The mass of the precipitate formed by mixing a solution containing 34 g of silver nitrate and an excess of sodium chloride solution is

n (silver nitrate) = m / M = 34/170 = 0.2 mol
the number of moles of silver nitrate = the number of moles of silver chloride, which means
m (silver chloride) = M (silver chloride) * n (silver nitrate) = 28.7 g

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