An excess of hydrochloric acid solution was added to 1000 g of a 3.4% silver nitrate solution.

An excess of hydrochloric acid solution was added to 1000 g of a 3.4% silver nitrate solution. Determine the mass of the sediment formed.

1. Let’s write down the reaction equation:

AgNO3 + HCl = AgCl + HNO3.

2. Find the amount of silver nitrate:

m (AgNO3) = m (solution) * ω (AgNO3) / 100% = 1000 g * 3.4% / 100% = 34 g.

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

3. According to the reaction equation, we find the amount of silver chloride, and then its mass:

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

m (AgCl) = n (AgCl) * M (AgCl) = 0.2 mol * 143.5 g / mol = 28.7 g.

Answer: m (AgCl) = 28.7 g.