\(Fe+H_2SO_{\text{ 4 }}\rightarrow FeSO_4+H_2\uparrow\\
2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\
Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\\
Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\
2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\
Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
xét
\(n_{H_2}=\dfrac{11,5}{22.4}=0,5\left(mol\right)\\
n_{Fe}=n_{H_2}=0,5\left(mol\right)\\
m_{Fe}=0,5.56=28\left(g\right)\left(1\right)\\
n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{3}\left(mol\right)\\
m_{Al}=\dfrac{1}{3}.27=9\left(g\right)\left(2\right)\\
n_{Zn}=n_{H_2}=0,5\left(mol\right)\\
m_{Zn}=0,5.65=32,5\left(g\right)\left(3\right)\)
từ (1) ; (2) và (3) => mAl bé nhất
a) Fe+2HCl-->FeCl2 + H2
2Al + 6HCl --> 2AlCl3 + 3H2
Zn+2HCl -->ZnCl2 + H2
Fe + H2SO4 --> FeSO4 + H2
Zn + H2SO4 --> ZnSO4 + H2
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) Chọn kim loại Al ( Nhôm)
Giải thích:
Do nAl=2/3nH2 --> nAl=2/3*(11,2/22,4)=1/3 (mol)
--> mAl=nAl*MAl=1/3 * 27 = 9 (g)