n_O={25,3.9,486\%}/{16}=0,15(mol)

Quy A về Na:x(mol),Na_2O:y(mol),Ba:z(mol)

->23x+62y+137z=25,3(1)

y=n_O=0,15(2)

2Na+2H_2O->2NaOH+H_2

Na_2O+H_2O->2NaOH

Ba+2H_2O->Ba(OH)_2+H_2

Theo PT: n_{H_2}=0,5x+z=0,15(3)

(1)(2)(3)->x=0,1;y=0,15;z=0,1

Theo pT: n_{Ba(OH)_2}=z=0,1(mol);n_{NaOH}=x+2y=0,4(mol)

m_{dd\ spu}=25,3+100-0,15.2=125(g)

C\%_{Ba(OH)_2}={0,1.171}/{125}.100\%=13,68\%

C\%_{NaOH}={0,4.40}/{125}.100\%=12,8\%