The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
Moles of silver = 16.2g / 107.8g/mol = 0.15mol
From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.
Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g
Hence required mass of silver nitrate is 25.48g.
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