The same amount of electricity was passed through electrolyzers with aqueous solutions of mercury (II)

The same amount of electricity was passed through electrolyzers with aqueous solutions of mercury (II) nitrate and silver nitrate. At the same time, mercury with a mass of 401.2 g was released. What is the mass of the released silver? Give the equations of electrode processes.

Given:
Hg (NO3) 2
AgNO3
m (Hg) = 401.2 g

To find:
m (Ag) -?

Decision:
1) Using the periodic system of D.I.Mendeleev, calculate the molar masses of Hg and Ag:
M (Hg) = Mr (Hg) = Ar (Hg) = 201 g / mol;
M (Ag) = Mr (Ag) = Ar (Ag) = 108 g / mol;
2) Calculate the mass of Ag:
m (substances) = (I * t / F) * (M / z);
I1 * t1 / F = I2 * t2 / F;
m (Hg) * z (Hg) / M (Hg) = m (Ag) * z (Ag) / M (Ag);
m (Ag) = m (Hg) * z (Hg) * M (Ag) / (M (Hg) * z (Ag)) = 401.2 * 2 * 108 / (201 * 1) = 431.1 g …

Answer: The mass of Ag is 431.1 g.