Đáp án:

$\rm a)\\n_{H_2}=\dfrac{2,479}{24,79}=0,1\ (mol)$

Phản ứng xảy ra:

$\rm  Zn+2HCl\to ZnCl_2+H_2\\ ZnO+2HCl\to ZnCl_2+H_2O$

Theo phản ứng:

$\rm n_{Zn}=n_{H_2}=0,1\ (mol)$

Khối lượng mỗi chất:

$\rm m_{Zn}=0,1×65=6,5\ (g)\\m_{ZnO}=10,55-6,5=4,05\ (mol)$

$\rm b)\\n_{ZnO}=\dfrac{4,05}{81}=0,05\ (mol)$

Theo phản ứng:

$\rm n_{HCl\ (pu)}=2×(n_{Zn}+n_{ZnO})=0,3\ (mol)$

$\rm n_{HCl\ (sd)}=0,3×120\%=0,36\ (mol)$

$\rm n_{ZnCl_2}=\dfrac{1}{2}n_{HCl\ (pu)}=0,15\ (mol)$

$\rm n_{HCl\ (du)}=0,36-0,3=0,06\ (mol)$

Bảo toàn khối lượng:

$\rm m_{dd\ sau\ pu}=10,55+\dfrac{0,36×36,5}{10\%}-0,1×2=141,75\ (g)$

$\rm C\%_{ZnCl_2}=\dfrac{0,15×136}{141,75}×100\%=14,39\%\\C\%_{HCl\ (du)}=\dfrac{0,06×36,5}{141,75}×100\%=1,545\%$

$\rm PTHH: Zn +2HCl \xrightarrow{} ZnCL_2 +H_2$  (1)

$ZnO + 2HCl \xrightarrow{} ZnCl_2 +H_2O$ 

$n_{H_2}=\dfrac{2,479}{24,79}=0,1(mol)$

theo pthh (1) 

$n_{Zn}=n_{H_2}=0,1(mol)$

`->`$ m_{Zn}=0,1.65=6,5(g)$ 

`=>$ m_{ZnO}=10,55-6,5= 4,05(g)$ 

`b)` 

$n_{ZnO}=\dfrac{4,05}{65+16}=0,05(mol)$ 

theo pthh 

$n_{HCl(pứ)}=2n_{Zn} +2n_{ZnO}=0,2+0,1=0,3(mol)$ 

theo đề ta có 

$n_{HCl(bd)}=0,3 .\dfrac{120}{100}=0,36(mol)$ 

`->n_{HCl(d ư)} =0,36-0,3=0,06(mol)`

$m_{dd-sp ứ)}=10,55+ \dfrac{0,36.36,5 }{10`%`}-0,1.2=141,75(g)$ 

`C%_{ZnCl_2}=\frac{0,15.136}{141,75} .100%=14,39%`

`C%_{HCl(d ư)} =\frac{0,06.36,5}{141,75}.100%=1,545%`