What is the mass of the precipitate obtained by the interaction of 1.7 g of silver nitrate

What is the mass of the precipitate obtained by the interaction of 1.7 g of silver nitrate with copper chloride, taken in excess?

Let’s implement the solution:

In accordance with the condition of the problem, we compose the equation of the process:
2AgNO3 + CuCl2 = 2AgCl + Cu (NO3) 2 – ion exchange, silver chloride was formed in the precipitate;

Let’s make the calculations:
M (AgNO3) = 169.8 g / mol;

M (AgCl) = 143.3 g / mol.

Determine the amount of the original substance:
Y (AgNO3) = m / M = 1.7 / 169.8 = 0.01 mol;

Y (AgCl) = 0.01 mol since the amount of substances according to the equation is equal to 2 mol.

We find the mass of the sediment:
m (AgCl) = Y * M = 0.01 * 143.3 = 1.43 g

Answer: a precipitate of silver chloride weighing 1.43 g was obtained.