a)\(n_{H_2SO_4}=\dfrac{200.9,8}{100.98}=0,2\left(mol\right)\)
Gọi x, y lần lượt là số mol Zn,Fe
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
1...............1...............1...............1(mol)
x................x...............x...............x(mol)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
1................1................1............1(mol)
y................y..................y...........y(mol)
Ta có\(\left\{{}\begin{matrix}65x+56y=12,1\\x+y=0,2\end{matrix}\right.\)
=>x=0,1,y=0,1
\(m_{Zn}=65.0,1=6,5\left(g\right)\)
\(\%m_{Zn}:\dfrac{6,5}{12,1}.100\%=53,7\%\)
\(\%m_{Fe}:100\%-53,7\%=46,3\%333\)
b) Khối lượng dung dịch sau phản ứng:
12,1+200-(0,1+0,1).2=211,7(g)
\(C\%\) dd \(ZnSO_4\):\(\dfrac{0,1.161}{211,7}.100\%=7,6\%\)
\(C\%\)dd \(FeSO_4:\)\(\dfrac{0,1.152}{211,7}.100\%=7,2\%\)
