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Question

6×103molK2Cr2O7reacts completely with 9×103 mol Xn to give XO3 and Cr+3 The valve of n is:


  1. 2

  2. 1

  3. 3

  4. 5


Solution

The correct option is B

1


K2Cr2O7 +Xn+ XO3Cr+3

Oxidation state of Chromium inK2Cr2O7=+6.Oxidation state of X inXO3=+5Change in oxidation state of X=(5n)

6×6×103 = (5-n)×9×103

4= 5-n

N =1

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