During the electrolysis of 624 g of a 10% barium chloride solution, the process was stopped when 4.48 liters of gas were released

During the electrolysis of 624 g of a 10% barium chloride solution, the process was stopped when 4.48 liters of gas were released at the cathode. A portion weighing 91.41 g was taken from the resulting solution. Calculate the mass of a 10% sodium carbonate solution required for the complete precipitation of barium ions from the selected portion of the solution

Let us write the equation for electrolysis of an aqueous solution of barium chloride:
BaCl2 + 2H2O → (electrolysis) H2 + Cl2 + Ba (OH) 2
Let’s find the mass and amount of the substance of the original barium chloride:
m (BaCl2) ref. = m (ВaCl2) solution ∙ ω (ВaCl2) = 624 g ∙ 0.1 = 62.4 g
n (BaCl2) ref. = m (BaCl2) ref. / M (BaCl2) = 62.4g / 208g / mol = 0.3 mol.
Let’s find the amount of the substance released at the cathode of hydrogen:
n (H2) = V (H2) / Vm = 4.48 L / 22.4 L / mol = 0.2 mol.
Let’s find the amount of substance and the mass of the formed Вa (OH) 2:
n (Ba (OH) 2) = n (H2) = 0.2 mol.
m (Вa (OH) 2) = n (Вa (OH) 2) ∙ М (Вa (OH) 2) = 0.2 mol ∙ 171 g / mol = 34.2 g.
Let’s find the amount of the substance and the mass of BaCl2 remaining in the solution:
n (BaCl2) reac. = n (H2) = 0.2 mol.
n (BaCl2) rest. = n (BaCl2) ref. – n (BaCl2) reac. = 0.3 mol – 0.2 mol = 0.1 mol.
m (BaCl2) rest. = n (ВaCl2) rest. ∙ М (ВaCl2) = 0.1 mol ∙ 208 g / mol = 20.8 g.
Find the mass of the final solution:
mkon.r-pa = m (BaCl2) solution – m (H2) –m (Cl2)
m (H2) = n (H2) ∙ M (H2) = 0.2 mol ∙ 2 g / mol = 0.4 g.
n (Cl2) = n (H2) = 0.2 mol.
m (Cl2) = n (Cl2) ∙ М (Cl2) = 0.2 mol ∙ 71 g / mol = 14.2 g.
mcon.r-pa = m (BaCl2) solution – m (H2) – m (Cl2) = 624g – 0.4g – 14.2g = 609.4g
ω (BaCl2) end. = m (BaCl2) /mcon.p-pa = 20.8g / 609.4g = 0.0341
ω (Ba (OH) 2) end. = m (Ba (OH) 2) /mcon.p-pa = 34.2g / 609.4g = 0.0561
Let’s find the mass and amount of barium hydroxide substance in the selected portion:
m (Ba (OH) 2) port. = mport solution. ∙ ω (Ba (OH) 2) end. = 91.41 g ∙ 0.0561 = 5.13 g
n (Ba (OH) 2) port. = m (Ba (OH) 2) port. / M (Ba (OH) 2) = 5.13g / 171g / mol = 0.03 mol.
Find the mass and amount of the barium chloride substance in the selected portion:
m (BaCl2) port. = mport solution. ∙ ω (ВaCl2) rest. = 91.41 g ∙ 0.0341 = 3.12 g
n (BaCl2) port. = m (BaCl2) port. / M (BaCl2) = 3.12 g / 208 g / mol = 0.015 mol.
Ba (OH) 2 + Na2CO3 → BaCO3 + 2NaOH (1)
BaCl2 + Na2CO3 → BaCO3 + 2NaCl (2)
Let us find the mass of sodium carbonate solution required for precipitation of Ba2 + ions:
Figures (1): n (Na2CO3) 1 = n (Ba (OH) 2) port. = 0.03 mol
Equations (2): n (Na2CO3) 2 = n (ВaCl2) port = 0.015 mol
n (Na2CO3) = n (Na2CO3) 1 + n (Na2CO3) 2 = 0.03 mol + 0.015 mol = 0.045 mol
m (Na2CO3) in – va = n (Na2CO3) ∙ M (Na2CO3) = 0.045 mol ∙ 106 g / mol = 4.77 g
m (Na2CO3) p-pa = m (Na2CO3) in-wa / ω (Na2CO3) = 4.77 g / 0.1 = 47.7 g.
Answer: m (Na2CO3) solution = 47.7 g.

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