Fe +2HCl----.FeCl2 +H2(1)
FexOy +2yHCl---->xFeCl2y/x +yH2O(2)
Ta có
n\(_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo pthh
n\(_{Fe}=n_{H2}=0,1\left(mol\right)\)
%m\(_{Fe}=\frac{0,1.56}{6,4}.100\%=87,5\%\)
%m\(_{FexOy}=100\%-87,5\%=12,5\%\)
b) Ta có
FexOy+ yH2---->xFe +yH2O(2)
Theo pthh
n\(_{FexOy}=\frac{1}{y}n_{H2O}=\frac{0,05}{y}\left(mol\right)\)
M\(_{FexOy}=3,2:\frac{0,05}{y}=64y\)
Ta có
y | FexOy |
1 | 64 |
2 | 128 |
3 | 192 |
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