Bài 1
Fe +2HCl---->FeCl2 +H2
Ta có
n\(_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
Theo pthh
n\(_{FeCl2}=n_{Fe}=0,1\left(mol\right)\)
m\(_{FeCl2}=0,1.127=12,7\left(g\right)\)
V\(_{H2}=0,1.22,4=2,24\left(l\right)\)
Bài 2 :
4Al + 6HCl-----> 2AlCl3 +3H2
ta có
n\(_{Al}=\frac{8,1}{27}=0,3\left(mol\right)\)
n\(_{HCl}=0,2.1=0,2\left(mol\right)\)
=> Al dư
Theo pthh
n\(_{H2}=\frac{1}{2}n_{HCl}=0,1\left(mol\right)\)
V\(_{H2}=0,1.22,4=2,24\left(l\right)\)
Theo pthh
n\(_{AlCl3}=\frac{1}{3}n_{HCl}=0,067\left(mol\right)\)
m\(_{AlCl3}=0,067.133,5=8,9445\left(g\right)\)
n = \(\frac{5,6}{56}=0,1mol\)
PTHH : \(Fe+2HCl\rightarrow FeCl2+H2\)
Mà nFeCl2= n Fe= 0,1 (mol)
=>mFeCl2= 12,7(g)
=>VH2= 0,1 . 22,4= 2,24l
