Ta co cac pthh
FeO+H2\(\rightarrow\)Fe+H2O
Fe2O3+3H2\(\rightarrow\)2Fe+3H2O
Fe +H2SO4\(\rightarrow\)FeSO4+H2
Theo de bai ta co
nong do mol cua dd H2SO4 la
CM=\(\dfrac{nH2SO4}{vH2SO4}\Rightarrow nH2SO4=CM.VH2SO4=2.\dfrac{100}{1000}=0,2mol\)
Theo pthh 3
nFe=\(nH2SO4=0,2=0,2mol\)
Goi x la so mol cua Fe tham gia vao Pthh 2
so mol cua Fe tham gia vao pthh 1 la \(0,2-x\)
Theo pthh 1 va 2 ta co
nFe2O3=\(\dfrac{1}{2}nfe=\dfrac{1}{2}xmol\)
nFeO=nFe=\(0,2-x\)mol
Theo de bai ta co he pt
160.\(\dfrac{1}{2}x+72.\left(0,2-x\right)=15,2\)
\(\Leftrightarrow\)80x +14,4-72x =15,2
\(\Leftrightarrow\)80x-72x=15,2-14,4
\(\Leftrightarrow\)8x=0,8
\(\Rightarrow\)x=\(\dfrac{0,8}{8}=0,1mol\)
\(\Rightarrow\)So mol cua FeO=0,2-x=0,2-0,1=0,1 mol
nFe2O3=\(\dfrac{1}{2}nFe=\dfrac{1}{2}0,1=0,05mol\)
a, Ta co
mFe2O3=160.0,05=8 g
mFeO=15,2-8=7,2 g
\(\Rightarrow\)% khoi luong cua moi Oxit la
%mFeO=\(\dfrac{7,2.100}{15,2}\approx47,37\%\)
%mFe2O3=100%-47,37%=52,63%
B, Theo pthh 1 va 2
nH2=nFe=0,1 mol
nH2=\(\dfrac{3}{2}nFe=\dfrac{2}{3}0,1=\dfrac{1}{15}mol\)
\(\Rightarrow\)VH2=(0,1+\(\dfrac{1}{15}\)).22,4=37,3 l
c, Ta co
MFeSO4.7H2O=152.30=4560 g/mol
Theo pthh
nFeSO4=nH2SO4=0,2 mol
\(\Rightarrow\)mFeSO4.H2O=0,2.4560=912g