Q.W rite chemical equations for combustion reaction of the following hydrocarbons:

The key idea is that complete combustion of a hydrocarbon always produces carbon dioxide and water. You balance the equation by first balancing carbon, then hydrogen, and finally oxygen.

(i) Butane ($C_4H_{10}$)

Balance C: $C_4H_{10} + O_2 \rightarrow 4CO_2 + H_2O$

Balance H: $C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O$

Balance O: $C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$

Multiply by 2: $2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$

(ii) Pentene ($C_5H_{10}$)

Balance C: $C_5H_{10} + O_2 \rightarrow 5CO_2 + H_2O$

Balance H: $C_5H_{10} + O_2 \rightarrow 5CO_2 + 5H_2O$

Balance O: $C_5H_{10} + \frac{15}{2}O_2 \rightarrow 5CO_2 + 5H_2O$

Multiply by 2: $2C_5H_{10} + 15O_2 \rightarrow 10CO_2 + 10H_2O$

(iii) Hexyne ($C_6H_{10}$)

Balance C: $C_6H_{10} + O_2 \rightarrow 6CO_2 + H_2O$

Balance H: $C_6H_{10} + O_2 \rightarrow 6CO_2 + 5H_2O$

Balance O: $C_6H_{10} + \frac{17}{2}O_2 \rightarrow 6CO_2 + 5H_2O$

Multiply by 2: $2C_6H_{10} + 17O_2 \rightarrow 12CO_2 + 10H_2O$ …

Combustion is a rapid reaction with oxygen, releasing heat and light. For any hydrocarbon (compound containing only carbon and hydrogen), complete combustion in excess oxygen yields only two products: carbon dioxide ($CO_2$) and water ($H_2O$). The general skeleton equation is:

The challenge is balancing the equation. The systematic method: first balance carbon atoms, then hydrogen atoms, and finally oxygen atoms. Oxygen is saved for last because it appears in both the reactant ($O_2$) and both products, so its coefficient often ends up as a fraction that we then clear by multiplying through.

Let’s apply this to each hydrocarbon.

1. Butane ($C_4H_{10}$)

   • Carbon: 4 C atoms on left → need 4 $CO_2$ on right.
   • Hydrogen: 10 H atoms on left → need 5 $H_2O$ on right (since each water has 2 H).
   • Oxygen: Count O atoms on right: $4 \times 2 = 8$ from $CO_2$, plus $5 \times 1 = 5$ from $H_2O$, total 13 O atoms. Since $O_2$ provides 2 O per molecule, we need $\frac{13}{2}$ $O_2$ molecules.
   • The equation so far: $C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$.
   • Multiply through by 2 to clear the fraction: $2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$.
   Tip

   A quick check: left side has $2 \times 4 = 8$ C, $2 \times 10 = 20$ H, $13 \times 2 = 26$ O. Right side has $8 \times 2 = 16$ O from $CO_2$ plus $10 \times 1 = 10$ O from $H_2O$, total 26 O. Balanced.

2. Pentene ($C_5H_{10}$)

   • Pentene is an alkene with one double bond, but the formula $C_5H_{10}$ tells us it has 5 carbons and 10 hydrogens. Combustion doesn’t care about the bond type — only the atom count matters.
   • Carbon: 5 C → 5 $CO_2$.
   • Hydrogen: 10 H → 5 $H_2O$.
   • Oxygen: Right side: $5 \times 2 = 10$ from $CO_2$, plus $5 \times 1 = 5$ from $H_2O$, total 15 O. So need $\frac{15}{2}$ $O_2$.
   • Equation: $C_5H_{10} + \frac{15}{2}O_2 \rightarrow 5CO_2 + 5H_2O$.
   • Multiply by 2: $2C_5H_{10} + 15O_2 \rightarrow 10CO_2 + 10H_2O$.
   Watch out

   A common mistake is to write pentene as $C_5H_8$ (confusing it with an alkyne) or to try to balance based on the double bond. Stick to the given molecular formula — combustion is purely stoichiometric.

3. Hexyne ($C_6H_{10}$)

   • Hexyne is an alkyne (triple bond), formula $C_6H_{10}$. Again, only the atom count matters.
   • Carbon: 6 C → 6 $CO_2$. …