Câu $1:$

$HCl:$Axit clohiđric 

$H_2SO_4:$Axit sunfuric 

$HNO_3:$Axit nitric 

$HMnO_4:$Axit pemanganic

$NaOH:$Natri hiđroxit 

$Fe(OH)_3:$Sắt $(III)$ hiđroxit 

$Al(OH)_3:$Nhôm hiđroxit 

 Câu $2:$

`n_(Zn)=(9,75)/65=0,15(mol)`

$PT:Zn+2HCl \to ZnCl_2+H_2$

$a)$

Theo `PT,n_(H_2)=n_(Zn)=0,15(mol)`

`=>V_(H_2)(đktc)=0,15.22,4=3,36(l)`

Theo `PT,n_(ZnCl_2)=n_(Zn)=0,15(mol)`

`=>m_(ZnCl_2)=0,15.136=20,4(g)`

$b)$

`m_(ddspứ)=9,75+300-0,15.2`

`=>m_(ddspứ)=309,45(g)`

`C%_(ZnCl_2)=(20,4)/(309,45) .100%=6,6%`

Đáp án + Giải thích các bước giải:

Câu 1:

`1) HCl:` Axit clohidric

`2) H_2SO_4:` Axit sunfuric

`3) HNO_3:` Axit nitric

`4) HMnO_4:` Axit permanganic

`5) NaOH:` Natri hidroxit

`6) Fe(OH)_3:` Sắt (III) hidroxit

`7) Al(OH)_3:` Nhôm hidroxit

____________________________________________________________

Câu 2:

`Zn+2HCl->ZnCl_2+H_2↑` 

`a) n_(Zn)=(9,75)/65=0,15 (mol)`

Theo phương trình:

`n_(H_2)=n_(ZnCl_2)=n_(Zn)=0,15 (mol)`

`=>V_(H_2)=0,15.22,4=3,36 (l)`

`=> m_(ZnCl_2)=0,15.136=20,4 (g)`

`b) m_(ZnCl_2(dd))=300+9,75-0,15.2=309,45 (g)`

`C%=(m_(ZnCl_2))/(m_(ZnCl_2 (dd))) .100%=(20,4)/(309,45) .100% ≈6,59%`