\(n_{H_2}=\dfrac{3,36}{22,4}0,15(mol)\\ a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ b,n_{Fe}=n_{H_2}=0,15(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,15.56}{14,8}.100\%=56,76\%\\ \Rightarrow \%_{Cu}=100\%-56,76\%=43,24\%\\ c,n_{H_2SO_4}=0,15(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,15.98}{20\%}=73,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{73,5}{1,4}=52,5(l)\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,15<--------------------0,15

=> m_Fe = 0,15.56 = 8,4 (g)

=> m_Cu = 15-8,4 = 6,6 (g)

c) Số nguyên tử Fe = 0,15.6.10²³ = 0,9.10²³

Ta có: 

n H2 = 0,05 ( mol )

1.PTHH

Fe + H2SO4 ====> FeSO4 + H2

FeO + H2SO4 ====> FeSO4 + H2O

theo pthh: n Fe = n H2 = 0,05 ( mol )

=> m Fe = 2,8 ( g )

=> m FeO = 7,2 ( g ) => n FeO = 0,1 ( mol )

2.

theo pthh: n H2SO4 = 0,05 + 0,1 = 0,15

  => m H2SO4 = 14,7 ( g )

  => m dd H2SO4 9,8% = 150 ( g )

\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

        1           1               1          1

      0,05      0,05          0,05     0,05

      \(FeO+H_2SO_4\rightarrow FeSO_4+H_2O|\)

        1             1                1           1

      0,1           0,1             0,1

1) \(n_{Fe}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(m_{Fe}=0,05.56=2,8\left(g\right)\)

\(m_{FeO}=10-2,8=7,2\left(g\right)\)

2) Có : \(m_{FeO}=7,2\left(g\right)\)

\(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

\(n_{H2SO4\left(tổng\right)}=0,05+0,1=0,15\left(mol\right)\)

\(m_{H2SO4}=0,15.98=14,7\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{14,7.100}{9,8}=150\left(g\right)\)

3) \(n_{FeSO4\left(tổng\right)}=0,05+0,1=0,15\left(mol\right)\)

⇒ \(m_{FeSO4}=0,15.152=22,8\left(g\right)\)

\(m_{ddspu}=10+150-\left(0,05.2\right)=159,9\left(g\right)\)

\(C_{FeSO4}=\dfrac{22,8.100}{159,9}=14,26\)⁰/₀

 Chúc bạn học tốt

Đáp án cần chọn là: B

Gọi số mol H₂, O₂ là a, b (mol)

=> \(\left\{{}\begin{matrix}a+b=\dfrac{22,4}{22,4}=1\\M_B=\dfrac{2a+32b}{a+b}=5,5.2=11\left(g/mol\right)\end{matrix}\right.\)

=> a = 0,7 (mol); b = 0,3 (mol)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

Xét tỉ lệ: \(\dfrac{0,7}{2}>\dfrac{0,3}{1}\) => H₂ dư, O₂ hết

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,6<--0,3------->0,6

=> \(\left\{{}\begin{matrix}m_{H_2O}=0,6.18=10,8\left(g\right)\\m_{H_2\left(dư\right)}=\left(0,7-0,6\right).2=0,2\left(g\right)\end{matrix}\right.\)

\(a,\\ Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)

Loại phản ứng: Phản ứng thế

\(b,n_{Fe}=2.n_{Fe_2O_3}+3.n_{Fe_3O_4}=2.\dfrac{32}{160}+3.0,15=0,85\left(mol\right)\\ m_{Fe}=0,85.56=47,6\left(g\right)\\ c,n_{H_2}=\dfrac{32}{160}.3+4.0,15=1,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=1,2.22,4=28\left(l\right)\)

Em xem sao oxit sắt lại hỏi KL nhôm nha! Vô lí!!!