The equivalent weight of HCl in the given reaction K2Cr2O7 +14HCl → 2KCl + 2CrCl3 + 3Cl2 + 7H2O is
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a
16.25
b
36.5
c
73
d
85.1
answer is D.
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Detailed Solution
K2Cr2O7+14H+1Cl-1→2kCl-1+2CrCl-13+3Cl20+7H2OTotal charge of 'Cl' on LHS = -14Total charge of 'Cl' on RHS = -8Increases in Oxidation number/14mole of HCl = 6MW of HCl = 36.5EHCl=14(36.5)6=85.1