Question

Q17. What volume of air (containing$20\%$${\mathrm{O}}_2$ by volume) will be required to burn completely $10c{m}^3$each of methane and acetylene?

Answer 1

• The balanced equation for ${\mathrm{CH}}_4$:

$\underset{\underset{1\mathrm{vol}}{\mathrm{methane}}}{{\mathrm{CH}}_4}\left(g\right)+\underset{\underset{2\mathrm{vol}}{\mathrm{oxygen}}}{2{\mathrm{O}}_2}\left(\mathrm{g}\right)\to \underset{\underset{1\mathrm{vol}}{\mathrm{carbon}\mathrm{dioxide}}}{{\mathrm{CO}}_2}\left(\mathrm{g}\right)+\underset{\underset{2\mathrm{vol}}{\mathrm{water}}}{2{\mathrm{H}}_2\mathrm{O}}\left(\mathrm{l}\right)$

• The balanced equation for ${\mathrm{C}}_2{\mathrm{H}}_2$:

$\underset{\underset{2\mathrm{vol}}{\mathrm{acetylene}}}{2{\mathrm{C}}_2{\mathrm{H}}_2}\left(g\right)+\underset{\underset{5\mathrm{vol}}{\mathrm{oxygen}}}{5{\mathrm{O}}_2}\left(\mathrm{g}\right)\to \underset{\underset{4\mathrm{vol}}{\mathrm{carbon}\mathrm{dioxide}}}{4{\mathrm{CO}}_2}\left(\mathrm{g}\right)+\underset{\underset{2\mathrm{vol}}{\mathrm{water}}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

• Calculate the volume of ${\mathrm{O}}_2$ for ${\mathrm{CH}}_4$

From the above equation it is clear that$1$ volume of ${\mathrm{CH}}_4$requires $2$vol of ${\mathrm{O}}_2$

Therefore, $10c{m}^3$of ${\mathrm{CH}}_4$ will require $$\begin{gathered} =10\times 2 \\ =20c{m}^3 \end{gathered}$$ of ${\mathrm{O}}_2$

• Calculate the volume of ${\mathrm{O}}_2$ for ${\mathrm{C}}_2{\mathrm{H}}_2$

Also $2$vol of ${\mathrm{C}}_2{\mathrm{H}}_2$ require $5$vol of ${\mathrm{O}}_2$

Therefore $10c{m}^3$ will require ${\mathrm{O}}_2$for complete combustion$$\begin{gathered} =\frac{5}{2}\times 10 \\ =25c{m}^3 \end{gathered}$$

Now total ${\mathrm{O}}_2$ required$$\begin{gathered} =20+25 \\ =45c{m}^3 \end{gathered}$$

Since air contains $20\%$of ${\mathrm{O}}_2$ by volume

• Thus the volume of air containing $45c{m}^3$of ${\mathrm{O}}_2$is $225c{m}^3$.