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