 7O_{2} ightarrow 4CO_{2} + 6H_{2}O $$
The balanced equation is: $$ 2C_{2}H_{6} + 7O_{2} ightarrow 4CO_{2} + 6H_{2}O $$

\( C _ { 2 } H _ { 6 } ( g ) + O _ { 2 } ( g ) \rightarrow CO _ { 2 } ( g ) + H _ { 2 } O ( g ) \)

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