HÚC BẠN HỌC TỐT!!!
Trả lời:
$Fe+H_2SO_4\rightarrow FeSO_4+H_2\\\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\\Mg+H_2SO_4\rightarrow MgSO_4+H_2\\\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2\,(mol)$
Đặt $a=n_{Fe};\,b=n_{Mg}$
Ta có hệ:
$\begin{cases}56a+24b=8\\a+b=0,2\end{cases}⇔\begin{cases}a=0,1\\b=0,1\end{cases}$
$⇔\begin{cases}m_{Fe}=0,1.56=5,6\,(g)\\m_{Mg}=0,1.24=2,4\,(g)\end{cases}⇔\begin{cases}\%m_{Fe}=70\%\\\%m_{Mg}=30\%\end{cases}$
Theo PT1: $n_{H_2SO_4}=n_{Fe}=0,1\,(mol)$
Theo PT2: $n_{H_2SO_4}=n_{Mg}=0,1\,(mol)$
$⇒∑n_{H_2SO_4}=0,1+0,1=0,2\,(mol)$
$⇒C_{MH_2SO_4}=\dfrac{n}{V}=\dfrac{0,2}{0,2}=1\,(M).$
