Question

How Would You Balance The Following Equation:${\mathrm{CH}}_4+{\mathrm{O}}_2\to {\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

Answer 1

Balancing the Chemical equation:

• When we balance chemical equations, we want the same number of each sort of atom on both sides of the equation.
• Only the coefficients are changed. The numbers in front of the molecule are called coefficients.
• Subscripts should never be changed.
• Metals must be balanced first, followed by nonmetals such as oxygen and hydrogen.
• The unbalanced chemical reaction provided is

$\underset{\mathrm{Methane}}{{\mathrm{CH}}_4\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}\mathrm{gas}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}\mathrm{vapours}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}$

• It is a combustion and redox process.
• The following table shows the number of atoms on each side:

Reactant side	Product side
1 Carbon atom from ${\mathrm{CH}}_4$	1 Carbon atom from ${\mathrm{CO}}_2$
4 Hydrogen atoms from ${\mathrm{CH}}_4$	2 Hydrogen atoms from ${\mathrm{H}}_2\mathrm{O}$
2 Oxygen atoms from ${\mathrm{O}}_2$	3 Oxygen atoms from ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$

• Now, as we can see, there are $4$ Hydrogen atoms on the left, so we must put $2$ on the right side before Water. $\underset{\mathrm{Methane}}{{\mathrm{CH}}_4\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}\mathrm{gas}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}\mathrm{vapours}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}$

Reactant side	Product side
1 Carbon atom from ${\mathrm{CH}}_4$	1 Carbon atom from ${\mathrm{CO}}_2$
4 Hydrogen atoms from ${\mathrm{CH}}_4$	4 Hydrogen atoms from ${\mathrm{H}}_2\mathrm{O}$
2 Oxygen atoms from ${\mathrm{O}}_2$	4 Oxygen atoms from ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$

• Second, the Oxygen must be balanced. As shown on the left, there are two Oxygen atoms in ${\mathrm{O}}_2$, and four on the product side, thus a $2$ must be inserted on the right side before Oxygen gas to balance.

$\underset{\mathrm{Methane}}{{\mathrm{CH}}_4\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}\mathrm{gas}}{2{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}\mathrm{vapours}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}$

• Finally, the Carbons must be balanced, but in this case, they are already at $1$ on each side; therefore, the equation is now balanced.

Reactant side	Product side
1 Carbon atom from ${\mathrm{CH}}_4$	1 Carbon atom from ${\mathrm{CO}}_2$
4 Hydrogen atoms from ${\mathrm{CH}}_4$	4 Hydrogen atoms from ${\mathrm{H}}_2\mathrm{O}$
4 Oxygen atoms from ${\mathrm{O}}_2$	4 Oxygen atoms from ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$

• Thus, the balanced chemical reaction is:

$\underset{\mathrm{Methane}}{{\mathrm{CH}}_4\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}\mathrm{gas}}{2{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}\mathrm{vapours}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}$