# 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

**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. To the resulting solution was added 53 g of a 10% sodium carbonate solution. Determine the mass fraction of copper (II) sulfate in the resulting 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 us find the amount of substance and the mass of CuSO4 remaining in the solution after electrolysis:

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 ∙ М (CuSO4) = 0.4 mol ∙ 160 g / mol = 64 g.

Let’s find the amount of the substance of the formed sulfuric acid:

n (Н2SO4) = n (CuSO4) reac. = 0.1 mol.

Find the mass and amount of the added sodium carbonate substance:

m (Na2CO3) = m (Na2CO3) solution ∙ ω (Na2CO3) = 53 g ∙ 0.1 = 5.3 g

n (Na2CO3) = m (Na2CO3) / M (Na2CO3) = 5.3g / 106g / mol = 0.05 mol.

When adding sodium carbonate, simultaneous reactions are possible:

2CuSO4 + 2Na2СО3 + Н2O → (СuОН) 2СО3 ↓ + СО2 ↑ + 2Na2SO4 (1)

Н2SO4 + Na2CO3 → СО2 ↑ + Н2О + Na2SO4 (2)

Because sulfuric acid in excess, then it immediately dissolves the basic copper carbonate formed by reaction (1) with the formation of CuSO4 and the release of CO2:

(СuОН) 2СО3 + 2Н2SO4 → 2CuSO4 + СО2 ↑ + 3Н2О (3)

Thus, the amount of CuSO4 in the solution remains unchanged, and the total amount of released CO2 in reactions (2) and (3) is determined by the amount of sodium carbonate:

n (Na2CO3) = n (CO2) = 0.05 mol

Find the mass of the final solution:

mcon.r-pa = m (CuSO4) solution – m (Cu) –m (О2) + m (Na2СО3) – m (СО2)

m (О2) = n (О2) ∙ М (О2) = 0.05 mol ∙ 32 g / mol = 1.6 g.

n (Cu) = 2 ∙ n (O2) = 2 ∙ 0.05 mol = 0.1 mol.

m (Cu) = n (Cu) ∙ M (Cu) = 0.1 mol ∙ 64 g / mol = 6.4 g.

m (СО2) = n (СО2) ∙ М (СО2) = 0.05 mol ∙ 44 g / mol = 2.2 g.

mcon.r-pa = m (CuSO4) solution – m (Cu) –m (O2) + m (Na2CO3) – m (CO2) = 500 g – 6.4 g – 1.6 g + 53 g – 2, 2 g = 542.8 g.

Find the mass fraction of copper (II) sulfate in the resulting solution:

ω (CuSO4) end solution = m (CuSO4) rest / m end solution = 64g / 542.8 g = 0.118 (11.8%).

Answer: ω (CuSO4) = 11.8%.