Đáp án:

$\rm a)\\n_{H_2}=\dfrac{4,48}{22,4}=0,2\ (mol)$

Phương trình phản ứng:

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

Theo phương trình:

$\rm n_{Zn}=n_{H_2}=0,2\ (mol)$

Phần trăm khối lượng mỗi chất:

$\rm\%m_{Zn}=\dfrac{0,2×65}{21,1}×100\%=61,61\%\\\%m_{ZnO}=100-61,61=38,39\%$

$\rm b)\\n_{ZnO}=\dfrac{21,1×38,39\%}{81}=0,1\ (mol)$

Theo phương trình:

$\rm n_{HCl}=2(n_{Zn}+n_{ZnO})=0,6\ (mol)$

$\rm\to m_{HCl}=\dfrac{0,6×36,5}{7,3\%}=300\ (g)$

`n_{H_2}={4,48}/{22,4}=0,2(mol)`

`PTHH:Zn+2HCl->ZnCl_2+H_2`

`(mol)`   `0,2`      `0,4`                         `←0,2`

          `ZnO+2HCl->ZnCl_2+H_2O`

`a.m_{Zn}=0,2.65=13(g)`

`m_{ZnO}=21,1-13=8,1(g)`

`%Zn=13/{21,1} .100% ~~61,61%`

`%ZnO=100-61,1=38,39%`

`b.n_{ZnO}={8,1}/81=0,1(mol)`

`=>n_{HCl}=0,4+2n_{ZnO}=0,4+2.0,1=0,6(mol)`

`m_{dd HCl}={0,6.36,5.100}/{7,3} =300(g)`