4HCl + PbO₂ ⇒ PbCl₂+ Cl₂ + 2H2O
2Cl⁻¹ -2e⁻ ⇒ Cl₂⁰ se oxideaza (ag.red) ||2|1
Pb⁺⁴ +2e⁻ ⇒ Pb⁺² se reduce (ag.ox) ||2|1
MPbO₂ = 239g/mol
n = m/M = 800/239 = 3,3472 moli PbO₂
4 moli HCl.............1 mol PbO₂
x = 13,389 moli........3,3472 moli
C = n/Vs => Vs =n/C = 13,389/6 = 2,23152 L sol. HCl
