0.5 g of pure iron wire was dissolved in excess of HCl in absence of air and was then heated with 0.25 g of `KNO_(3)`. The following reaction takes place,
`3FeCl_(2)+KNO_(2)+4"HCl"to3FeCl_(3)+KCl+2H_(2)O+NO`
When the reaction was over, the resulting solution was titrated against 0.1 N `K_(2)Cr_(2)O_(7)`. What volume of dichromate would be consumed? ATomic weight of `Fe=55.585` atomic weight of `K=39`?

Explanation

0.5 g of iron wire dissolved in HCl in the absence of air would give `(0.5)/(55.85)=8.95xx10^(-3) " mol of "FeCl_(2)`
`0.25 g of KNO_(3), i.e., (0.25)/(101)=2.475xx10^(-3)` " mol of "`KNO_(3)`
would oxidise `7.425xx10^(-3)` " mol of "`FeCl_(2)`.
`FeCl_(2)` remaining unoxidised is `(8.95-7.425)xx10^(-3)mol-=1.525xx10^(-3)` equivalent. If V " mL of " 0.1 N dichromate is consumed, then `(V)/(1000)xx0.1=1.525xx10^(-3)`
`V=(1.525xx10^(-3)xx10^(3))/(0.1)=15.25mL`