200 mL of x times 10^{-3} M potassium dichr | vedclass

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Question

(C) The balanced redox reaction between dichromate ion and ferrous ion in acidic medium is: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6F

Vedclass · Accepted Answer

$375$

Vedclass · Answer

(C) The balanced redox reaction between dichromate ion and ferrous ion in acidic medium is: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} ightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ From the stoichiometry, $1 	ext{ mole of } K_2Cr_2O_7$ reacts with $6 	ext{ moles of } Fe^{2+}$. Number of moles of Mohr's salt = $	ext{Molarity} 	imes 	ext{Volume (in L)} = 0.6 	imes 0.750 = 0.45 	ext{ moles}$. Number of moles of $K_2Cr_2O_7$ required = $\frac{0.45}{6} = 0.075 	ext{ moles}$. Given $200 \ mL$ $(0.2 \ L)$ of $x 	imes 10^{-3} \ M$ solution contains $0.075 	ext{ moles}$. $0.2 	imes x 	imes 10^{-3} = 0.075$ $x 	imes 10^{-3} = \frac{0.075}{0.2} = 0.375$ $x = 0.375 	imes 10^3 = 375$.

$200 \ mL$ of $x \times 10^{-3} \ M$ potassium dichromate $(K_2Cr_2O_7)$ is required to oxidise $750 \ mL$ of $0.6 \ M$ Mohr's salt $(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O)$ solution in acidic medium. Here $x =$ ?

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