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Balance the combustion reaction between butane and oxygen.

[tex]2C_4H_{10} + \quad \vee O_2 \rightarrow \quad \vee CO_2 + \quad \vee H_2O[/tex]


Sagot :

To balance the combustion reaction of butane (C₄H₁₀) with oxygen (O₂), follow these steps:

1. Write the unbalanced equation:

[tex]\[ \text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \][/tex]

2. Balance the carbon (C) atoms first:

Butane has 4 carbon atoms, so we need 4 CO₂ molecules to balance the carbons on the product side.

[tex]\[ \text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4 \text{CO}_2 + \text{H}_2\text{O} \][/tex]

3. Balance the hydrogen (H) atoms next:

Butane has 10 hydrogen atoms, so we need 5 H₂O molecules to balance the hydrogens on the product side (since each H₂O molecule contains 2 hydrogen atoms).

[tex]\[ \text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \][/tex]

4. Balance the oxygen (O) atoms last:

- On the product side, we have oxygen atoms from 4 CO₂ (4 2 = 8 oxygens) plus oxygen atoms from 5 H₂O (5 1 = 5 oxygens), for a total of 13 oxygen atoms.
- These 13 oxygen atoms must come from O₂ molecules. Since each O₂ molecule contains 2 oxygen atoms, we need [tex]\( \frac{13}{2} = 6.5 \)[/tex] O₂ molecules.

[tex]\[ \text{C}_4\text{H}_{10} + 6.5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \][/tex]

5. Adjust the coefficients to make all numbers whole:

Multiply through by 2 to clear the fraction.

[tex]\[ 2 \text{C}_4\text{H}_{10} + 13 \text{O}_2 \rightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \][/tex]

The balanced equation 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]

Therefore, the coefficients are:
- Butane (C₄H₁₀): 2
- Oxygen (O₂): 13
- Carbon dioxide (CO₂): 8
- Water (H₂O): 10
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