AgNO3 solution was added to the HCl solution. A precipitate of 0.287 grams was formed.

AgNO3 solution was added to the HCl solution. A precipitate of 0.287 grams was formed. Determine the amount of AgNO3 substance that has reacted.

The formed precipitate is AgCl, silver chloride.

Let’s write the reaction equation:

AgNO3 + HCl = AgCl ↓ + HNO3

Let’s find the amount of silver chloride substance:

v (AgCl) = m (AgCl) / M (AgCl) = 0.287 / 143.5 = 0.002 (mol).

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

v (AgNO3) = v (AgCl) = 0.002 (mol).

Thus, the amount of silver nitrate substance that has reacted is:

v (AgNO3) = 0.002 (mol).

Answer: 0.002 mol.