Cho 5,6g Fe tdụng với 100ml dd HCl 1M , sản phẩm là FeCl𝟸 và H𝟸 a.Sau pứ chất nào hết?chất nào còn dư ? và dư bao nhiêu?? b.Tính khối lượng FeCl𝟸 và thể tích H₂(đktc) sau pứ
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\
n_{HCl}=0,5.1=0,5mol\\
Fe+2HCl\rightarrow FeCl_2+H_2\\
\Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\
Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95g\\ b)m_{FeCl_2}=0,1.127=12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl\left(dư\right)}=\dfrac{10,95}{200}\cdot100=5,475\%\\ C_{\%HCl\left(pư\right)}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,5.1=0,5\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{2}\), ta được HCl dư.
Theo PT: \(n_{HCl\left(pư\right)}=2n_{Fe}=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,5-0,2=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)
b, \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
c, \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
d, \(m_{HCl}=0,5.36,5=18,25\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)
\(a.n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{HCl}=2n_{Fe}=0,2mol\\ m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95\%\\ b)n_{Fe}=n_{FeCl_2}=n_{H_2}=0,1mol\\ m_{FeCl_2}=0,1.12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\), ta được Fe dư.
a, Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
⇒ VH2 = 0,05.22,4 = 1,12 (l)
b, Sau pư, Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
⇒ mFe (dư) = 0,05.56 = 2,8 (g)
c, Theo PT: nFeCl2 = nFe (pư) = 0,05 (mol)
\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,05}{0,1}=0,5M\)
Bạn tham khảo nhé!
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\dfrac{n_{Fe}}{1}=0.1>\dfrac{n_{HCl}}{2}=\dfrac{0.1}{2}=0.05\)
\(\Rightarrow Fedư\)
\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(m_{Fe\left(dư\right)}=\left(0.1-0.05\right)\cdot56=2.8\left(g\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0.05}{0.1}=0.5\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
PTHH :
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Trc p/u: 0,1 0,1
p/u : 0,05 0,1 0,05 0,05
Sau p/u : 0,05 0 0,05 0,05
-> Fe dư sau p/u
a) \(m_{H_2}=0,05.2=0,1\left(g\right)\)
b) sau p/ư Fe dư
\(m_{Fedư}=0,05.2,8\left(g\right)\)
c) \(m_{FeCl_2}=0,05.\left(56+35,5.2\right)=6,35\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)
\(n_{HCl}=0,1\cdot1=0,1mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,1 0 0
0,05 0,1 0,05 0,05
0,05 0 0,05 0,05
\(V_{H_2}=0,05\cdot22,4=1,12k\)
Sau phản ứng Fe còn dư và dư \(m=0,05\cdot56=2,8g\)
\(C_{M_{FeCl_2}}=\dfrac{0,05}{0,1}=0,5M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\), ta được Fe dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b, Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(\Rightarrow m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\)
c, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
PTHH :
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
trc p/u : 0,1 0,1
p/u : 0,05 0,1 0,05 0,05
sau p/u: 0,05 0 0,05 0,05
---> sau p/ư : Fe dư
\(a,V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b, \(m_{Fedư}=0,05.56=2,8\left(g\right)\)
\(c,_{FeCl_2}=0,05.127=6,35\left(g\right)\)
\(m_{ddFeCl_2}=5,6+\left(0,1.36,5\right)-\left(0,05.1\right)=9,2\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{6,35}{9,2}.100\%\approx69\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{KOH}=0,2\cdot1=0,2\left(mol\right)\\n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) Axit còn dư 0,1 mol
\(\Rightarrow n_{H_2}=0,1\left(mol\right)\) \(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
`Fe + 2HCl -> FeCl_2 + H_2`
`0,05` `0,1` `0,05` `0,05` `(mol)`
`a)n_[Fe]=[5,6]/56=0,1(mol)`
`n_[HCl]=1.0,1=0,1(mol)`
Ta có: `[0,1]/1 > [0,1]/2`
`=>Fe` dư.
`=>m_[Fe(dư)]=(0,1-0,05).56=2,8(g)`
`b)m_[FeCl_2]=0,05.127=6,35(g)`
`V_[H_2]=0,05.22,4=1,12(l)`