6 - 10-3 molK 2 Cr 2 O 7 reacts completely with 9 - 10-3 mol X n to give XO 3-and Cr -3 The valve of n is-A- 2B- 1C- 3D- 5

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Introduction

6 × 10-3 molK...

Question

6 $\times$ $10^{- 3}$ mol $K_{2} C r_{2} O_{7}$ reacts completely with 9 $\times$ $10^{- 3}$ mol $X^{n}$ to give $X O_{3}^{-}$ and $C r^{+ 3}$ The valve of n is:

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Solution

The correct option is A
1

$K_{2} C r_{2} O_{7}$ + $X^{n +}$ $⟶$ $X O^{-}_{3}$ $C r^{+ 3}$

$Oxidation state of Chromium in K_{2} C r_{2} O_{7} = + 6. Oxidation state of X in X O_{3}^{-} = + 5 Change in oxidation state of X = (5 - n)$

$6 \times$ 6 $\times$ $10^{- 3}$ = (5-n) $\times$ 9 $\times$ $10^{- 3}$

4= 5-n

N =1

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Similar questions

6 $\times$ $10^{- 3}$ mol $K_{2} C r_{2} O_{7}$ reacts completely with 9 $\times$ $10^{- 3}$ mol of $X^{n} +$ to give $X O_{3}^{-}$ and $C r^{+ 3}$ The valve of n is:

6 \times 10^{- 3}

moles of

K_{2} C r_{2} O_{7}

reacts completely with

9 \times 10^{- 3}

moles of

X^{n +}

to give

X O_{3}^{-}

and

C r^{3 +}

. The value of n is:

6 $\times$ $10^{- 3}$ mol $K_{2} C r_{2} O_{7}$ reacts completely with 9 $\times$ $10^{- 3}$ mol of $X^{n} +$ to give $X O_{3}^{-}$ and $C r^{+ 3}$ The valve of n is: