Calculate the mass of the precipitate formed by the interaction of 102.5 g of zinc chloride with silver nitrate.

Let’s compose the reaction equation for the interaction of zinc chloride and silver nitrate:
ZnCl2 + 2AgNO3 = Zn (NO3) 2 + 2AgCl;

Let’s find the molecular weight of zinc chloride:
65 + 35.5 * 2 = 136 g / mol;

Let’s find the amount of substance in zinc chloride:
102.5 / 136 = 0.754 mol;

Silver chloride precipitates. Its amount of substance is 1.508 mol, since it has a coefficient 2 times greater than that of zinc chloride. The molecular weight of silver chloride is 143.5 g / mol. Let’s find the mass of the sediment:
143.5 * 1.508 = 216.398 grams;

Answer: The mass of the sediment is 216.398 grams.