170 g of a silver nitrate solution was mixed with an excess of sodium chloride solution. A precipitate with a mass of 8.61 g

170 g of a silver nitrate solution was mixed with an excess of sodium chloride solution. A precipitate with a mass of 8.61 g fell out. Calculate the mass fraction of salt in the silver nitrate solution.

1. Silver cations precipitate chlorine anions from the solution, that is, as a result of the reaction, a precipitate of silver chloride precipitates:

AgNO3 + NaCl = AgCl ↓ + NaNO3;

2. Calculate the chemical amount of the precipitate formed:

n (AgCl) = m (AgCl): M (AgCl);

M (AgCl) = 108 + 35.5 = 143.5 g / mol;

n (AgCl) = 8.61: 143.5 = 0.06 mol;

3. From the ratio of substances in the equation, we find the number of moles of silver nitrate:

n (AgNO3) = n (AgCl) = 0.06 mol;

4. Set the mass of the original silver salt:

m (AgNO3) = n (AgNO3) * M (AgNO3);

M (AgNO3) = 108 + 14 + 3 * 16 = 170 g / mol;

m (AgNO3) = 0.06 * 170 = 10.2 g;

5. Then the mass fraction of nitrate in the initial solution:

w (AgNO3) = m (AgNO3): m (solution AgNO3) = 10.2: 170 = 0.06 or 6%.

Answer: 6%.