Question

Oxygen oxidises ethyne to carbon dioxide and water as shown by the equation:
$2{\mathrm{C}}_2{\mathrm{H}}_2+5{\mathrm{O}}_2\to 4{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$
What volume of ethyne gas at s.t.p. is required to produce $8.4d{m}^3$ of carbon dioxide at s.t.p.?
$[H=1,C=12,O=16]$

Answer 1

• According to the question :

$$\begin{gathered} 2{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \left(\mathrm{Ethyne}\right)(\mathrm{Carbon}\mathrm{dioxide}) \\ 2\mathrm{vol}5\mathrm{vol}4\mathrm{vol}2\mathrm{vol} \\ ?\mathrm{vol}8.4{\mathrm{dm}}^3 \end{gathered}$$

• $\text{If}4\text{vol. of}{\mathrm{CO}}_2\to 2\text{vol. of}{\mathrm{C}}_2{\mathrm{H}}_2$
• $\text{Then,}1\text{vol. of}{\mathrm{CO}}_2=\frac{2}{4}=\frac{1}{2}\text{vol. of}{\mathrm{C}}_2{\mathrm{H}}_2$
  $$\begin{gathered} 8.4{\mathrm{dm}}^3\text{of}{\mathrm{CO}}_2=\frac{1}{2}\times 8.4{\mathrm{dm}}^3\text{of}{\mathrm{C}}_2{\mathrm{H}}_2 \\ =4.2{\mathrm{dm}}^3\text{of}{\mathrm{C}}_2{\mathrm{H}}_2 \end{gathered}$$
• Therefore, the volume of ethyne gas is $4.2d{m}^3$.