C2:

1. Base tan trong nước là `NaOH, KOH, Ba(OH)_2`

Base không tan trong nước là `Zn(OH)_2, Fe(OH)_3`

2.

`NaOH: `Sodium Hydroxide

`KOH:` Potassium Hydroxide

`Ba(OH)_2:` Barium Hydroxide

`Zn(OH)_2:`Zinc Hydroxide

`Fe(OH)_3:` Iron `(III)` Hydroxide

C3:

PTHH:

`Fe + 2HCl -> FeCl_2 + H_2`

`n_{H_2} = (3,7185)/(24,79) = 0,15` mol

`n_{Fe} = n_{H_2} = 0,15` mol

`=> m_(Fe) = 0,15 . 56 = 8,4g`

`n_{HCl} = 2n_{H_2} = 2 . 0,15 = 0,3` mol

`200ml = 0,2l`

`=> C_{M dd HCl} = (0,3)/(0,2) = 1,5M`

Câu 2

`1.`

Base tan được trong nước: $\rm NaOH$, $\rm KOH$, $\rm Ba(OH)_2$

Base k tan trong nước: $\rm Zn(OH)_2$, $\rm Fe(OH)_3$

`2.`

$\rm Zn(OH)_2$ : Zinc hydroxide

$\rm NaOH$ : Sodium hydroxide

$\rm Fe(OH)_3$: Iron (III) hydroxide

$\rm KOH$: Potassium hydroxide

$\rm Ba(OH)_2$: Barium hydroxide

Câu `3.`

$\rm Fe+2HCl \xrightarrow{} FeCl_2+H_2$

$\rm n_{H_2}=\dfrac{3.7185}{24.79}=0.15$ (mol)

$\rm PT: n_{Fe}=n_{H_2}=0.15$ (mol)

$\rm \rightarrow m_{Fe}=0.15 . 56=8.4$ (g)

$\rm PT: n_{HCl}=2n_{Fe}=2 . 0.15=0.3$ (mol)

$\rm C_M HCl=\dfrac{0.3}{0.2}=1.5$ (M)

`#` NSD