The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is
$$2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_6O \rightarrow 2Cr_2(SO_4)_3 + 3C_2H_4O_2 + 2K_2SO_4 + 11H_2O$$
If the rate of appearance of $$Cr_2(SO_4)_3$$ is 2.67 mol min$$^{-1}$$ at a particular time, the rate of disappearance of $$C_2H_6O$$ at the same time is _________ mol min$$^{-1}$$. (Nearest integer)

We have the balanced chemical equation for the breath analyser reaction:

$$2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_6O \;\longrightarrow\; 2Cr_2(SO_4)_3 + 3C_2H_4O_2 + 2K_2SO_4 + 11H_2O$$

The concept of rate in chemical kinetics tells us that for any reactant or product, the rate of reaction is obtained by dividing the time-rate of change of its concentration by its stoichiometric coefficient. Mathematically, for a general reaction

$$aA + bB \rightarrow cC + dD,$$

the rate $$r$$ can be written as

$$r \;=\; -\frac{1}{a}\,\frac{d[A]}{dt} \;=\; -\frac{1}{b}\,\frac{d[B]}{dt} \;=\; \frac{1}{c}\,\frac{d[C]}{dt} \;=\; \frac{1}{d}\,\frac{d[D]}{dt}.$$

Applying this definition to the given equation, we correlate the rates of ethanol disappearance and chromic sulfate appearance:

$$r \;=\; \frac{1}{2}\,\frac{d[Cr_2(SO_4)_3]}{dt} \;=\; -\frac{1}{3}\,\frac{d[C_2H_6O]}{dt}.$$

We are told that the rate of appearance of $$Cr_2(SO_4)_3$$ is

$$\frac{d[Cr_2(SO_4)_3]}{dt} \;=\; +2.67\;\text{mol min}^{-1}.$$

Substituting this value into the expression for the rate gives

$$r \;=\; \frac{1}{2}\times 2.67 \;=\; 1.335\;\text{mol min}^{-1}.$$

Now we equate this same rate to the ethanol term:

$$1.335 \;=\; -\frac{1}{3}\,\frac{d[C_2H_6O]}{dt}.$$

Multiplying both sides by $$-3$$ yields

$$\frac{d[C_2H_6O]}{dt} \;=\; -3 \times 1.335 \;=\; -4.005\;\text{mol min}^{-1}.$$

The negative sign simply indicates disappearance. The magnitude of the disappearance rate is therefore

$$4.005\;\text{mol min}^{-1}.$$

Rounding to the nearest integer, we obtain $$4\;\text{mol min}^{-1}$$ as required.

So, the answer is $$4$$.