KClO$$_3$$ + 6FeSO$$_4$$ + 3H$$_2$$SO$$_4$$ $$\to$$ KCl + 3Fe$$_2$$(SO$$_4$$)$$_3$$ + 3H$$_2$$O
The above reaction was studied at 300 K by monitoring the concentration of FeSO$$_4$$ in which initial concentration was 10 M and after half an hour became 8.8 M. The rate of production of Fe$$_2$$(SO$$_4$$)$$_3$$ is _______ $$\times 10^{-6}$$ mol L$$^{-1}$$ s$$^{-1}$$ (Nearest integer)

The balanced reaction is:

$$\text{KClO}_3 + 6\text{FeSO}_4 + 3\text{H}_2\text{SO}_4 \rightarrow \text{KCl} + 3\text{Fe}_2(\text{SO}_4)_3 + 3\text{H}_2\text{O}$$

Now, find rate of disappearance of FeSO$$_4$$.

Change in concentration = 10 - 8.8 = 1.2 M

Time = 30 min = 1800 s

$$-\frac{d[\text{FeSO}_4]}{dt} = \frac{1.2}{1800} = \frac{2}{3000} = 6.667 \times 10^{-4} \text{ mol L}^{-1}\text{s}^{-1}$$

Next, relate to rate of production of Fe$$_2$$(SO$$_4$$)$$_3$$.

From stoichiometry: 6 mol FeSO$$_4$$ produces 3 mol Fe$$_2$$(SO$$_4$$)$$_3$$

$$\frac{d[\text{Fe}_2(\text{SO}_4)_3]}{dt} = \frac{3}{6} \times \frac{d[\text{FeSO}_4]}{dt} = \frac{1}{2} \times 6.667 \times 10^{-4}$$

$$= 3.333 \times 10^{-4} = 333 \times 10^{-6} \text{ mol L}^{-1}\text{s}^{-1}$$

The rate of production is $$333 \times 10^{-6}$$ mol L$$^{-1}$$ s$$^{-1}$$.