@$\text{Ly}$

$a)$ PTHH: $Zn$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2↑$

$($ $Cu$ không tác dụng với $HCl$ $)$

$b)$ $n_{H_2}$ $=$ $\dfrac{6,72}{22,4}$ $=$ $0,3$ $mol$

Theo PT trên, $n_{Zn}$ $=$ $n_{H_2}$ $=$ $0,3$ $mol$

$m_{Zn}$ $=$ $0,3.65=19,5$ $(g)$

`%m_(Zn` $=$ $\dfrac{19,5}{25,9}$`.100%` $≈$ $75,29$`%`

`%m_(Cu` $=$ $100-75,29=24,71$`%`

$c)$ $n_{CuO}$ $=$ $\dfrac{28}{80}$ $=$ $0,35$ $mol$

PTHH: $CuO$ $+$ $H_2$ $\xrightarrow[]{t^o}$ $Cu$ $+$ $H_2O$

Ban đâu, $n_{CuO}$ $=$ $0,35$ $mol$ và $n_{H_2}$ $=$ $0,3$ $mol$

Mà $\dfrac{0,35}{1}$ $>$ $\dfrac{0,3}{1}$

$⇒$ $CuO$ dư, $H_2$ hết và tính theo $n_{H_2}$

Theo PT, $n_{Cu}$ $=$ $n_{H_2}$ $=$ $0,3$ $mol$

$m_{Cu}$ $=$ $0,3.64=19,2$ $(g)$

$n_{CuO(dư)}$ $=$ $0,35-0,3=0,05$ $mol$

$m_{CuO(dư)}$ $=$ $0,05.80=4$ $(g)$

$m_{CR}$ $=$ $4+19,2=23,2$ $(g)$

Đáp án:

`↓` 

Giải thích các bước giải:

 `a)`

`Zn+2HCl->ZnCl_2+H_2`

`b)`

`n_{Zn}=n_{H_2}=(6,72)/(22,4)=0,3(mol)`

`=>m_{Zn}=0,3xx65=19,5(g)`

`=>%m_{Zn}=(19,5xx100)/(25,9)~~75,3(%)`

`=>%m_{Cu}=100-75,3=24,7(%)`

`c)`

`CuO+H_2` $\xrightarrow{t^o}$ `Cu+H_2O`

`n_{CuO}=(28)/80=0,35(mol)`

`n_{H_2}=0,3(mol)`

`=>H_2` hết và `CuO` dư

`=>n_{CuO}(dư)=0,35-0,3=0,05(mol)`

`=>m_{CuO}=0,05xx80=4(g)`

`m_{Cu}=0,3xx64=19,2(g)`

`=>m_{Crắn}=19,2+4=23,2(g)`

`text{# Yor}`