A 5.1% solution of silver nitrate weighing 1000 g was subjected to electrolysis. At the same time, 10.8 g of the substance

A 5.1% solution of silver nitrate weighing 1000 g was subjected to electrolysis. At the same time, 10.8 g of the substance was released at the cathode. Then, 500 g of a 13.5% solution of copper (II) chloride was added to the electrolytic cell, and the solution was again subjected to electrolysis until 8.96 L (n.o.) gas was released at the anode. What are the mass fractions of substances in the final solution?

Let’s compose the equation for electrolysis of a silver nitrate solution:
4AgNO3 + 2H2O = 4Ag + O2 ↑ + 4HNO3
Let’s calculate the amount of AgNO3 in the initial solution (n0), electrolyzed (n1) and remaining in the solution (n2):
n0 (AgNO3) = (1000 ∙ 0.051) / 170 = 0.3 mol
n1 (AgNO3) = n (Ag) = 10.8 / 108 = 0.1 mol
n2 (AgNO3) = 0.3 – 0.1 = 0.2 mol
The remaining AgNO3 reacts with the added copper (II) chloride:
2AgNO3 + СuCl2 = 2AgCl ↓ + Cu (NO3) 2
Let us calculate the amount of CuCl2 that was added to the AgNO3 (n0) solution, reacted with AgNO3 (n1), and remained in the solution after the (n2) reaction:
n0 (CuCl2) = (500 ∙ 0.135) / 135 = 0.5 mol
n1 (CuCl2) = 0.5n (AgNO3) = 0.5 ∙ 0.2 = 0.1 mol
n2 (CuCl2) = 0.5 – 0.1 = 0.4 mol
Let’s compose the equation for electrolysis of a CuCl2 solution:
CuCl2 = Cu + Cl2 ↑
Suppose that during electrolysis all the remaining copper (II) chloride, i.e. 0.4 mol, decomposed. This produces 0.4 mol of chlorine. Let’s determine the volume of gas:
V (Cl2) = 0.4 ∙ 22.4 = 8.96 mol
This corresponds to the condition of the problem, therefore, other electrolytic processes (electrolysis of Cu (NO3) 2, water) did not occur.
We determine the amount of substance and the mass of substances in solution after the second electrolysis:
n (НNO3) = n1 (AgNO3) = 0.1 mol
m (НNO3) = 0.1 ∙ 63 = 6.3 g
n (Cu (NO3) 2) = 0.5n2 (AgNO3) = 0.1 mol
m (Cu (NO3) 2) = 0.1 ∙ 188 = 18.8 g
We calculate the mass of the solution and the mass fractions of substances in it:
m (solution) = m (solution AgNO3) + m (solution CuCl2) –m (Ag) –
m (Сu) –m (Cl2) – m (AgCl) – m (О2)
m (Ag) = 0.1 ∙ 108 = 10.8 g
m (О2) = 0.025 ∙ 32 = 0.8 g
m (Cu) = 0.4 ∙ 64 = 25.6 g
m (Cl2) = 0.4 ∙ 71 = 28.4 g
m (AgCl) = 0.2 ∙ 143.5 = 28.7 g
m (solution) = 1000 + 500 – 10.8 – 0.8 – 25.6 – 28.4 – 28.7 = 1405.7 g
ɷ (НNO3) = (6.3 ∙ 100%) / 1405.7 = 0.45%
ɷ (Cu (NO3) 2) = (18.8 ∙ 100%) / 1405.7 = 1.34%
Answer: ɷ (НNO3) = 0.45%; ɷ (Cu (NO3) 2) = 1.34%