Moles of K_2Cr_2O_7 used to oxidise 1 mole of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In $Fe_{0.92}O$,let the oxidation state of $Fe$ be $x$. Since the oxidation state of $O$ is $-2$,we have $0.92x + (-2) = 0$,so $x = \frac{2}{0.92}

Vedclass · Accepted Answer

$\frac{0.76}{6}$

Vedclass · Answer

(C) In $Fe_{0.92}O$,let the oxidation state of $Fe$ be $x$. Since the oxidation state of $O$ is $-2$,we have $0.92x + (-2) = 0$,so $x = \frac{2}{0.92}$.When $1$ $mole$ of $Fe_{0.92}O$ is oxidized to $Fe^{3+}$,the change in oxidation state of $Fe$ is $(3 - \frac{2}{0.92}) 	imes 0.92 = 3 	imes 0.92 - 2 = 2.76 - 2 = 0.76$.$K_2Cr_2O_7$ acts as an oxidizing agent where $Cr$ changes from $+6$ to $+3$,a change of $3$ per $Cr$ atom,or $6$ per $K_2Cr_2O_7$ molecule.Equating the total change in oxidation states: $n 	imes 6 = 0.76$.Therefore,$n = \frac{0.76}{6}$ $moles$.

Moles of $K_2Cr_2O_7$ used to oxidise $1$ $mole$ of $Fe_{0.92}O$ to $Fe^{3+}$ are -

Similar Questions

Identify the correct statement for the reaction: $I_2 + KClO_3 \xrightarrow{\Delta} ICl + KIO_3$

Assertion : When a metal is treated with conc. $HNO_3$,it generally yields a nitrate,$NO_2$,and $H_2O$.
Reason : Conc. $HNO_3$ reacts with a metal and first produces a metal nitrate and nascent hydrogen. The nascent hydrogen then further reduces $HNO_3$ to $NO_2$.

$KMnO_4$ acts as an oxidising agent in alkaline medium. When alkaline $KMnO_4$ is treated with $KI$,iodide ion is oxidized to

Assign $A, B, C, D$ from the given type of reaction for the following chemical equation:
$3Fe_{(s)} + 4H_2O_{(g)} \xrightarrow{\Delta} Fe_3O_{4(s)} + 4H_{2(g)}$

Which of the following reactions involves oxidation-reduction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module