The ration of coefficients of `HNO_3,Fe(NO_3) and NH_4NO_3` in the following redox equation.
`Fe + HNO_3 rarr Fe(NO_3)_2+NH_4NO_3 " in " +H_2O` the balance form will be

Options

A. `1:10:4`
B. `10:4:1`
C. `4:10:1`
D. `10:1:4`

Answer

B

Explanation

`Fe+HNO_(3)rarrFe(NO_(3))_(2)+NH_(4)NO_(3)+H_(2)O`
Half reaction method :
`Fe rarr Fe^(2+)+2e^(-)..(1)xx4`
`NO_(3)^(-)+10H^(o+)+8e^(Theta)rarrNH_(4)^(o+)+3H_(2)O.....(2)`
Equation `(1)xx4+(2)`
`4Fe+NO_(3)^(-)+10H^(o+)rarr4fe^(2+)+NH_(4)^(o+)+3H_(2)O`
`"Add"^(n)` of spectator ion
`4Fe+10HNO_(3)rarr4Fe(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O`