`#Ly`

$a)$

PTHH:

$(1):$ $Zn$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2↑$

$(2):$ $ZnO$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2O$

Đổi $200$ $ml$ $=$ $0,2$ $l$

`n_(HCl` `=` `C_M.V=2.0,2=0,4` $(mol)$

`n_(H_2` `=` `V/(22,4)` `=` `(1,12)/(22,4)=0,05` $(mol)$

Theo PT $(1)$, `n_(Zn` `=` `n_(H_2` `=` `0,05` $(mol)$

`m_(Zn` `=` $n.M=0,05.65=3,25$ $(g)$

$b)$

Theo PT $(1)$, `n_(HCl` `=` `2n_(H_2` `=` `0,1` $(mol)$

`n_(HCl(2)` `=` `0,4-0,1=0,3` $(mol)$

Theo PT $(2)$, `n_(ZnO` `=` `(1)/(2)n_(HCl` `=` `0,15` $(mol)$

`m_(ZnO` `=` `m.M=0,15.81=12,15` $(g)$

`m_(hhCR` `=` `3,25+12,15=15,4` $(g)$

\(\begin{array}{l}
a)\\
n_{H_2}=\frac{1,12}{22,4}=0,05(mol)\\
Zn+2HCl\to ZnCl_2+H_2\\
ZnO+2HCl\to ZnCl_2+H_2O\\
Theo\,PT:\,n_{Zn}=n_{H_2}=0,05(mol)\\
\to m_{Zn}=0,05.65=3,25(g)\\
b)\\
200ml=0,2l\\
n_{HCl}=0,2.2=0,4(mol)\\
Theo\,PT:\,n_{ZnO}=\frac{n_{HCl}-2n_{H_2}}{2}=0,15(mol)\\
\to m=0,15.81+3,25=15,4(g)
\end{array}\)