Redox and Equivalent ConceptMedium
Question
The number of moles of Cr2O72− needed to oxidize 0.136 equivalent of N2H5+ through the reaction
N2H5+ + Cr2O72− → N2 + Cr3+ + H2O is
Options
A.0.023
B.0.091
C.0.136
D.0.816
Solution
$n_{eq}Cr_{2}O_{7}^{2 -} = n_{eq}N_{2}H_{5}^{+}$
$\Rightarrow n \times 6 = 0.136 \Rightarrow n = 0.0226$
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