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.
Hg (NO3) 2
m (Hg) = 401.2 g
m (Ag) -?
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.
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