`#Ly`

Tóm tắt:

`∑``m_(CuO;ZnO` `=` `12,1` $(g)$

`V_(ddHCl` `=` `150` `(ml)`

`C_MHCl` `=` `2` $(M)$

$a)$ `%m_(CuO` $=$ $?$; `%m_(ZnO` `=` `?`

$b)$ `C%_(H_2SO_4` `=` `20%`; `m_(ddH_2SO_4` `=` `?`

Bài giải:

Đổi $150$ $ml$ $=$ $0,15$ $l$

`n_(HCl` `=` `C_M.V=2.0,15=0,3` $(mol)$

PTHH:

$(1):$ $CuO$ $+$ $2HCl$ `->` $CuCl_2$ $+$ $H_2O$

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

Gọi số $mol$ $CuO,ZnO$ pư là $x,y$

Theo PT $(1)$, `n_(HCl` `=` `2n_(CuO` `=` `2x` $(mol)$

Theo PT $(2)$, `n_(HCl` `=` `2n_(ZnO` `=` `2y` $(mol)$

`∑``n_(HCl` `=` $2x+2y=0,3$ $(mol)$

`m_(hh` `=` `80x+81y=12,1` $(g)$

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

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

`%m_(CuO` `=` `(4)/(12,1).100%≈33,06%`

`%m_(ZnO` `=` `100-33,06=66,94%`

$b)$

PTHH:

$CuO$ $+$ $H_2SO_4$ `->` $CuSO_4$ $+$ $H_2O$

$ZnO$ $+$ $H_2SO_4$ `->` $ZnSO_4$ $+$ $H_2O$

Theo $2$ PT trên, `n_(H_2SO_4` `=` `n_(CuO` $+$ `n_(ZnO` `=` `0,05+0,1=0,15` $(mol)$

`m_(H_2SO_4` `=` `n.M=0,15.98=14,7` $(g)$

`m_(ddH_2SO_4` `=` `(m_{H_2SO_4}.100)/(C%)` `=` `(14,7.100)/(20)=73,5` $(g)$

Tóm tắt:

\(\begin{array}{l}
Hon\,hop+HCl\\
m_{hon\,hop}=m_{CuO}+m_{ZnO}=12,1(g)\\
V_{dd\,HCl}=150ml\\
C_{M\,HCl}=2M\\
a)\%m_{CuO}=?;\%m_{ZnO}=?\\
b)Hon\,hop+H_2SO_4\\
C\%_{H_2SO_4}=20\%\\
m_{dd\,H_2SO_4}=?\\
---------------\\
a)\\
CuO+2HCl\to CuCl_2+H_2O\\
ZnO+2HCl\to ZnCl_2+H_2O\\
150ml=0,15l\\
n_{HCl}=2.0,15=0,3(mol)\\
Dat\,n_{CuO}=x(mol);n_{ZnO}=y(mol)\\
\to 80x+81y=12,1(1)\\
Theo\,PT:\,2x+2y=0,3(2)\\
(1)(2)\to x=0,05;y=0,1\\
\%m_{CuO}=\frac{0,05.80}{12,1}.100\%\approx 33,06\%\\
\%m_{ZnO}=100-33,06=66,94\%\\
b)\\
CuO+H_2SO_4\to CuSO_4+H_2O\\
ZnO+H_2SO_4\to ZnSO_4+H_2O\\
Theo\,PT:\,\sum n_{H_2SO_4}=x+y=0,15(mol)\\
\to m_{dd\,H_2SO_4}=\frac{0,15.98}{20\%}=73,5(g)
\end{array}\)