Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:
\(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
a---->2a------------------>a------------------>a
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
b---->2b------------------>b------------------>b
Theo bài ra, ta có hệ: \(\left\{{}\begin{matrix}56a+24b=8\\a+b=0,2\end{matrix}\right.\)
\(\Leftrightarrow a=b=0,1\left(TM\right)\) hay \(n_{Mg}=n_{Fe}=0,1\left(mol\right)\)
\(m_{CH_3COOH}=\left(0,1.2+0,1.2\right).60=24\left(g\right)\\ \rightarrow C\%_{CH_3COOH}=\dfrac{24}{200}.100\%=12\%\)
\(m_{dd}=200+8-0,2.2=207,6\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{0,1.174}{207,6}.100\%=8,38\%\\C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{0,1.142}{207,6}.100\%=6,84\%\end{matrix}\right.\)
