KClO _ 3 + 6 FeSO _ 4 + 3 H _ 2 SO _ 4 → KCl + 3 Fe _ 2 ( SO _ 4 )...

{KClO}_{3} + 6{FeSO}_{4} + 3{H}_{2}{SO}_{4} \to KCl + 3{Fe}_{2} ( {SO}_{4}{)}_{3} + 3{H}_{2}O$</p><p>From the above equation, rate of decomposition${FeSO}_{4}$and rate of production of${Fe}_{2}{\left({SO}_{4}\right)}_{3} can be written as follows,

$\dfrac{1}{6} \times Rateof{decomposition}_{{FeSO}_{4}} = \dfrac{1}{3} \times Rateof{production}_{{Fe}_{2}{\left({SO}_{4}\right)}_{3}}$

Rate of decomposition of ${FeSO}_{4}$

$\dfrac{Changeinconcentration}{timeinseconds} = \dfrac{( 10 - 8 . 8 )}{30 \times 60}$

$= \dfrac{1 . 2}{1800}mol{L}^{- 1}{s}^{- 1}$

Rate of production of ${Fe}_{2}{\left({SO}_{4}\right)}_{3}$

$= \dfrac{0 . 6}{1800} = 3 . 33 \times {10}^{- 4}$

$= 333 . 3 \times {10}^{- 6}{mol}^{- 1}{s}^{- 1}$