goi x, y lần lượt là so mol cua Al, Fe
\(n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
de: x \(\rightarrow\) 3x \(\rightarrow\) 1,5x
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
de: y \(\rightarrow\) 2y \(\rightarrow\) y
Ta co: 27x + 56y = 4,4
1,5x + y = 0,16
=> x = 0,08 ; y = 0,04
a, \(m_{Al}=0,08.27=2,16g\)
\(\%m_{Al}=\dfrac{2,16}{4,4}.100\%\approx49,1\%\)
\(\%m_{Fe}=100-49,1\approx50,9\%\)
b, 2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2 (1)
Fe + H2SO4 \(\rightarrow\) FeSO4 + H2 (2)
Ta thấy: \(n_{H_2SO_4\left(2\right)}=n_{Fe}=0,04\)
\(n_{H_2SO_4\left(1\right)}=\dfrac{n_{Al}.3}{2}=0,12\)
\(m_{H_2SO_4}=98.\left(0,12+0,04\right)=15,68g\)
nH2=V/22,4=3,584/22,4=0,16(mol)
gọi x,y lần lượt là sô mol của Al và Fe
Pt1: 2Al + 6HCl-> 2AlCl3 + 3H2
cứ:: 2...........6.............2............3 (mol)
vậy: x-------->3x------->x-------->1,5x(mol)
pt2: Fe + 2HCl -> FeCl2 +H2
cứ:: 1............2............1...........1 (mol)
vậy: y------->2y-------->y-------->y (mol)
Từ 2pt và đề ta có:
27x + 56y=4,4
1,5x+y=0,16
=> x=0,08(mol) ,y =0,04(mol)
=> mAl=n.M=0,08.27=2,16(g)
mFe=n.M=0,04.56=2,24(g)
\(\Rightarrow\%m_{Al}=\dfrac{m_{Al}.100\%}{m_{hh}}=\dfrac{2,16.100}{4,4}\approx49,1\left(\%\right)\)
\(\Rightarrow\%m_{Fe}=100\%-\%m_{Al}=100\%-49,1\%=50,9\left(\%\right)\)