Question

number of moles of K2 cr2 O7 reduced by 1 mole of SN 2 + ion is

Answer 1

Answer:  K2Cr2O7 ---------> 2K+ + CrO72-  

Cr2O72−(aq) + 14H+ + 6e− → 2Cr3+(aq) + 7H2O (source wiki)  

Sn2+ ------------> Sn4+ +2e-  

You can see each Cr2O72- ion gains 6 electrons and Sn+2 loses two electrons each to form Sn4+  

therefore you need 3 moles of Sn+2 to reduce 1 mole of K2Cr2O7  

( so your answer is a 1/3 mole of K2Cr2O7)  

full equation looks like this  

Cr2O72- + 3Sn2+ +14H+ -------------------> 3Sn4+ + 2Cr3+ + 7H2O  

also K2Cr2O7 = potassium dichromate and Sn = tin