When decane undergoes complete combustion, a mixture of carbon dioxide and water is formed - Edexcel - GCSE Chemistry Combined Science - Question 4 - 2016 - Paper 1

To balance the combustion of decane, the reaction starts with the formula for decane, which is $C_{10}H_{22}$. The general form of hydrocarbon combustion can be expressed as:

$C_xH_y + O_2 \rightarrow CO_2 + H_2O$

1. Count the Carbon and Hydrogen Atoms:

   • Decane ($C_{10}H_{22}$) has 10 carbon atoms and 22 hydrogen atoms.
   • This means we will produce 10 molecules of $CO_2$ (since each carbon produces one molecule of carbon dioxide) and 11 molecules of $H_2O$ (since two hydrogen atoms produce one molecule of water).

2. Write the Products:

   • The balanced equation looks like:

   $C_{10}H_{22} + O_2 \rightarrow 10CO_2 + 11H_2O$

3. Balance the Oxygen Atoms:

   • From the products, calculate oxygen atoms:
     • $10CO_2$ contributes $10 \times 2 = 20$ oxygen atoms.
     • $11H_2O$ contributes $11 \times 1 = 11$ oxygen atoms.
   • Total: $20 + 11 = 31$ oxygen atoms.

4. Calculate Required O2 Molecules:

   • Since one molecule of $O_2$ contains 2 oxygen atoms, the number of $O_2$ molecules required is:

   $\frac{31}{2} = 15.5$

   • To eliminate the fraction, multiply the entire equation by 2:

   $2C_{10}H_{22} + 31O_2 \rightarrow 20CO_2 + 22H_2O$

This balanced chemical equation shows the complete combustion of decane.