: Cho 11,2 gam hỗn hợp bột gồm Zn và Cu vào 250ml dung dịch AgNO3 1M. Sau khi phản ứng xảy ra hoàn toàn, thu được dung dịch X và 30,2gam chất rắn. Tính khối lượng mỗi kim loại trong hỗn hợp ban đầu

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Đáp án:

$m_{Zn}=3,25g$

$m_{Cu}=8g$

Giải thích các bước giải:

 Có $n_{AgNO_3}$ $=CM.V=1.0,25=0,25$ mol

Vì $Zn$ có tính khử mạnh hơn $Cu$ nên tác dụng với $AgNO_3$ trước

1) $Zn+2AgNO_3$ $\rightarrow$ $Zn(NO_3)_2+2Ag$

2) $Cu+2AgNO_3$ $\rightarrow$ $Cu(NO_3)_2+2Ag$

__________________________________________________________

Bảo toàn nguyên tố nhận thấy:

$n_{Ag}$ tạo thành $=$ $n_{AgNO_3}$ tham gia $=0,25$ mol

Vậy $m_{Ag}$ tạo thành $=0,25.108=27g$

$\rightarrow$ $m_{\text{rắn dư}}$ $=30,2-27=3,2g$

$\rightarrow$ $m_{Cu}$ dư $=3,2g$ vì $Zn$ đã phản ứng hết còn $Cu$ chưa phản ứng hoàn toàn.

Có $n_{Cu}$ dư $=$ $\dfrac{3,2}{64}$ $=0,05$ mol

Vậy $n_{AgNO_3}$ không phản ứng với $Cu$ dư cũng là $0,05$ mol

Gọi $x,y$ là số mol của $Zn,Cu$

$+)$ $m_{Zn}$ $+$ $m_{Cu}$ $=11,2g$ hay $65x+64y=11,25$ (*)

$+)$ Số mol $AgNO_3$ phản ứng là $0,25$ hay $2x+2y-0,05.2=0,25$ (**)

Từ (*) và (**) ta thiết lập được hệ phương trình sau:

$\begin{cases} 65x+64y=11,25\\2x+2y=0,35\ \end{cases}$⇔$\begin{cases} x=0,05\\y=0,125\ \end{cases}$

Vậy $m_{Zn}$ $=0,05.65=3,25g$

Vậy $m_{Cu}$ $=0,125.64=8g$

___________________________________________________

Sửa đề: Cho $11,25$ gam hỗn hợp...

***Nếu là $11,2$ khi thay vào giải hệ sẽ ra số mol của $Zn$ là $0$ và không thể tiếp tục tính

Đáp án:

↓↓↓↓

Giải thích các bước giải:

$\textit{ Zn a mol, Cu b mol ,Ag 0,25 mol}$

$\textit{Sau phản ứng thu được 30,2 gam chất rắn gồm có : }$

$\textit{0,25 . 108  = 27 gam Ag }$

$\textit{Còn lại 30,2 - 27 =3,2 gam Cu }$

$\textit{Từ đây có hệ phương trình }$

$\textit{65a + 64b = 11 ,25 }$

$\textit{2a + (b - 0,05). 2 = 0,25 }$

$\textit{⇒ a= 0,25 ; b= 0,125 }$

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