Đáp án:

$\rm a)\\n_{Zn}=\dfrac{15,6}{65}=0,24\ (mol)$

Gọi hydrohalic là HR.

Phản ứng xảy ra:

$\rm Zn+2HR\to ZnR_2+H_2$

Theo phương trình:

$\rm\kern,8cm n_{Zn}=n_{ZnR_2}\\⇔\ 0,24=\dfrac{54}{65+2M_R}\\ ⇔\ M_R=80\ (g/mol)$

Vậy HR là HBr.

$\rm n_{H_2}=n_{Zn}=0,24\ (mol)$

$\rm V=0,24×24,79=5,9496\ (L)$

$\rm b)$

$\rm n_{KOH}=0,2×0,2=0,04\ (mol)$

Phản ứng trung hoà:

$\rm HBr+KOH\to KBr+H_2O$

$\rm \sum n_{HBr}=n_{KOH}+2n_{Zn}=0,52\ (mol)$

$\rm C\%_{HBr}=\dfrac{0,52×81}{200}×100\%=21,06\%$

`a)` 

Acid: `HX` 

`Zn + 2HX→ ZnX_2 + H_2` 

Theo pt: 

`n_{Zn}= n_{ZnX_2}` 

`<=> {15,6}/{65}= {54}/{65+ 2X}`

`→ X=80`$(g/mol)$ 

`→ X: Br` vậy hydrohalic acid là `HBr` 

`b)` 

`HBr + KOH → KBr + H_2O`

`n_{KOH}={200}/{1000}. 0,2=0,04(mol)`

Theo pt: 

`n_{HBr}= n_{KOH}+ 2n_{Zn}`

`= 0,04 + {15,6}/{65}.2=0,52(mol)` 

`x% ={52. 81}/{200}. 100%=21,06%`