Calculate the mass of sodium chloride that reacts with a 170 g silver nitrate solution with a mass fraction of 5%.

1. Let’s compose the equation of interaction of sodium chloride with silver nitrate:

NaCl + AgNO3 = AgCl ↓ + NaNO3;

2.Calculate the mass of silver nitrate:

m (AgNO3) = w (AgNO3) * m (solution) = 0.05 * 170 = 8.5 g;

3.Calculate the chemical amount of AgNO3:

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

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

n (AgNO3) = 8.5: 170 = 0.05 mol;

4. find the chemical amount of sodium chloride:

n (NaCl) = n (AgNO3) = 0.05 mol;

5.Calculate the mass of NaCl:

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

M (NaCl) = 23 + 35.5 = 58.5 g / mol;

m (NaCl) = 0.05 * 58.5 = 2.925 g.

Answer: 2.925 g.