Gọi số mol Fe và Fe2O3 là x, y
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
(mol) 1 2 1 1
(mol) x 2x x x
\(PTHH:Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
(mol) 1 6 2 3
(mol) y 6y 2y 3y
Ta có hpt:
\(\left\{{}\begin{matrix}56x+160y=21,6\\22,4x=2,24\end{matrix}\right.\Leftrightarrow x=y=0,1\left(mol\right)\)
\(\%m_{Fe}=\frac{0,1.56}{21,6}.100\%\approx26\left(\%\right)\)
\(\%m_{Fe_2O_3}=100\%-26\%=74\left(\%\right)\)
Fe + 2HCl → FeCl2 + H2 (1)
Fe2O3 + 6HCl → 2FeCl3 + 3H2O (2)
\(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo pT1: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,1\times56=5,6\left(g\right)\)
\(\Rightarrow m_{Fe_2O_3}=21,6-5,6=16\left(g\right)\)
\(\Rightarrow\%m_{Fe}=\frac{5,6}{21,6}\times100\%=25,93\%\)
\(\%m_{Fe_2O_3}=\frac{16}{21,6}\times100\%=74,07\%\)
nH2 = 2.24/22.4=0.1 mol
Fe + 2HCl --> FeCl2 + H2
0.1________________0.1
mFe = 5.6 g
mFe2O3 = 21.6 - 5.6 = 16 g
%Fe = 25.93%
%Fe2O3 = 74.07%
dd gì bạn?
HCl , H2SO4 ,.....................
