\(\quad \mathrm{CH}_{4}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) Answer: Because oxygen appears as a pure elemental substance, we save it for last. Starting with carbon (an arbitrary choice), we note that there is one carbon on each side of the equation, meaning \(\mathrm{C}\) is balanced. There are four hydrogen atoms on the left and only two on the right. To fix this, we put a balancing coefficient \(2 \mathrm{in}\) front of the \(\mathrm{H}_{2} \mathrm{O}\) : \(\mathrm{CH}_{4}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) This balances the hydrogen. Now it's time to balance the oxygen. There are two oxygen atoms on the left and four on the right (two from the \(\mathrm{CO}_{2}\) molecule and two from the two \(\mathrm{H}_{2} \mathrm{O}\) molecules). To balance the equation, we put a \(2 \mathrm{in}\) front of the \(\mathrm{O}_{2}\). \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)

Step by step solution

01

Balancing Carbon

As there is one carbon atom in both methane (CH4) and carbon dioxide (CO2), carbon is already balanced in the given equation. So, no changes are needed for carbon atoms.

02

Balancing Hydrogen

There are four hydrogen atoms in methane (CH4) and two hydrogen atoms in water (H2O). To balance the hydrogen, we need to place a coefficient of 2 in front of the H2O, resulting in the following equation: CH4 + O2 → CO2 + 2 H2O Now, there are four hydrogen atoms on both sides of the equation, so the hydrogen is balanced.

03

Balancing Oxygen

We now have two oxygen atoms in O2 and four oxygen atoms on the right side (two from CO2 and two more from 2 H2O). To balance the oxygen atoms, we need to place a coefficient of 2 in front of the O2, resulting in this balanced equation: CH4 + 2 O2 → CO2 + 2 H2O Now, there are four oxygen atoms on both sides of the equation, so all elements are balanced in the chemical equation.

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