Assuming Fe to be Fe(II):
Fe + AgNo3 ---> Fe(No3)2 + Ag
Fe + 2AgNo3 ---> Fe(No3)2 + 2Ag
Assuming Fe to be Fe(III):
Fe + AgNo3 ---> Fe(No3)3 + Ag
Fe + 3AgNo3 ---> Fe(No3)3 + 3Ag
I am not sure which form of iron (Fe II or Fe III) you need to answer this question but I am assuming it's iron (II). This will give you an general sense of how to solve these types of equations.
This is an oxidaton-reduction equation: Oxidation reagent is the one that accepts the electrons and the reduction is one that donates the electrons. Based on a standard reduction potential table, iron is the oxidation reagent because as you go up the table the easier the element accepts the electrons. Therefore, Ag or silver is the reducing agent.
Fe2+ + 2e- = Fe
2Ag = 2Ag + e-
Combined this equations to get the reactants and to balance the equation.
So we the formula for iron (II) and silver nitrate which is:
Fe(II) + AgNO3
Now we add the products:
Fe(II) + AgNO3 ----> FeNO3 + Ag(s)
Now to balance the equation, must consider the following Ag has two moles (Remember: 2Ag = 2Ag + e-) meaning that Ag needs two electrons from nitrate to from a stable molecule. Iron (II) is also 2+ meaning that it also needs two electrons from nitrate in the products. Therefore, the equation looks like so:
Fe (II) + 2AgNO3 -----> Fe(NO3)2 + 2Ag
For iron (III) the equations looks like this:
Fe (III) + 3AgNO3 -----> Fe(NO3)3 + 3Ag