Q. \[ \mathrm{CH}_4 + \mathrm{O}_2 \]

We are asked to balance the combustion of methane given by the unbalanced equation \( \mathrm{CH_4} + \mathrm{O_2} = \mathrm{CO_2} + \mathrm{H_2O} \).

Step 1. Write the unbalanced equation and identify coefficients to determine. Let the coefficients be shown explicitly as integers in front of each species. For now use implied coefficient 1 where none is written. The skeleton equation is \( 1\,\mathrm{CH_4} + 1\,\mathrm{O_2} = 1\,\mathrm{CO_2} + 1\,\mathrm{H_2O} \).

Step 2. Count the number of each type of atom on both sides. Display the counts in a small table. The table lists atom types and counts on reactant and product sides.

\[
\begin{array}{l c c}
\text{Atom} & \text{Reactants} & \text{Products} \\
\mathrm{C:} & 1 & 1 \\
\mathrm{H:} & 4 & 2 \\
\mathrm{O:} & 2 & 3 \\
\end{array}
\]

Interpretation of the table. Carbon is already balanced, with 1 carbon atom on each side. Hydrogen is not balanced, with 4 hydrogen atoms on the reactant side and 2 on the product side. Oxygen is not balanced, with 2 oxygen atoms on the reactant side and 3 on the product side.

Step 3. Balance hydrogen next. There are 4 hydrogen atoms in \( \mathrm{CH_4} \). Each \( \mathrm{H_2O} \) molecule contains 2 hydrogen atoms. To get 4 hydrogens on the product side, place coefficient 2 in front of \( \mathrm{H_2O} \). The equation becomes \( \mathrm{CH_4} + \mathrm{O_2} = \mathrm{CO_2} + 2\,\mathrm{H_2O} \).

Step 4. Recount oxygen after adjusting hydrogen. On the product side, \( \mathrm{CO_2} \) contributes 2 oxygen atoms, and \( 2\,\mathrm{H_2O} \) contributes \(2 \times 1 = 2\) oxygen atoms, for a total of 4 oxygen atoms on the product side. The reactant \( \mathrm{O_2} \) supplies 2 oxygen atoms per molecule. To supply 4 oxygens on the reactant side, we need 2 molecules of \( \mathrm{O_2} \). Place coefficient 2 in front of \( \mathrm{O_2} \). The equation becomes \( \mathrm{CH_4} + 2\,\mathrm{O_2} = \mathrm{CO_2} + 2\,\mathrm{H_2O} \).

Step 5. Final check. Count atoms with the chosen coefficients.

\[
\begin{array}{l c c}
\text{Atom} & \text{Reactants} & \text{Products} \\
\mathrm{C:} & 1 & 1 \\
\mathrm{H:} & 4 & 2 \times 2 = 4 \\
\mathrm{O:} & 2 \times 2 = 4 & 2 + 2 \times 1 = 4 \\
\end{array}
\]

All atom counts match on both sides. Therefore the balanced combustion equation for methane is

\[ \mathrm{CH_4} + 2\,\mathrm{O_2} = \mathrm{CO_2} + 2\,\mathrm{H_2O}. \]