Mg + H2SO4 → MgSO4 + H2 (1)
Fe + H2SO4 → FeSO4 + H2 (2)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,2\times2=0,4\left(g\right)\)
a) Gọi \(x,y\) lần lượt là số mol của Mg và Fe
Ta có: \(\left\{{}\begin{matrix}24x+56y=8\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\Rightarrow\%Mg=\dfrac{2,4}{8}\times100\%=30\%\)
\(m_{Fe}=0,1\times56=5,6\left(g\right)\)
\(\Rightarrow\%Fe=\dfrac{5,6}{8}\times100\%=70\%\)
b) \(m_X=m_{hh}+m_{ddH_2SO_4}-m_{H_2}=8+72,4-0,4=80\left(g\right)\)
Theo PT1: \(n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgSO_4}=0,1\times120=12\left(g\right)\)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{12}{80}\times100\%=15\%\)
Theo PT2: \(n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,1\times152=15,2\left(g\right)\)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{15,2}{80}\times100\%=19\%\)