What mass of metal is formed by the interaction of iron with silver nitrate (1) in a solution

What mass of metal is formed by the interaction of iron with silver nitrate (1) in a solution weighing 320 g with a mass fraction of salt of 0.2?

1. Iron is a more active metal in comparison with silver, therefore it displaces it from the salt solution:

Fe + 3AgNO3 = Fe (NO3) 3 + 3Ag;

2. Calculate the mass of silver nitrate:

m (AgNO3) = w (AgNO3) * m (solution) = 0.2 * 320 = 64 g;

3. Let’s calculate the chemical amount of silver nitrate:

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

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

n (AgNO3) = 64: 170 = 0.3765 mol;

4. Let’s set the amount of the formed metal:

n (Ag) = n (AgNO3) = 0.3765 mol;

5. Find the mass of silver:

m (Ag) = n (Ag) * M (Ag) = 0.3765 * 108 = 40.66 g.

Answer: 40.66 g.