When conducting electrolysis of 500 g of a 16% solution of copper (II) sulfate, the process was stopped when 1.12 liters

When conducting electrolysis of 500 g of a 16% solution of copper (II) sulfate, the process was stopped when 1.12 liters of gas were released at the anode. A portion weighing 98.4 g was taken from the resulting solution. Calculate the mass of a 20% sodium hydroxide solution required for the complete precipitation of copper ions from the selected portion of the solution.

Let us write down the electrolysis equation for an aqueous solution of copper (II) sulfate:
2CuSO4 + 2H2O → (electrolysis) 2Cu + O2 + 2H2SO4
Let’s find the mass and amount of the substance of the initial copper (II) sulfate:
m (CuSO4) ref. = m (CuSO4) solution ∙ ω (CuSO4) = 500 g ∙ 0.16 = 80 g
n (CuSO4) ref. = m (CuSO4) ref. / M (CuSO4) = 80 g / 160 g / mol = 0.5 mol.
Let’s find the amount of substance released at the oxygen anode:
n (O2) = V (O2) / Vm = 1.12 L / 22.4 L / mol = 0.05 mol.
Let’s find the amount of substance and the mass of CuSO4 remaining in the solution:
n (CuSO4) reac. = 2 ∙ n (О2) = 2 ∙ 0.05 mol = 0.1 mol.
n (CuSO4) rest. = n (CuSO4) ref. – n (CuSO4) reac. = 0.5 mol – 0.1 mol = 0.4 mol.
m (CuSO4) rest. = n (CuSO4) rest ∙ M (CuSO4) = 0.4 mol ∙ 160 g / mol = 64 g.
Find the mass of the final solution:
mcon.r-pa = m (CuSO4) solution – m (O2) – m (Cu)
m (О2) = n (О2) ∙ М (О2) = 0.05 mol ∙ 32 g / mol = 1.6 g.
n (Cu) = n (CuSO4) reac. = 0.1 mol.
m (Cu) = n (Cu) ∙ M (Cu) = 0.1 mol ∙ 64 g / mol = 6.4 g.
mcone solution = m (CuSO4) solution – m (O2) – m (Cu) = 500 g – 1.6 g – 6.4 g = 492 g
n (H2SO4) = n (CuSO4) reac. = 0.1 mol.
m (H2SO4) = n (H2SO4) ∙ М (H2SO4) = 0.1 mol ∙ 98 g / mol = 9.8 g.
ω (CuSO4) end. = m (CuSO4) rest. / mc. p-pa = 64 g / 492 g = 0.13
ω (H2SO4) end. = m (H2SO4) /mcon.p-pa = 9.8g / 492g = 0.02
Let’s find the mass and amount of copper (II) sulfate substance in the selected portion:
m (CuSO4) port. = mport solution. ∙ ω (CuSO4) end. = 98.4 g ∙ 0.13 = 12.8 g
n (CuSO4) port. = m (CuSO4) port. / M (CuSO4) = 12.8 g / 160 g / mol = 0.08 mol.
Let’s find the mass and amount of barium hydroxide substance in the selected portion:
m (H2SO4) port. = mport solution. ∙ ω (H2SO4) end. = 98.4 g ∙ 0.02 = 1.968 g
n (H2SO4) port. = m (H2SO4) port. / M (H2SO4) = 1.968g / 98g / mol = 0.02 mol.
CuSO4 + 2NaOH → Cu (OH) 2 + Na2SO4 (1)
H2SO4 + 2NaOH → Na2SO4 + 2H2O (2)
Let us find the mass of sodium hydroxide required to precipitate Cu2 + ions:
From equation (1): n (NaOH) 1 = 2 ∙ n (CuSO4) porc. = 2 ∙ 0.08 mol = 0.16 mol.
From equation (2): n (NaOH) 2 = 2 ∙ n (H2SO4) port. = 2 ∙ 0.02 mol = 0.04 mol.
n (NaOH (reactive)) = n (NaOH (reactive) 1) + n (NaOH (reactive) 2) = 0.16 mol + 0.04 mol = 0.2 mol
m (NaOH) in islands = n (NaOH) ∙ М (NaOH) = 0.2 mol ∙ 40 g / mol = 8g.
m (NaOH) solution = m (NaOH) in-islands / ω (NaOH) = 8g / 0.2 = 40g
Answer: m (NaOH) solution = 40g.

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