`a)`

`Zn+ 2HCl→ ZnCl_2+ H_2` 

`ZnO + 2HCl → ZnCl_2+ H_2O`

`n_{H_2O} = {0,9}/{18} = 0,05(mol)`

`n_{ZnO} = n_{H_2O} = 0,05(mol)`

`n_{Zn}= {6,5 - 0,05 .81}/{65} = 0,038(mol)`

`n_{H_2} = n_{Zn}= 0,038(mol)`

`V= 0,038 .22,4 = 0,8512(l)`

`b)`

`%m_{ZnO}= {0,05 . 81}/{6,5} .100= 62,3%`

`%m_{Zn}= 100 - 62,3 = 37,7%`

`c)`

`n_{HCl} = 2(n_{Zn}+ n_{ZnO})`

`= 2. (0,038 + 0,05) = 0,176(mol)`

`mdd = 6,5 + {0,176 .36,5}/{16%}- 0,038 .2= 46,574(g)`

`C%_{ZnCl_2}= {(0,038+0,05) . 136}/{46,574}. 100= 25,7%`

$\rm a)$ Ta có: $\rm n_{H_2O}=\dfrac{0,9}{18}=0,05\ (mol)$

Phản ứng xảy ra:

$\rm\quad\quad\;\;\; 2H_2+O_2\xrightarrow{\quad t^o\quad}2H_2O\\(mol)\;0,05\hspace{1cm}\leftarrow\hspace{1,2cm}0,05$

$\rm ⇒V_{H_2}=0,05×22,4=1,12\ (lit)$

$\rm b)$

Phản ứng xảy ra:

$\rm \quad\quad\;\;\; Zn+2HCl\to ZnCl_2+H_2\\(mol)\;0,05\hspace{,5cm}\leftarrow\hspace{,5cm}\;\;\;\;0,05\;\;\;\;0,05$

$\rm \quad\quad\;\;\; ZnO+2HCl\to ZnCl_2+H_2O\\(mol)\;\:0,04\hspace{,5cm}\to\hspace{,75cm}\;\;\;\;0,04$

Ta có: $\rm n_{ZnO}=\dfrac{6,5-0,05×65}{65+16}=0,04\ (mol)$

Vậy $\rm\%m_{Zn}=\dfrac{0,05×65}{6,5}=50\%$

$\rm\quad\;\;\%m_{ZnO}=100\%-50\%=50\%$

$\rm c)$ Từ phản ứng:

$\rm n_{HCl}=2×(0,05+0,04)=0,18\ (mol)$

$\rm n_{ZnCl_2}=0,05+0,04=0,09\ (mol)$

Bảo toàn khối lượng:

$\rm m_{dd\ ZnCl_2}=6,5+\dfrac{0,18×36,5}{16\%}-0,05×2=47,4625\ (gam)$

Vậy $\rm C\%_{ZnCl_2}=\dfrac{0,09×136}{47,4625}×100\%=25,79\%$