When electrolysis was carried out on 360 g of a 15% solution of copper (II) chloride, the process was stopped when 4.48 liters of gas were released at the anode. A portion weighing 66.6 g was taken from the resulting solution. Calculate the mass of a 10% sodium hydroxide solution required for the complete precipitation of copper ions from the selected portion of the solution.
Let us write the equation for electrolysis of an aqueous solution of copper (II) chloride:
CuCl2 → (electrolysis) Cu + Cl2
Let’s find the mass and amount of the substance of the initial copper (II) chloride:
m (CuCl2) ref. = m (CuCl2) solution ∙ ω (CuCl2) = 360 g ∙ 0.15 = 54 g
n (CuCl2) ref. = m (CuCl2) ref. / M (CuCl2) = 54 g / 135 g / mol = 0.4 mol.
Let’s find the amount of chlorine released at the anode:
n (Cl2) = V (Cl2) / Vm = 4.48 L / 22.4 L / mol = 0.2 mol.
Find the amount of substance and the mass of CuCl2 remaining in the solution:
n (CuCl2) reac. = n (Cl2) = 0.2 mol.
n (CuCl2) rest. = n (CuCl2) ref. – n (CuCl2) reac. = 0.4 mol – 0.2 mol = 0.2 mol.
m (CuCl2) rest. = n (CuCl2) rest ∙ M (CuCl2) = 0.2 mol ∙ 135 g / mol = 27 g
Find the mass of the final solution:
mkon.r-pa = m (CuCl2) solution – m (Cl2) – m (Cu)
m (Cl2) = n (Cl2) ∙ М (Cl2) = 0.2 mol ∙ 71 g / mol = 14.2 g.
m (Cu) = n (Cu) ∙ M (Cu) = 0.2 mol ∙ 64 g / mol = 12.8 g.
mcone solution = m (CuCl2) solution – m (Cl2) – m (Cu) = 360 g – 14.2 g – 12.8 g = 333 g
ω (CuCl2) end. = m (CuCl2) rest / mcon.r-pa = 27 g / 333 g = 0.0811
Let us find the mass and amount of copper (II) chloride substance in the selected portion:
m (CuCl2) port. = mport solution. ∙ ω (CuCl2) end. = 66.6 g ∙ 0.0811 = 5.4 g
n (CuCl2) port. = m (CuCl2) port. / M (CuCl2) = 5.4 g / 135 g / mol = 0.04 mol.
CuCl2 + 2NaOH → Cu (OH) 2 + 2NaCl
n (NaOH) = 2 ∙ n (CuCl2) port. = 2 ∙ 0.04 mol = 0.08 mol.
m (NaOH) in islands = n (NaOH) ∙ М (NaOH) = 0.08 mol ∙ 40 g / mol = 3.2 g.
m (NaOH) solution = m (NaOH) in-va / ω (NaOH) = 3.2 g / 0.1 = 32 g.
Answer: m (NaOH) solution = 32 g.
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