\( \square C _ { 4 } H _ { 10 } + \square O _ { 2 } + \rightarrow CO _ { 2 } + \square H _ { 2 } O \)

To balance the combustion reaction of \( C_4H_{10} \) (butane), we start with the unbalanced equation: \[ C_4H_{10} + O_2 \rightarrow CO_2 + H_2O \] 1. **Count the number of carbon (C), hydrogen (H), and oxygen (O) atoms:** - In \( C_4H_{10} \), there are 4 carbon atoms and 10 hydrogen atoms. - On the product side, each \( CO_2 \) molecule contains 1 carbon atom, and each \( H_2O \) molecule contains 2 hydrogen atoms. 2. **Balance the carbon atoms:** - Since there are 4 carbon atoms in \( C_4H_{10} \), we need 4 \( CO_2 \) molecules: \[ C_4H_{10} + O_2 \rightarrow 4 CO_2 + H_2O \] 3. **Balance the hydrogen atoms:** - There are 10 hydrogen atoms in \( C_4H_{10} \), which means we need 5 \( H_2O \) molecules (since each \( H_2O \) has 2 hydrogen atoms): \[ C_4H_{10} + O_2 \rightarrow 4 CO_2 + 5 H_2O \] 4. **Count the total number of oxygen atoms needed on the product side:** - From \( 4 CO_2 \), we get \( 4 \times 2 = 8 \) oxygen atoms. - From \( 5 H_2O \), we get \( 5 \times 1 = 5 \) oxygen atoms. - Total oxygen atoms needed = \( 8 + 5 = 13 \). 5. **Determine the number of \( O_2 \) molecules required:** - Since each \( O_2 \) molecule contains 2 oxygen atoms, we need \( \frac{13}{2} = 6.5 \) \( O_2 \) molecules. 6. **To avoid fractions, multiply the entire equation by 2:** \[ 2 C_4H_{10} + 13 O_2 \rightarrow 8 CO_2 + 10 H_2O \] Now, the balanced equation is: \[ \boxed{2 C_4H_{10} + 13 O_2 \rightarrow 8 CO_2 + 10 H_2O} \]