KClO_3 + 6FeSO_4 + 3H_2SO_4 rightarrow KCl + | vedclass

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Question

(D) The balanced chemical equation is: $KClO_3 + 6FeSO_4 + 3H_2SO_4 \rightarrow KCl + 3Fe_2(SO_4)_3 + 3H_2O$.The rate of reaction $(ROR)$ is given by:

Vedclass · Accepted Answer

$333$

Vedclass · Answer

(D) The balanced chemical equation is: $KClO_3 + 6FeSO_4 + 3H_2SO_4 ightarrow KCl + 3Fe_2(SO_4)_3 + 3H_2O$.The rate of reaction $(ROR)$ is given by: $ROR = -\frac{1}{6} \frac{\Delta[FeSO_4]}{\Delta t} = \frac{1}{3} \frac{\Delta[Fe_2(SO_4)_3]}{\Delta t}$.Therefore,the rate of production of $Fe_2(SO_4)_3$ is: $\frac{\Delta[Fe_2(SO_4)_3]}{\Delta t} = -\frac{3}{6} \frac{\Delta[FeSO_4]}{\Delta t} = -\frac{1}{2} \frac{\Delta[FeSO_4]}{\Delta t}$.Given $\Delta[FeSO_4] = 8.8 \ M - 10 \ M = -1.2 \ M$ and $\Delta t = 30 \ 	ext{min} = 1800 \ s$.Rate of production $= -\frac{1}{2} 	imes \frac{-1.2 \ M}{1800 \ s} = \frac{0.6}{1800} \ M \ s^{-1} = 0.000333 \ M \ s^{-1}$.Converting to the required units: $0.000333 	imes 10^6 	imes 10^{-6} \ mol \ L^{-1} \ s^{-1} = 333 	imes 10^{-6} \ mol \ L^{-1} \ s^{-1}$.

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