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Question

Number of moles of K2Cr2O7 in acidic medium required to oxidise one mole of Cu3P to CuSO4 and H3PO4 is:

A
611
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B
53
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C
35
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D
116
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Solution

The correct option is D 116
Chemical reaction involved:

11 K2Cr2O7+6 Cu3P+62 H2SO4
11 Cr2(SO4)3+18 CuSO4+6 H3PO4+
11 K2SO4+53 H2O

Given moles of Cu3P=1

Salts react in such a way that two or more atoms in the salt undergo a change in oxidation state as a result of either oxidation or reduction. In such a case, we can calculate the n - factor of the salt as the total increase or decrease in oxidation state per mole of the salt.

The oxidation state of Cu changes from +1 to +2 and the oxidation state of P changes from -3 to +5 (both Cu and P are oxidised). n-factor of Cu3P=3|21|+1|5(3)|
=11

For the given reaction,
Gram equivalents of K2Cr2O7= Gram equivalent of Cu3P

Number of moles × n-factor = Number of moles × n-factor.
n×6=1×11

n=116

Thus ,116 moles of K2Cr2O7 are required.
So, option (A) is the correct answer.

Additional Information:

Reaction Involved:

Cr2O27+Cu3PCr3++PO24+Cu2+

O.S Cr=+6 Cu=+1 Cr=+3 P=+5 Cu=+2
O=2 P=3 O=2

n-factor based on Cr2O27=(ΔO.S)×atoms of Cr in Cr2O27=(63)×2=6

n-factor based on Cu=(ΔO.S)×atoms of Cu in Cu3P=(21)×3=3

n-factor based on P=(ΔO.S)×atoms of P in Cu3P=[5(3)×1]=8

Thus, n-factor of Cu3P=11

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