What mass of iron is required to replace over from 8.00 g of silver nitrate dissolved in water

Assume that Fe(II) is used
The chemical equation will be as follows:
2 AgNO3 (aq) + Fe2+ ---> Fe(NO3)2 (aq) + 2 Ag 

From the periodic table:
molar mass of silver = 107.87 gm
molar mass of nitrogen = 14 gm
molar mass of oxygen = 16 gm

molar mass of silver nitrate = 107.87 + 14 + 3(16) = 169.87 gm

number of moles in 8 gm = mass / molar mass = 8 / 169.87 = 0.047 moles

From the balanced chemical equation, we can see that two moles of silver nitrate reacts with one mole of iron(II), therefore, the ratio is 2:1

number of moles of Fe(II) required to react with 0.047 moles of silver nitrate can be calculated as follows:
number of moles of Fe(II) = 0.047 x (1/2) = 0.0235 moles

To get the mass, multiply the number of moles by the molar mass of iron as follows:
mass of required iron = 0.0235 x 55.845 = 1.312 gm

So3 that’s what i think