`n_{H_2} = (2,479)/(24,79) = 0,1` mol

Đặt `n_K = n_{Na} = x` mol; `n_{Na_2O} = y` mol

`=> 39x + 23x + 62y = 18,6`

`=> 62x + 62y = 18,6g(1)`

`2K + 2H_2O -> 2KOH + H_2(1)`

`2Na + 2H_2O -> 2NaOH + H_2(2)`

`Na_2O + H_2O -> 2NaOH(3)`

Theo PTHH `(1),(2)`

`n_{H_2} = 1/2n_K + 1/2n_{Na} = 1/2x + 1/2x = 0,1Mol`

`=> x = 0,1`

Thay `x = 0,1` vào PT(1)

`=> 62 . 0,1 + 62y = 18,6`

`=> y = 0,2`

Bảo Toàn Khối Lượng

`=> m_{dd sau} = 18,6 + 181,6 - 0,1. 2 = 200g`

Theo PTHH (1)

`n_{KOH} = n_{K} = x = 0,1` mol

Theo PTHH (2), (3)

`n_{NaOH} = n_{Na} + 2n_{Na_2O} = x + 2y = 0,1 + 2 . 0,2 = 0,5` mol

`C%_{dd KOH} = (0,1 . 56)/(200) . 100% = 2,8%`

`C%_{dd NaOH} = (0,5 . 40)/(200) . 100% = 10%`

`Hh:K:x(mol);Na:x(mol);Na_2O:z(mol)`

`->39x+23x+62z=18,6(1)`

`2K+2H_2O->2KOH+H_2`

`2Na+2H_2O->2NaOH+H_2`

`Na_2O+H_2O->2NaOH`

Theo pt:

`n_{H_2}=0,5x+0,5x=x={2,479}/{24,79}=0,1(2)`

`(1)(2)->x=0,1;y=0,2`

`->n_{KOH}=x=0,1(mol);n_{NaOH}=x+2y=0,5(mol)`

`m_{dd\ A}=18,6+181,6-0,1.2=200(g)`

`C\%_{KOH}={0,1.56}/{200}.100\%=2,8\%`

`C\%_{NaOH}={0,5.40}/{200}.100\%=10\%`