Answer:
Approximately [tex]2.5\; \rm L[/tex].
Explanation:
Start by finding the balanced equation for the reaction between [tex]\rm CO[/tex] and [tex]\rm O_2[/tex].
When [tex]\rm O_2[/tex] is in excess, [tex]\rm CO[/tex] reacts with [tex]\rm O_2\![/tex] to produce [tex]\rm CO_2[/tex].
[tex]\rm ?\; CO + ?\; O_2 \to ?\; CO_2[/tex].
Assume that the coefficient of [tex]\rm CO_2[/tex] is [tex]1[/tex]. ([tex]\rm CO_2\![/tex] is the only product of this reaction.)
[tex]\rm ?\; CO + ?\; O_2 \to 1\; CO_2[/tex].
Apply the conservation of atoms to find the coefficient of the other two species.
[tex]\rm 1\; CO + ?\; O_2 \to 1\; CO_2[/tex].
[tex]\displaystyle \rm 1\; CO + \frac{1}{2}\; O_2 \to 1\; CO_2[/tex].
Multiply all coefficients in this equation by [tex]2[/tex] to eliminate the fraction.
[tex]\displaystyle \rm 2\; CO + 1\; O_2 \to 2\; CO_2[/tex].
Note the ratio between the coefficient of [tex]\rm O_2[/tex] and [tex]\rm CO[/tex] in this balanced equation:
[tex]\displaystyle \frac{n(\mathrm{O_2})}{n(\mathrm{CO})} = \frac{1}{2}[/tex].
Assume that both [tex]\rm CO[/tex] and [tex]\rm O_2[/tex] behave like ideal gases. Under the same temperature and pressure, the ratio between the volume of these two gases will match the ratio between the quantity of their particles:
[tex]\displaystyle \frac{V(\mathrm{O_2})}{V(\mathrm{CO})} = \frac{n(\mathrm{O_2})}{n(\mathrm{CO})} = \frac{1}{2}[/tex].
In other words, in this reaction, every one unit volume of [tex]\rm O_2[/tex] will react with two unit volume of [tex]\rm CO[/tex].
[tex]V(\mathrm{CO}) = 1\; \rm L[/tex].
Under these assumptions:
[tex]\displaystyle V(\mathrm{O}_2) = \frac{V(\mathrm{O_2})}{V(\mathrm{CO})}\cdot V(\mathrm{CO}) = \frac{1}{2} \times 1\; \rm L = 0.5\; \rm L[/tex].
The question stated that when measured by volume, one-fifth of the air is oxygen. That is:
[tex]\displaystyle \frac{V(\mathrm{O_2})}{V(\text{air})} = \frac{1}{5}[/tex].
Rearrange to obtain:
[tex]\displaystyle \frac{V(\text{air})}{V(\mathrm{O_2})} = 5[/tex].
Therefore:
[tex]\displaystyle V(\text{air}) = \frac{V(\text{air})}{V(\mathrm{O_2})} \cdot V(\mathrm{O_2}) = 5 \times 0.5\; \rm L = 2.5\; \rm L[/tex]
