\(Fe\left(0,1\right)+2HCl\left(0,2\right)--->FeCl_2\left(0,1\right)+H_2\left(0,1\right)\)\(\left(1\right)\)
\(Fe_2O_3\left(0,05\right)+6HCl\left(0,3\right)--->2FeCl_3\left(0,1\right)+3H_2O\)\(\left(2\right)\)
\(n_{H_2}=0,1\left(mol\right)\)
Theo (1) \(n_{Fe}=0,1\left(mol\right)\)\(\Rightarrow m_{Fe}=5,6\left(g\right)\)
\(\Rightarrow m_{Fe_2O_3}=13,6-5,6=8\left(g\right)\)\(\Rightarrow n_{Fe_2O_3}=0,05\left(mol\right)\)
=> Rainbow có: \(\left\{{}\begin{matrix}\%m_{Fe}=41,18\%\\\%m_{Fe_2O_3}=58,82\%\end{matrix}\right.\)
Phản ứng vừa đủ => Dung dịch thu được sau phản ứng : \(\left\{{}\begin{matrix}FeCl_2:0,1\left(mol\right)\\FeCl_3:0,1\left(mol\right)\end{matrix}\right.\)
Theo (1) và (2) \(\sum n_{HCl}=0,5\left(mol\right)\)
\(\Rightarrow m_{HCl}=18,25\left(g\right)\)\(\Rightarrow m_{ddHCl}=625\left(g\right)\)
\(m dd sau =13,6+625-0,1.2=638,4(g)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,1.127.100}{638,4}=1,99\%\\C\%_{FeCl_3}=\dfrac{0,1.162,5.100}{638,4}=2,55\%\end{matrix}\right.\)
Ta co pthh
Fe + 2HCl \(\rightarrow\) FeCl2 + H2 (1)
Fe2O3 + 6HCl \(\rightarrow\) 2FeCl3 + 3H2O (2)
Ta thấy trong 2 pt thì pt 1 sinh ra khí H2 nên
ta có
nH2=\(\dfrac{2,24}{22,4}=0,1mol\)
Theo pthh 1
nFe=nH2=0,1 mol
\(\rightarrow\) mFe=0,1.56=5,6 g
mFe2O3=13,6-5,6=8 g
a, Ta co
%mFe=\(\dfrac{5,6.100\%}{13,6}\approx41,18\%\)
%mFe2O3=100%-41,18%=58,2%
b, Theo pthh 1
nHCl=2nH2=2.0,1=0,2 mol
nFeCl2=nH2=0,1 mol -> mFeCl2=0,1.127=12,7 g
Theo pthh 2
nFe2O3=\(\dfrac{8}{160}=0,05mol\)
nFeCl3=2nFe2O3=2.0,05 = 0,1 mol -> mFeCl3=0,1.162,5 =16,25 g
nHCl=6nFe2O3=6.0,05=0,3 mol
-> số mol của HCl (1) và (2) sau phản ứng là :
nHCl(sau-phan-ung) = 0,2+0,3 = 0,5 mol
-> mct=mHCl=0,5 .36,5 = 18,25 g
mddHCl=\(\dfrac{mct.100\%}{C\%}=\dfrac{18,25.100\%}{2,92\%}=625\left(g\right)\)
-> mdd(sau-phan-ung) = m(hon-hop-ban-dau) + mddHCl(sau-phan-ung) - mH2 = 13,6 + 625 - (0,1.2) = 638,4 g
\(\Rightarrow\) C%\(_{\text{dd}FeCl2}=\dfrac{12,7}{638,4}.100\%\approx1,99\%\)
C%\(_{\text{dd}FeCl3}=\dfrac{16,25}{638,4}.100\%\approx2,55\%\)
