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\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 < 0,4 ( mol )
0,1 0,1 ( mol )
\(V_{H_2}=0,1.22,4=2,24l\)
\(n_{ZnCl_2}=\dfrac{0,1.1}{1}=0,1mol\)
`a)PTHH:`
`Fe + 2HCl -> FeCl_2 + H_2`
`0,3` `0,6` `0,3` `0,3` `(mol)`
`n_[Fe]=[22,4]/56=0,4(mol)`
`n_[HCl]=0,3.2=0,6(mol)`
Ta có:`[0,4]/1 > [0,6]/2`
`=>Fe` dư
`b)m_[FeCl_2]=0,3.127=38,1(g)`
`c)m_[Fe(dư)]=(0,4-0,3).56=5,6(g)`
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(n_{HCl}=0,3.2=0,6\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Xét: \(\dfrac{0,4}{1}>\dfrac{0,6}{2}\) ( mol )
0,3 0,6 0,3 ( mol )
\(m_{FeCl_2}=0,3.127=38,1\left(g\right)\)
\(m_{Fe\left(dư\right)}=\left(0,4-0,3\right).56=5,6\left(g\right)\)
1)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15->0,3--->0,15-->0,15
=> VH2 = 0,15.22,4 = 3,36 (l)
c) mdd sau pư = 8,4 + 250 - 0,15.2 = 258,1 (g)
=> \(C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%=7,38\%\)
2)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4---->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
mZnCl2 = 0,2.136 = 27,2 (g)
c) \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,2}=2M\)
d)
PTHH: A + 2HCl --> ACl2 + H2
0,2<--0,4
=> \(M_A=\dfrac{4,8}{0,2}=24\left(g/mol\right)\)
=> A là Mg(Magie)
a) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5-------1---------0,5------0,5
b) \(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)
c) \(H_2+CuO\rightarrow Cu+H_2O\)
0,5-----0,5------0,5----0,5
Khối lượng đồng tạo thành: \(m_{Cu}=n_{Cu}.64=0,5.64=32\left(g\right)\)
a) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
0,5-------------------------->0,5`
b) `V_{H_2} = 0,5.22,4 = 11,2 (l)`
c) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,5---->0,5
`=> m_{Cu} = 0,5.64 = 32 (g)`
\(Fe+2HCl\underrightarrow{t^o}FeCl_2+H_2\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(V_{H_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)
\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(m_{Cu}=n.M=0,5.64=32\left(g\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)
ti le 1 : 2 : 1 : 1
n(mol) 0,5-->1--------->0,5------>0,5
\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)
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: \(n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\)
\(m_{FeCl_2}=n_{FeCl_2}\times M_{FeCl_2}=0,2.127=25,4\left(g\right)\)
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
\(\dfrac{27}{280}\)----------->\(\dfrac{27}{280}\)-->\(\dfrac{27}{280}\)
=> VH2 = \(\dfrac{27}{280}.22,4=2,16\left(l\right)\)
c) \(n_{FeCl_2}=\dfrac{27}{280}\left(mol\right)\)
a. \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b. \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5-------1---------0,5-----0,5
Theo PTHH: \(\Rightarrow n_{H_2}=n_{Fe}=0,5\left(mol\right)\)
\(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)
c. \(H_2+CuO\rightarrow Cu+H_2O\)
0,5-------0,5-----0,5----0,5
\(\Rightarrow m_{Cu}=n_{Cu}.M_{Cu}=0,5.64=32\left(g\right)\)
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)