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(C) From the stoichiometry of the reaction: $2 K_{2}Cr_{2}O_{7} + 8 H_{2}SO_{4} + 3 C_{2}H_{6}O$ $\rightarrow 2 Cr_{2}(SO_{4})_{3} + 3 C_{2}H_{4}O_{2}

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) From the stoichiometry of the reaction: $2 K_{2}Cr_{2}O_{7} + 8 H_{2}SO_{4} + 3 C_{2}H_{6}O$ $ightarrow 2 Cr_{2}(SO_{4})_{3} + 3 C_{2}H_{4}O_{2} + 2 K_{2}SO_{4} + 11 H_{2}O$.The rate of reaction is given by: $	ext{Rate} = -\frac{1}{3} \frac{d[C_{2}H_{6}O]}{dt} = \frac{1}{2} \frac{d[Cr_{2}(SO_{4})_{3}]}{dt}$.Given,$\frac{d[Cr_{2}(SO_{4})_{3}]}{dt} = 2.67 \ mol \ min^{-1}$.Therefore,$-\frac{d[C_{2}H_{6}O]}{dt} = \frac{3}{2} 	imes \frac{d[Cr_{2}(SO_{4})_{3}]}{dt}$.$-\frac{d[C_{2}H_{6}O]}{dt} = \frac{3}{2} 	imes 2.67 = 4.005 \ mol \ min^{-1}$.The nearest integer is $4$.

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