Question

How many moles of k2cr2o7 can be reduced by 1 mole of sn+2?

Answer 1

The balanced chemical reactions (ionic reactions)for reduction of K2Cr2O7 by Sn2+are:
Cr2O72- + 14H++ 6e-→2Cr3+ + 7H2O
( ​Sn2+ → Sn4+ + 2e- )    × 3

Balanced Net reaction : 3Sn2++ Cr2O72- + 14H+ → 3 Sn4+ + 2Cr3+ + 7H2O​

Thus according to the balanced reaction : 1 mole of Cr2O72- will be reduced by 3 moles of Sn2+
Thus 1 mole of Sn2+  will reduce = 1/3 moles of Cr2O72-
                                                            = 0.33 moles of Cr2O72-

Answer 2

$\frac{1}{3}$ = 0.333 moles of $K_2Cr_2O^{2-}_{7}$ can be reduced by 1 mole of $Sn^2^+$

First we have to write the  balanced net chemical equations for reduction of $K_2Cr_2O_7$ by $Sn^2^+$ is:

$K_2Cr_2O^{2-}_{7} + 14H^+ +3Sn^{2+} = 2CR^{3+} + Sn^{4+} + 7H_2O$

​Thus from this equation it is clear that,

1 mole of $K_2Cr_2O^{2-}_{7}$ will be reduced by 3 moles of $Sn^{2+}$

i.e, 1 mole of  $Sn^2^+$ will reduce = $\frac{1}{3}$ moles of  $K_2Cr_2O^{2-}_{7}$

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