\(n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(m_{hh}=56a+27b=11\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(a.........a..............a\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(b.........1.5b................1.5b\)
\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(\%Fe=\dfrac{0.1\cdot56}{11}\cdot100\%=50.91\%\)
\(\%Al=100\%-50.91\%=49.09\%\)
\(n_{H_2SO_4}=0.4\cdot2=0.8\left(mol\right)\)
\(C\%_{H_2SO_4}=\dfrac{0.8\cdot98}{200}\cdot100\%=39.2\%\)
Đúng 1
Bình luận (1)
