Question

From the balanced equation ${\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$, compute the mass of Oxygen combined with $8\mathrm{g}$ of methane.

Answer 1

Step 1: Give a balanced chemical equation:

According to the given balanced equation,

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

Step 2: Moles produced:

$1$mole of Methane combines with $2$ moles of oxygen.

• The molecular mass of $1$ mol of Methane (${\mathrm{CH}}_4$) is $12+(4\times 1)=12+4=16g$
• The molecular mass of $1$ mol of oxygen (${\mathrm{O}}_2$) is $2\times 16=32g$

So, the volume of $2$ moles of ${\mathrm{O}}_2$ is $2\times 32=64g$

Step 3: Expression for reactant side:

The reactant side of the equation can be expressed as,

Thus,$16\mathrm{g}$ of ${\mathrm{CH}}_4$combines with $64\mathrm{g}$ of ${\mathrm{O}}_2$

So, $8\mathrm{g}$ of ${\mathrm{CH}}_4$will combine with $\mathrm{x}\mathrm{g}$ of ${\mathrm{O}}_2$

Step 4: Putting the equation in a corresponding ratio:

We get;

$$\begin{gathered} \frac{16}{8}=\frac{64}{x} \\ x=\frac{64\times 8}{16}=32g \end{gathered}$$

Thus, the mass of Oxygen required to combine with $8\mathrm{g}$ of methane is $32\mathrm{g}$.