nFeO=48 : 72 = 0,67 (mol) 
pthh : FeO + H2 -t-> Fe + H2O
          0,67-->0,67--->0,67 (mol ) 
VH2 = 0,67 . 22,4 = 14,93 (l) 
mFe = 0,67 . 56 = 37,52 (g) 

Bài 3:

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

a, PTHH: Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

                0,1<------0,4

              Zn + 2HCl ---> ZnCl₂ + H₂

             0,4<-------------------------0,4

b, m_Fe3O4 = 0,1.232 = 23,2 (g)

c, m_Zn = 0,4.65 = 26 (g)

Bài 4:

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

a, PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

              0,1---->0,2---------------->0,1

b, V_H2 = 0,1.22,4 = 2,24 (l)

c, \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)

a.b.c.\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{9,8}{98}=0,1mol\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

 0,1       0,1                          0,1 ( mol )

\(V_{H_2}=n.22,4=0,1.22,4=2,24l\)

\(m_{Zn}=n.M=0,1.65=6,5g\)

d.\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{1,6}{160}=0,01mol\)

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

 0,01 <   0,1                                        ( mol )

 0,01                          0,02               ( mol )

\(m_{Fe}=n.M=0,02.56=1,12g\)

a,PTHH:Zn+H2SO4-->ZnSO4+H2

b,nH2SO4=9,8÷(2×32+16×4)=0,1

PTHH:Zn+H2SO4--->ZnSO4+H2

           0,1    0,1            0,1          0,1

-->Vh2=0,1×22,4=2,24

c,nZn=0,1

mZn=0,1×65=6,5

d,nH2=0,1,nfe2o3=1,6+(56×2+16×3)=0,01

PTHH:3H2+Fe2O3--->2Fe+3H2O

              0,1      0,01

 Ta thấy 0,1/3>0,1/1 nên sau pư h2 dư,fe2o3 hết.tính theo fe2o3

nfe=2nfe2o3=0,02

--->mfe=0,02×56=1,12       

PT: Fe2O3+3H2to→2Fe+3H2O

CuO+H2to→Cu+H2O

a, Ta có:  mFe2O3=20.60%=12(g)

⇒nFe2O3=\(\dfrac{12}{160}\)=0,075(mol

mCuO=20−12=8(g

⇒nCuO=\(\dfrac{8}{80}\)=0,1(mol)

Theo pT: 

nFe=2nFe2O3=0,15(mol)

nCu=nCuO=0,1(mol)

⇒mFe=0,15.56=8,4(g)

mCu=0,1.64=6,4(g)

b, Theo PT: nH2=3nFe2O3+nCuO=0,325(mol)

⇒VH2=0,325.22,4=7,28(l)

c. Zn+2HCl->ZnCl2+H2

               0,65----------0,325

=>m HCl=0,65.36,5=23,725g

Ủa bạn cái câu a . 20x60% ( 20 ở đâu vậy bạn

\(nFeO=\dfrac{16}{56+16}=\dfrac{2}{9}\left(mol\right)\)

\(FeO+H_2\rightarrow Fe+H_2O\)

2/9       2/9      2/9    2/9

\(mFe=\dfrac{2}{9}.56=12,4\left(g\right)\)

\(VH_2=\dfrac{2}{9}.22,4=4,98\left(l\right)\)

a) Fe₂O₃  + 3 H₂ → 2Fe + 3H₂O

b) nFe₂O₃  = \(\dfrac{39}{160}\)= 0,24375 mol

=> nFe = 2nFe₂O₃ = 0,24375.2 = 0,4875 mol

c) Theo pt phản ứng nH₂ = 3nFe

=> nH₂ = 0,24375. 3 =0.73125 mol

<=> V_H2 = 0.73125 . 22,4 = 16,38 lít

a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2\left(mol\right)\Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\)

\(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

c, \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\), ta được H₂ dư.

Theo PT: \(n_{Cu}=n_{CuO}=0,2\left(mol\right)\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)

\(a/Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\\ b/n_{H_2}=n_{FeCl_2}=0,1mol\\ m_{FeCl_2}=0,1.127=12,7\left(g\right)\\ c/V_{H_2}=0,1.22,4=2,24\left(l\right)\\ d/n_{HCl}=0,1.2=0,2\left(mol\right)\\ V_{HCl\left(pư\right)}=\dfrac{0,2}{2}=0,1\left(l\right)\)

\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

Còn lại giống câu dưới nha 

\(n_{FeO}=\dfrac{32}{72}=\dfrac{4}{9}\left(mol\right)\)

\(FeO+H_2\underrightarrow{^{t^0}}Fe+H_2O\)

\(\dfrac{4}{9}.....\dfrac{4}{9}....\dfrac{4}{9}\)

\(V_{H_2}=\dfrac{4}{9}\cdot22.4=10\left(l\right)\)

\(m_{Fe}=\dfrac{4}{9}\cdot56=24.89\left(g\right)\)

\(a) n_{FeO} = \dfrac{32}{72} = \dfrac{4}{9}(mol)\\ FeO + H_2 \xrightarrow{t^o} Fe + H_2O\\ n_{Fe} = n_{H_2} = n_{FeO} = \dfrac{4}{9}(mol)\\ V_{H_2} = \dfrac{4}{9}.22,4 = 9,95(lít)\\ b) m_{Fe}= \dfrac{4}{9}.56 = 24,8(gam)\)