What mass is the precipitate formed when the solution containing 34 g of silver nitrate and

What mass is the precipitate formed when the solution containing 34 g of silver nitrate and the sodium chloride solution is poured?

The equation for the reaction of silver nitrate with sodium chloride:

AgNO3 + NaCl = AgCl + NaNO3

Amount of silver nitrate substance:

v (AgNO3) = m (AgNO3) / M (AgNO3) = 34/170 = 0.2 (mol).

According to the reaction equation, 1 mol of AgCl is formed per 1 mol of AgNO3, therefore:

v (AgCl) = v (AgNO3) = 0.2 (mol).

Thus, the mass of the precipitate of silver chloride formed is:

m (AgCl) = v (AgCl) * M (AgCl) = 0.2 * 143.5 = 28.7 (g).

Answer: 28.7 (d).