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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, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
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
\(V_{FeCl_2}=0,1.2=0,2\left(l\right)\)
\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0.1}{0,2}=0,5\left(M\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)
\(Fe(0,1)+2HCl(0,2)--->FeCl_2(0,1)+H_2(0,1)\)
\(nFe=0,2(mol)\)
\(nHCl=0,2\left(mol\right)\)
So sánh: \(\dfrac{nFe}{1}=0,2>\dfrac{nHCl}{2}=0,1\)
=> Fe còn dư sau phản ứng, chọn nHCl để tính
\(a)\)
Theo PTHH: \(nH_2=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}\left(đktc\right)=2,24\left(l\right)\)
\(b)\)
Theo PTHH: \(nFe\left(pứ\right)=0,1\left(mol\right)\)
\(\Rightarrow nFe\left(dư\right)=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow mFe\left(dư\right)=0,1.56=5,6\left(g\right)\)
\(c)\)
Dung dich sau phản ứng là FeCl2
Theo PTHH: \(nFeCl_2=0,1\left(mol\right)\)
Nồng độ mol/l các chất sau phản ứng
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
Ta co pthh
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
Theo pthh
nFe=\(\dfrac{11,2}{56}=0,2mol\)
nHCl=\(\dfrac{100.2}{1000}=0,2mol\)
Theo pthh
nFe=\(\dfrac{0,2}{1}mol>nHCl=\dfrac{0,2}{2}mol\)
\(\Rightarrow\) Fe du sau phan ung
a, Theo pthh
nH2=\(\dfrac{1}{2}nHCl=\dfrac{1}{2}.0,2=0,1mol\)
\(\Rightarrow\)VH2=0,1.22,4=2,24 l
b, Theo pthh
nFe=\(\dfrac{1}{2}nHCl=\dfrac{1}{2}.0,2=0,1mol\)
\(\Rightarrow\) So gam Fe du sau phan ung la
mFe=(0,2-0,1).56=5,6 g
c, Theo pthh
nFeCl2=\(\dfrac{1}{2}nHCl=\dfrac{1}{2}.0,2=0,1mol\)
\(\Rightarrow\)CM=\(\dfrac{n}{V}=\dfrac{0,1}{0,1}=1M\)
\(a,n_{HCl}=0,1.1=0,1\left(mol\right)\\ n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
LTL: \(0,1>\dfrac{0,1}{2}\) => Fe dư
Theo pthh: \(n_{H_2}=n_{FeCl_2}=n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> VH2 = 0,05.22,4 = 1,12 (l)
b, Chất dư là Fe
mFe (dư) = (0,1 - 0,05).56 = 2,8 (g)
c, \(C_{M\left(FeCl_2\right)}=\dfrac{0,05}{0,1}=0,5M\)
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)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=C_M.V=2.0,1=0,2\left(mol\right)\)
PTHH: Fe + 2HCl \(\xrightarrow[]{}\) FeCl2 + H2
...............1.......2..................1................1
...........0,1...0,2............0,1..........0,1(mol)
\(\dfrac{n_{Fe}}{1}>\dfrac{n_{HCl}}{2}\)
\(\dfrac{0,2}{1}>\dfrac{0,2}{2}\)
=> Fe dư
V\(H_2\) =n.22,4=0,1.22,4=2,24(l)
b) -Fe dư
n\(Fe_{dư}\)=n\(Fe_{đầu}\) - n \(Fe_{pứ}\) =0,2-0,1=0,1 (mol)
m \(Fe_{dư}\) =n.M=0,1.56=5,6(g)
c)C\(M_{FeCl_2}\) =\(\dfrac{n}{V}=\dfrac{0,1}{0,1}=1M\)
C\(M_{H_2}\) =\(\dfrac{n}{V}=\dfrac{0,1}{0,1}=1M\)
nFe = \(\dfrac{11,2}{56}\)= 0,2mol
V = 100 ml = 0,1 (l)
nHCl = 0,1 .2 = 0,2mol
Fe + 2HCl -> FeCl2 + H2
0,2(dư);0,2(hết) -> 0,1mol
=> VH2 = 0,1 . 22,4 = 2,24 (l)
=>mdư = 0,1 .56 = 5,6 g
D = \(\dfrac{0,2}{0,1}\) = 2mol/l
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\C_{M_{FeCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
n H2 = n Fe = 11,2/56 = 0,2(mol)
V H2 = 0,2.22,4 = 4,48(lít)
b)
n HCl = 2n Fe = 0,2.2 = 0,4(mol)
=> CM HCl = 0,4/0,4 = 1M
c)
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
Ta thấy :
n CuO = 64/80 = 0,8 > n H2 = 0,2 nên CuO dư
Theo PTHH :
n CuO pư = n Cu = n H2 = 0,2(mol)
n Cu dư = 0,8 - 0,2 = 0,6(mol)
Vậy :
%m Cu = 0,2.64/(0,2.64 + 0,6.80) .100% = 21,05%
%m CuO = 100% -21,05% = 78,95%