Find the mass of silver chloride, which was obtained by the interaction of 115 g of silver nitrate with calcium chloride.

1. The interaction of silver nitrate with calcium chloride occurs according to the equation:

2AgNO3 + CaCl2 = 2AgCl + Ca (NO3) 2;

2.Calculate the chemical amount of silver nitrate:

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

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

n (AgNO3) = 115: 170 = 0.6765 mol;

3.According to the chemical reaction equation, the amounts of silver nitrate and chloride will be the same:

n (AgCl) = n (AgNO3) = 0.6765 mol;

4. find the mass of the obtained silver chloride:

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

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

m (AgCl) = 0.6765 * 143.5 = 97.08 g.

Answer: 97.08 g.