When merging a solution of silver nitrate with an excess of a solution of sodium chloride, a precipitate of 28.6 g mass formed. The mass of silver nitrate in the initial solution is
AgNO3 + NaCl = NaNO3 + AgCl
n (silver chloride) = m / M = 28.6 / 143.5 = 0.2 mol
the number of moles of silver nitrate = the number of moles of silver chloride, which means
m (silver nitrate) = M (silver nitrate) * n (silver chloride) = 170 * 0.2 = 34 g
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