An equation for the reaction of silver nitrate + iron(ii) sulphate + sulphuric acid?

Iron (II) sulfate and concentrated sulfuric acid can be used to test for contamination by nitrates.  This is called the (brown) ring test.  In your example, the sulfuric acid simultaneously reacts with both silver nitrate and iron (II) sulfate.  Iron (II) becomes oxidized to iron (III):

4 Ag(N03) + 4 H2SO4 + 4 FeSO4 --> 2Ag2SO4 + 4 HNO3 + 2 Fe2(SO4)3 + 2H2

(the coefficients appear to be doubled, but this is necessary for...

Iron (II) sulfate and concentrated sulfuric acid can be used to test for contamination by nitrates.  This is called the (brown) ring test.  In your example, the sulfuric acid simultaneously reacts with both silver nitrate and iron (II) sulfate.  Iron (II) becomes oxidized to iron (III):

4 Ag(N03) + 4 H2SO4 + 4 FeSO4 --> 2Ag2SO4 + 4 HNO3 + 2 Fe2(SO4)3 + 2H2

(the coefficients appear to be doubled, but this is necessary for balancing subsequent reactions)

The nitric acid that is formed (HNO3) then reacts:

4 HNO3 --> 2 H20 + 3 O2 + 4 NO

The iron (III) sulfate then reacts with the NO formed in the previous step:

2 Fe2(SO4)3 + 2 H2 + 4 NO --> 4[Fe(H2O)5NO]SO4 + 2 H2SO4

The [Fe(H20)5NO]SO4 causes the brown ring (indicating the presence of nitrates).

The overall reaction is:

4AgNO3 + 4H2SO4 + 4FeSO4 + 18H2O --> 4[Fe(H20)5NO]SO4 + 2Ag2SO4 + 3O2 + 2H2SO4