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a. Đổi 200 ml = 0,2 lít
\(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=2.0,2=0,2\left(mol\right)\)
PTHH : Fe + 2HCl -> FeCl2 + H2
0,1 0,2 0,1 0,1
Ta thấy : \(\dfrac{0.2}{1}>\dfrac{0.2}{2}\) => Fe dư , HCl đủ
\(m_{Fe\left(dư\right)}=\left(0,2-0,1\right).56=5,6\left(g\right)\)
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. Sau phản ứng chất tan là FeCl2
\(V_{FeCl_2}=0,1.2=0,2\left(l\right)\)
\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0.1}{0,2}=0,5\left(M\right)\)
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=0,1.2=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\), ta được Fe dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe\left(dư\right)}=0,1.56=5,6\left(g\right)\)
c, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=n_{FeCl_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ a,m_{Fe}=0,2.56=11,2\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,V_{ddFeCl_2}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{HCl}=1.0,1=0,1\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,1}{1}\Rightarrow Fe.dư\\ n_{H_2}=n_{FeCl_2}=n_{Fe\left(p.ứ\right)}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b,n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\\ c,V_{ddFeCl_2}=V_{ddHCl}=0,1\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
nFe = 0,1 mol
nHCl = 0,3 mol
Fe + 2HCl ---> FeCl2 + H2
0,1 < 0,3/2 .....=> HCl dư sau phản ứng
nFeCl2 = 0,1 mol => CM = 0,1/0,2 = 0,5M
nHCl(dư) = 0,1 mol => CM = 0,1/0,2 = 0,5M
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{HCl}=0,2\cdot1,5=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) \(\Rightarrow\) Fe p/ứ hết, HCl còn dư
\(\Rightarrow n_{HCl\left(dư\right)}=0,1\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\)
c) Theo PTHH: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\)
\(\Rightarrow C_{M_{FeCl_2}}=C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
`n_[Fe]=[11,2]/56=0,2(mol)`
`n_[HCl]=0,3.2=0,6(mol)`
`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`
`0,2` `0,4` `0,2` `0,2` `(mol)`
`a)` Ta có:`[0,2]/1 < [0,6]/2`
`=>HCl` dư
`=>V_[H_2]=0,2.22,4=4,48(l)`
`b)HCl` còn dư sau p/ứ
`=>m_[HCl(dư)]=(0,6-0,4).36,5=7,3(g)`
`c)C_[M_[FeCl_2]]=[0,2]/[0,3]~~0,67(M)`
`C_[M_[HCl(dư)]=[0,6-0,4]/[0,3]~~0,67(M)`
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=0,3.2=0,6\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 < 0,6 ( mol )
0,2 0,4 0,2 0,2 ( mol )
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
Chất dư là HCl
\(m_{HCl\left(dư\right)}=\left(0,6-0,4\right).36,5=7,3\left(g\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,3}=0,66\left(M\right)\)
\(C_{M_{HCl\left(dư\right)}}=\dfrac{0,2}{0,3}=0,66\left(M\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)
\(n_{HCl}=2,5.0,2=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 < 0,5 ( mol )
0,2 0,4 0,2 0,2 ( mol )
Chất dư là HCl
\(m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65g\)
\(V_{H_2}=0,2.22,4=4,48l\)
\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,2}=1M\)
nFe = 5.6/56 = 0.1 (mol)
nHCl = 0.2*2 = 0.4 (mol)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
LTL : 0.1/1 < 0.4/2 => HCl dư
mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g)
VH2 = 0.2*22.4 = 4.48 (l)
CM FeCl2 = 0.1/0.2 = 0.5(M)
CM HCl dư = 0.2 / 0.2 = 1(M)
CM FeCl2 = n/v = 0,1 / 2,24 = 0,446 chứ nhỉ