a.\(Fe+2HCl->FeCl_2+H_2\)
b.\(FeO+2HCl->FeCl_2+H_2O\)
\(n_{H_2}=0,2\left(mol\right)\)
\(\%mFe=\dfrac{0,2.56}{20}.100\)
\(\%mFe=56\%\)
\(=>\%mFeO=100-56=44\%\)
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