\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)

\(0.05........0.15......0.1\)

\(\%Fe_2O_{3\left(bk\right)}=\dfrac{0.05\cdot160}{20}\cdot100\%=40\%0\%\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_X=20-0.05\cdot160+5.6=17.6\left(g\right)\)

a) \(n_{H_2O}=\dfrac{2,4.10^{23}}{6.10^{23}}=0,4\left(mol\right)\)

=> \(n_{H_2\left(pư\right)}=0,4\left(mol\right)\)

Theo ĐLBTKL

=> \(m=28,4+0,4.18-0,4.2=34,8\left(g\right)\)

b) \(n_{Fe\left(X\right)}=\dfrac{28,4.59,155\%}{56}=0,3\left(mol\right)\)

n_O = n_H2O = 0,4 (mol)

=> n_Fe : n_O = 3:4

=> CTHH: Fe₃O₄

c) \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

 \(n_{Fe_3O_4}=\dfrac{34,8}{232}=0,15\left(mol\right)\)

\(n_{H_2\left(bd\right)}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,8}{4}\) => Hiệu suất tính theo Fe₃O₄

n_Fe(X) = 0,3 (mol)

=> n_{Fe3O4 (bị khử)} = 0,1 (mol)

=> \(\dfrac{0,1}{0,15}.100\%=66,67\%\)

\(n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\\ n_{H_2O}=\dfrac{2,4.10^{23}}{6.10^{23}}=0,4\left(mol\right)\\ Đặt.oxit.sắt:Fe_xO_y\left(x,y:nguyên,dương\right)\\ Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\\ Vì:\dfrac{0,4}{y}< \dfrac{0,8}{y}\\ \Rightarrow H_2dư\\ \Rightarrow n_{H_2\left(p.ứ\right)}=n_{H_2O}=n_{O\left(mất\right)}=0,4\left(mol\right)\\ a,m=m_{oxit}=m_{rắn}+m_O=28,4+0,4.16=34,8\left(g\right)\\b,m_{Fe}=28,4.59,155\%=16,8\left(g\right)\\ \Rightarrow n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ \Rightarrow x:y=0,3:0,4=3:4\\ \Rightarrow CTHH:Fe_3O_4\\ c,n_{Fe_3O_4\left(bđ\right)}=\dfrac{34,8}{232}=0,15\left(mol\right)\\ \Rightarrow n_{Fe\left(LT\right)}=3.0,15=0,45\left(mol\right)\\ n_{Fe\left(TT\right)}=0,3\left(mol\right)\)

\(\Rightarrow H=\dfrac{0,3}{0,45}.100=66,667\%\)

n_H2=\(\dfrac{V}{22,4}\)=\(\dfrac{1,12}{22,4}\)=0,05(mol)

n Fe3O4=\(\dfrac{m}{M}\)=\(\dfrac{6,9}{232}\)=0,03(mol)

pthh:

Fe₃O₄ + 4H₂ -> 3Fe + 4H₂O

1mol.............4mol.......3mol......4mol

0,0125mol<-0,05mol->0,0375mol

*lập tỉ lệ và so sánh:\(\dfrac{0,05}{4}\)<\(\dfrac{0,03}{1}\)=>hirdro pứ hết, Fe3O4 dư

a) khối lượng Fe3O4 phản ứng là:

m(pứ)= n*M=0,0125*232=2,9(g)

phần trăm oxit sắt có bị khử là:

\(\dfrac{m\left(pứ\right)\cdot100}{m\left(bđ\right)}\)= \(\dfrac{2,9\cdot100}{6,9}\)=42%

khối lượng chất rắn sau phản ứng là:

m Fe= 0,0375*56=2,1(g)

m(Fe3O4 dư)= m(bđ)- m(pứ)=6,9-2,9=4(g)

$CO + O_{oxit} \to CO_2$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{O(oxit)} = n_{CaCO_3} = \dfrac{8}{100} = 0,08(mol)$

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{Fe} = n_{H_2} = \dfrac{1,344}{22,4} = 0,06(mol)$

Ta có : 

$n_{Fe} : n_O = 0,06 : 0,08 = 3 : 4$

Vậy oxit là $Fe_3O_4$

Công thức oxit sắt có dạng: \(Fe_xO_y\)

\(Fe_xO_y+yCO\rightarrow xFe+yCO_2\uparrow\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

\(\Rightarrow n_{Fe}=n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

\(\Rightarrow n_{CO}=n_{CO_2}=n_{CaCO_3}=0,08\left(mol\right)\)

\(\Rightarrow n_{O\left(Fe_xO_y\right)}=n_{O\left(CO_2\right)}-n_{O\left(CO\right)}=2n_{CO_2}-n_{CO}=0,08\left(mol\right)\)

\(\Rightarrow n_{Fe}:n_O=0,06:0,08=3:4\)

\(\Rightarrow Fe_3O_4\)

\(Co+\left\{{}\begin{matrix}Fe:amol\\O:bmol\end{matrix}\right.\)\(\rightarrow\left[{}\begin{matrix}Fe\rightarrow H_2SO_4\\Co_2\rightarrow Ca\left(OH\right)_2\rightarrow caCo_3\end{matrix}\right.\)

\(\cdot m\downarrow=m_{CaCo_3}=8\Rightarrow n_{CaCo_3}=n_{Co_2}=\dfrac{8}{100}=0,08\)

\(\Rightarrow n_O=n_{CO}=n_{CO_2}=0,08\)

\(\cdot Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

0,06                                     0,06

\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\Rightarrow n_{Fe}=n_{H_2}=0,06\)

\(\Rightarrow\dfrac{n_{Fe}}{n_O}=\dfrac{0,06}{0,08}=\dfrac{3}{4}\Rightarrow Fe_3O_4\)

Giả sử : \(m_{dd_{H_2SO_4\left(bđ\right)}}=100\left(g\right)\) , \(n_{H_2O\left(\text{hấp thụ}\right)}=a\left(mol\right)\)

Khi đó : 

\(m_{H_2SO_4\left(bđ\right)}=98\%\cdot100=98\left(g\right)\)

Sau khi hấp thụ :

\(m_{dd_{H_2SO_4}}=18a+100\left(g\right)\)

\(C\%_{H_2SO_4}=\dfrac{98}{18a+100}\cdot100\%=92.979\%\)

\(\Rightarrow a=0.3\)

\(n_{O\left(oxit\right)}=n_{H_2O}=0.3\left(mol\right)\) \(\Rightarrow m_O=0.3\cdot16=4.8\left(g\right)\)

\(m_{Fe}=32-4.8=27.2\left(g\right)\) \(\Rightarrow n_{Fe}=\dfrac{27.2}{56}=\dfrac{17}{35}\left(mol\right)\)

\(n_{Fe}:n_O=\dfrac{17}{35}:0.3=34:21\)

Tới đây em xem lại khối lượng của oxit sắt ban đầu nha. 

 \(a)Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ b)n_{H_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ n_{Fe} = \dfrac{3}{4}n_{H_2} = 0,3(mol)\\ n_{Fe_3O_4\ pư} = \dfrac{1}{4}n_{H_2} = 0,1(mol)\\ \Rightarrow m_{chất\ rắn\ sau\ phản\ ứng} = 0,3.56 + (34,8 -0,1.232)=28,4(gam)\\ c) \%m_{Fe_3O_4\ bị\ khử} = \dfrac{0,1.232}{34,8}.100\% = 66,67\%\)

Đáp án : D

n_Fe = 0,105 mol

Fe₃O₄ => m_oxit = 8,12g

Đáp án D

nFe = 0,105 => nFe₃O₄ = 0,035 => m = 0,035.232 = 8,12 => Chọn D.