Câu 3 :
\(n_{Mg}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(m_{hh}=24a+27b=7.8\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+H_2\)
\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(\%Mg=\dfrac{0.1\cdot24}{7.8}\cdot100\%=30.77\%\)
\(\%Al=100\%-30.77=69.23\%\)
\(m_{Muối}=m_{MgCl_2}+m_{AlCl_3}=0.1\cdot95+0.2\cdot133.5=36.2\left(g\right)\)
Câu 2 :
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2.......0.2.....................0.2\)
\(C_{M_{HCl}}=\dfrac{0.2}{0.1}=2\left(M\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
