Question

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Answer 1

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