Sure, let's go through the process of balancing the equation for the reaction of butane (C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex]) with oxygen (O[tex]\(_2\)[/tex]) to form carbon dioxide (CO[tex]\(_2\)[/tex]) and water (H[tex]\(_2\)[/tex]) step-by-step:
Step 1: Write the unbalanced chemical equation.
[tex]\[ \text{C}_4 \text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2 \text{O} \][/tex]
Step 2: Balance the carbon (C) atoms.
- There are 4 carbon atoms in butane (C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex]), so we need 4 CO[tex]\(_2\)[/tex] molecules on the right-hand side to balance the carbon atoms.
[tex]\[ \text{C}_4 \text{H}_{10} + \text{O}_2 \rightarrow 4 \text{CO}_2 + \text{H}_2 \text{O} \][/tex]
Step 3: Balance the hydrogen (H) atoms.
- There are 10 hydrogen atoms in butane (C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex]), so we need 5 H[tex]\(_2\)[/tex]O molecules on the right-hand side to balance the hydrogen atoms.
[tex]\[ \text{C}_4 \text{H}_{10} + \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2 \text{O} \][/tex]
Step 4: Balance the oxygen (O) atoms.
- Now, count the oxygen atoms on the right-hand side.
- From 4 CO[tex]\(_2\)[/tex]: [tex]\(4 \times 2 = 8\)[/tex] oxygen atoms.
- From 5 H[tex]\(_2\)[/tex]O: [tex]\(5 \times 1 = 5\)[/tex] oxygen atoms.
- Total: [tex]\(8 + 5 = 13\)[/tex] oxygen atoms needed on the left-hand side.
- Since we have O[tex]\(_2\)[/tex] molecules (each molecule has 2 oxygen atoms), we need [tex]\(\frac{13}{2}\)[/tex] O[tex]\(_2\)[/tex] molecules.
[tex]\[ \text{C}_4 \text{H}_{10} + \frac{13}{2} \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2 \text{O} \][/tex]
Step 5: To eliminate the fractional coefficient, multiply the entire equation by 2.
[tex]\[ 2 \text{C}_4 \text{H}_{10} + 13 \text{O}_2 \rightarrow 8 \text{CO}_2 + 10 \text{H}_2 \text{O} \][/tex]
So, the balanced chemical equation for the reaction of butane with oxygen to form carbon dioxide and water is:
[tex]\[ 2 \text{C}_4 \text{H}_{10} + 13 \text{O}_2 \rightarrow 8 \text{CO}_2 + 10 \text{H}_2 \text{O} \][/tex]
