When an excess of sodium chloride was added to the silver nitrate solution

When an excess of sodium chloride was added to the silver nitrate solution, a precipitate weighing 1.44 g was formed. Find the mass of silver nitrate.

Let’s implement the solution:
1. We compose the reaction equation:
AgNO3 + NaCl = AgCl + NaNO3 – ion exchange reaction, silver chloride is released in the precipitate;
2. Let’s make calculations using the formulas:
M (AgCl) = 107.8 + 35.5 = 143.3 g / mol;
M (AgNO3) = 107.8 + 14 + 16 * 3 = 169.8 g / mol;
3. Determine the number of moles of silver chloride, if its mass is known:
Y (AgCl) = m / M = 1.44 / 143.3 = 0.01 mol;
4. Determine the amount of moles of silver nitrate by the formula:
Y (AgNO3) = 1 * 0.01 / 1 = 0.01 mol;
M (AgNO3) = Y * M = 0.01 * 169.8 = 1.69 g.
Answer: silver nitrate weighing 1.69 g enters the reaction.