What is the mass of sediment formed when 200 g of 33.8% solution of silver nitrate with sodium chloride are poured?

Find the mass of silver nitrate AgNO3 in solution.

W = m (substance): m (solution) × 100%, hence

m (substance) = (m (solution) × W): 100%.

m (substance) = (200 g × 33.8%): 100% = 67.6 g.

Find the amount of magnesium nitrate AgNO3 by the formula:

n = m: M.

M (AgNO3) = 170 g / mol.

n = 67.6 g: 170 g / mol = 0.398 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

NaCl + AgNO3 = AgCl ↓ + NaNO3.

According to the reaction equation, there is 1 mol of AgCl per mole of AgNO3. Substances are in quantitative ratios 1: 1.

The amount of substance will be equal.

n (AgCl) = n (AgNO3) = 0.398 mol.

Let us find the mass of AgCl.

M (AgCl) = 143.5 g / mol.

m = n × M.

m = 143.5 g / mol × 0.398 mol = 57.11 g.

Answer: 57.11 g.