Question

13. What is the volume of air required to completely burn 1 L of CO? Assume

that all volumes are measured at the same temperature and pressure and
also that air contains one-fifth by volume of oxygen​

Answer 1

Approximately
`2.5\; \rm L`.

Step-by-step explanation:

Start by finding the balanced equation for the reaction between
`\rm CO` and
`\rm O_2`.

When
`\rm O_2` is in excess,
`\rm CO` reacts with
`\rm O_2\!` to produce
`\rm CO_2`.

`\rm ?\; CO + ?\; O_2 \to ?\; CO_2`.

Assume that the coefficient of
`\rm CO_2` is
`1`. (
`\rm CO_2\!` is the only product of this reaction.)

`\rm ?\; CO + ?\; O_2 \to 1\; CO_2`.

Apply the conservation of atoms to find the coefficient of the other two species.

`\rm 1\; CO + ?\; O_2 \to 1\; CO_2`.

`\displaystyle \rm 1\; CO + (1)/(2)\; O_2 \to 1\; CO_2`.

Multiply all coefficients in this equation by
`2` to eliminate the fraction.

`\displaystyle \rm 2\; CO + 1\; O_2 \to 2\; CO_2`.

Note the ratio between the coefficient of
`\rm O_2` and
`\rm CO` in this balanced equation:

`\displaystyle \frac{n(\mathrm{O_2})}{n(\mathrm{CO})} = (1)/(2)`.

Assume that both
`\rm CO` and
`\rm O_2` 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:

`\displaystyle \frac{V(\mathrm{O_2})}{V(\mathrm{CO})} = \frac{n(\mathrm{O_2})}{n(\mathrm{CO})} = (1)/(2)`.

In other words, in this reaction, every one unit volume of
`\rm O_2` will react with two unit volume of
`\rm CO`.

`V(\mathrm{CO}) = 1\; \rm L`.

Under these assumptions:

`\displaystyle V(\mathrm{O}_2) = \frac{V(\mathrm{O_2})}{V(\mathrm{CO})}\cdot V(\mathrm{CO}) = (1)/(2) * 1\; \rm L = 0.5\; \rm L`.

The question stated that when measured by volume, one-fifth of the air is oxygen. That is:

`\displaystyle \frac{V(\mathrm{O_2})}{V(\text{air})} = (1)/(5)`.

Rearrange to obtain:

`\displaystyle \frac{V(\text{air})}{V(\mathrm{O_2})} = 5`.

Therefore:

`\displaystyle V(\text{air}) = \frac{V(\text{air})}{V(\mathrm{O_2})} \cdot V(\mathrm{O_2}) = 5 * 0.5\; \rm L = 2.5\; \rm L`