a) nFe=0,1(mol); nHCl=0,4(mol)

PTHH: Fe + 2 HCl -> FeCl2 + H2

Ta có: 0,1/1 < 0,4/2

=> Fe hết, HCl dư, tish theo nFe.

b) nH2=nFeCl2=Fe=0,1(mol)

=> V(H2,đktc)=0,1.22,4=2,24(l)

c) mFeCl2=127.0,1=12,7(g)

a) nFe=0,1(mol); nHCl=0,4(mol) PTHH: Fe + 2 HCl -> FeCl2 + H2 Ta có: 0,1/1 < 0,4/2 => Fe hết, HCl dư, tish theo nFe. b) nH2=nFeCl2=Fe=0,1(mol) => V(H2,đktc)=0,1.22,4=2,24(l) c) mFeCl2=127.0,1=12,7(g)

\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1  < 0,4                              ( mol )

0,1                                    0,1   ( mol )

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

\(n_{ZnCl_2}=\dfrac{0,1.1}{1}=0,1mol\)

`n_[FeO]=[14,4]/72=0,2(mol)`

`a)PTHH:`

`FeO + 2HCl -> FeCl_2 + H_2 O`

`0,2`       `0,4`            `0,2`                          `(mol)`

`b)m_[FeCl_2]=0,2.127=25,4(g)`

`c)m_[dd HCl]=[0,4.36,5]/10 .100=146(g)`

`d)C%_[FeCl_2]=[25,4]/[14,4+146].100~~15,84%`

\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

ti le        1      :    2       :      1      :    1

n(mol)     0,5-->1--------->0,5------>0,5

\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)

\(a,\text{Sơ đồ p/ứ: }Fe+HCl\to FeCl_2+H_2\\ b,PTHH:Fe+2HCl\to FeCl_2+H_2\\ c,\text{Bảo toàn KL: }m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}+56=150+8=158\\ \Rightarrow m_{HCl}=102(g)\)

\(a,n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

          0,05<-0,1<------0,05<---0,05

\(b,m_{Fe}=0,05.56=2,8\left(g\right)\\ c,m_{ddHCl}=\dfrac{0,1.36,5}{14,6\%}=25\left(g\right)\\ m_{dd}=25+2,8-0,05.2=27,7\left(g\right)\\ \rightarrow C\%_{FeCl_2}=\dfrac{0,05.127}{27,7}.100\%=22,92\%\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{5,4}{56}=\dfrac{27}{280}\left(mol\right)\)

Fe + 2HCl --> FeCl₂ + H₂

Xét tỉ lệ: \(\dfrac{\dfrac{27}{280}}{1}< \dfrac{0,4}{2}\) => Fe hết, HCl dư

Fe + 2HCl --> FeCl₂ + H₂

\(\dfrac{27}{280}\)----------->\(\dfrac{27}{280}\)-->\(\dfrac{27}{280}\)

=> V_H2 = \(\dfrac{27}{280}.22,4=2,16\left(l\right)\)

c) \(n_{FeCl_2}=\dfrac{27}{280}\left(mol\right)\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

tỉ lệ:          1      :    2     :     1        :     1

n(mol)      0,2--->0,4------->0,2----->0,2

\(m_{FeCl_2}=n\cdot M=0,2\cdot\left(56+35,5\cdot2\right)=25,4\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\\ V_{H_2\left(dkc\right)}=n\cdot24,79=4,958\left(l\right)\)

a) \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

b) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo phương trình hóa học: \(n_{H_2}=n_{Fe}=o,2\left(mol\right)\)

\(V_{H_2\left(dktc\right)}=n_{H_2}\times22,4=0,2\times22,4=4,48\left(l\right)\)

\(V_{H_2\left(dkc\right)}=n_{H_2}\times24,79=0,2\times24,79=4,96\left(l\right)\)

c) Theo phương trình hóa học: \(n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\)

\(m_{FeCl_2}=n_{FeCl_2}\times M_{FeCl_2}=0,2.127=25,4\left(g\right)\)

BTKL: \(m_{Fe}+m_{HCl}=m_{muối}+m_{H_2}\)

 \(\Rightarrow m_{H_2}=5,6+7,3-12,7=0,2\left(g\right)\)