`#Ly`

$a)$ 

Gọi `n_{CuO}=x` $(mol)$; `n_{ZnO}=y` $(mol)$

`m_(\text{oxit}` `=` `80x+81y=12,1` $(g)$

Bảo toàn $O:$ `n_(H_2O` `=` `n_(CuO` `=` `x` $(mol)$

Bảo toàn $O:$ `n_(H_2O` `=` `n_(ZnO` `=` `y` $(mol)$

`m_(H_2O` `=` `18x+18y=2,7` $(g)$

Ta có hệ PT:

`80x+81y=12,1`

`18x+18y=2,7`

`=>` $\begin{cases} x=0,05(mol)\\y=0,1(mol)\end{cases}$

`m_(CuO` `=` `0,05.80=4` $(g)$

`m_(ZnO` `=` `0,1.81=8,1` $(g)$

$b)$

Bảo toàn $H_2:$ `n_(H_2` `=` `∑``n_(H_2O` `=` `0,05+0,1=0,15` $(mol)$

`V_(H_2` `=` `0,15.22,4=3,36` $(l)$

$c)$ 

`m_(hh` `=` `16,2:2=8,1` $(g)$

Ta có:

Cứ $16,2$ $g$ hh thì có $0,05$ $mol$ $CuO$ và $0,1$ $mol$ $ZnO$

Vậy $8,1$ $g$ hh thì có $0,025$ $mol$ $CuO$ và $0,05$ $mol$ $ZnO$

PTHH:

$CuO$ $+$ $2HCl$ `->` $CuCl_2$ $+$ $H_2O$

$ZnO$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2O$

`->` `n_(HCl` `=` `2(n_{CuO}+n_{ZnO})=2.(0,025+0,05)=0,15` $(mol)$

`V_(ddHCl` `=` `(0,15)/(2)=0,075` $(l)$

Đổi $0,075$ $l$ $=$ $75$ $ml$

Em tham khảo nha:

Em xem lại đề nha

\(\begin{array}{l}
a)\\
ZnO + {H_2} \xrightarrow{t^0} Zn + {H_2}O\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
hh:ZnO(a\,mol),CuO(b\,mol)\\
{n_{{H_2}O}} = \dfrac{{2,7}}{{18}} = 0,15\,mol\\
\left\{ \begin{array}{l}
81a + 80b = 12,1\\
a + b = 0,15
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,05
\end{array}\)

\(\begin{array}{l}
{m_{CuO}} = 0,05 \times 80 = 4g\\
{m_{ZnO}} = 12,1 - 4 = 8,1g\\
b)\\
{n_{{H_2}}} = {n_{{H_2}O}} = 0,15\,mol\\
{V_{{H_2}}} = 0,15 \times 22,4 = 3,36l\\
c)\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{HCl}} = 0,05 \times 2 + 0,025 \times 2 = 0,15\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,15}}{2} = 0,075l = 75\,ml
\end{array}\)