IGCSE Chemistry 0620: Catalytic Converter Limiting Reactant and Mass Calculation

A research team fits an experimental catalytic converter to a test engine. Instead of removing nitrogen monoxide, this converter is designed to remove nitrogen dioxide from the exhaust gas, converting it together with carbon monoxide into carbon dioxide and nitrogen gas according to the equation: $4\text{CO} + 2\text{NO}_2 \rightarrow 4\text{CO}_2 + \text{N}_2$

A sample of exhaust gas passing over the catalyst contains $8.4\ \text{g}$ of carbon monoxide and $13.8\ \text{g}$ of nitrogen dioxide. ( $M_r$ : $\text{CO} = 28$ , $\text{NO}_2 = 46$ , $\text{N}_2 = 28$ .)

(a) Calculate the number of moles of carbon monoxide and the number of moles of nitrogen dioxide in the sample. [2]

(b) Show that carbon monoxide is the limiting reactant, and calculate the maximum mass, in g, of nitrogen gas that could be produced from this sample. [3]

(c) State one adverse effect of nitrogen dioxide that is removed by this reaction, and explain why the carbon dioxide produced is still a cause for environmental concern, even though it is less immediately harmful than the original exhaust gases. [2]

Part (a): Moles of carbon monoxide and nitrogen dioxide

Using $n = \dfrac{\text{mass}}{M_r}$ :

$n(\text{CO}) = \frac{8.4}{28} = 0.30\ \text{mol}$

$n(\text{NO}_2) = \frac{13.8}{46} = 0.30\ \text{mol}$

Part (b): Limiting reactant and mass of nitrogen produced

The balanced equation is:

$4\text{CO} + 2\text{NO}_2 \rightarrow 4\text{CO}_2 + \text{N}_2$

This requires carbon monoxide and nitrogen dioxide in a mole ratio of $4$ to $2$ , i.e. $2$ to $1$ .

Dividing the available moles by their coefficients in the equation:

$\frac{n(\text{CO})}{4} = \frac{0.30}{4} = 0.075 \qquad \frac{n(\text{NO}_2)}{2} = \frac{0.30}{2} = 0.15$

Carbon monoxide gives the smaller value, so carbon monoxide is the limiting reactant (nitrogen dioxide is in excess). Equivalently: fully reacting $0.30\ \text{mol}$ of $\text{NO}_2$ would need $0.60\ \text{mol}$ of $\text{CO}$ (using the $2$ to $1$ ratio), but only $0.30\ \text{mol}$ of $\text{CO}$ is available.

Using the $4$ to $1$ ratio of $\text{CO}$ to $\text{N}_2$ , and the limiting reactant $\text{CO}$ :

$n(\text{N}_2) = \frac{0.30}{4} = 0.075\ \text{mol}$

Converting to mass:

$\text{mass} = n \times M_r = 0.075 \times 28 = 2.1\ \text{g}$

So the maximum mass of nitrogen gas produced is $\boxed{2.1}\ \text{g}$ .

Part (c): Environmental significance of the products

Before the reaction, nitrogen dioxide in the exhaust would contribute to acid rain, photochemical smog and respiratory problems. This reaction converts it, together with carbon monoxide, into carbon dioxide and nitrogen. Neither of which causes these specific problems.

However, carbon dioxide is still released, and carbon dioxide is a greenhouse gas: higher levels of carbon dioxide in the atmosphere cause increased global warming, which leads to climate change. So although the immediate toxic and acid-rain/smog-forming hazards are removed, the reaction does not eliminate the exhaust’s environmental impact altogether.

Final answers

(a) $n(\text{CO}) = 0.30\ \text{mol}$ ; $n(\text{NO}_2) = 0.30\ \text{mol}$
(b) Carbon monoxide is the limiting reactant; maximum mass of nitrogen produced $= \boxed{2.1}\ \text{g}$
(c) The reaction removes nitrogen dioxide’s contribution to acid rain, photochemical smog and respiratory problems, but the carbon dioxide produced remains a greenhouse gas that contributes to global warming and climate change.