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\(Fe\left(0,1\right)+CuSO_4\left(0,1\right)\rightarrow FeSO_4\left(0,1\right)+Cu\left(0,1\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\frac{8,4}{56}=0,15\)
\(m_{ddđ}=100.1,08=108\)
\(n_{CuSO_4}=0,1.1=0,1\)
Ta thấy \(\frac{0,15}{1}>\frac{0,1}{1}\) nên Fe còn dư CuSO4 hết
\(\Rightarrow a=m_{Cu}=0,1.64=6,4\)
\(\Rightarrow m_{Fe\left(pứ\right)}=0,1.56=5,6\)
\(\Rightarrow m_Y=108+5,6-6,4=107,2\)
\(m_{FeSO_4}=0,1.152=15,2\)
\(\Rightarrow C\%\left(FeSO_4\right)=\frac{15,2}{107,2}.100\%=14,18\%\)
\(\left\{{}\begin{matrix}n_{HCl}=0,1\left(mol\right)\\n_{AlCl3}=0,2\left(mol\right)\\n_{Al\left(OH\right)3}=0,1\left(mol\right)\end{matrix}\right.\)
Tạo kết tủa => HCl bị trung hoà hết
\(NaOH+HCl\rightarrow NaCl+H_2O\)
\(\Rightarrow n_{NaOH\left(trung.hoa\right)}=0,1\left(mol\right)\)
- TH1: dư AlCl3
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)
\(\Rightarrow n_{NaOH}=0,3\left(mol\right)\)
\(\Rightarrow\Sigma n_{NaOH}=0,4\left(mol\right)\)
\(\Rightarrow V=0,4\left(l\right)=400\left(ml\right)\)
- TH2: dư NaOH
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)
=> 0,2 mol AlCl3 tạo 0,2 mol Al(OH)3. Có 0,6 mol NaOH phản ứng
=> 0,2-0,1= 0,1 mol Al(OH)3 tan
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
\(\Rightarrow n_{NaOH}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{NaOH}=0,8\left(mol\right)\)
\(\Rightarrow V=0,8\left(l\right)=800\left(ml\right)\)
a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(m_{CuCl_2}=270.10\%=27\left(g\right)\Rightarrow n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)
Ta có: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\) ⇒ Fe hết, CuCl2 dư
PTHH: Fe + CuCl2 ---> FeCl2 + Cu
Mol: 0,15 0,15 0,15 0,15
\(a=m_{Cu}=0,15.64=9,6\left(g\right)\)
b, \(m_{dd.sau.pứ}=8,4+270-9,6=268,8\left(g\right)\)
\(m_{CuCl_2dư}=\left(0,2-0,15\right).135=6,75\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{CuCl_2dư}=\dfrac{6,75.100\%}{268,8}=2,51\%\\C\%_{FeCl_2}=\dfrac{0,15.127.100\%}{268,8}=7,09\%\end{matrix}\right.\)
c, \(V_{ddCuCl_2}=\dfrac{270}{1,35}=200\left(ml\right)=0,2\left(l\right)\)
\(\left\{{}\begin{matrix}C_{M_{CuCl_2dư}}=\dfrac{0,2-0,15}{0,2}=0,25M\\C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75M\end{matrix}\right.\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15mol\)
\(m_{CuCl_2}=\dfrac{270\cdot10\%}{100\%}=27g\Rightarrow n_{CuCl_2}=0,2mol\)
\(Fe+CuCl_2\rightarrow FeCl_2+Cu\)
0,15 0,2 0,15 0,15
\(a=m_{Cu}=0,15\cdot64=9,6g\)
\(m_{FeCl_2}=0,15\cdot127=19,05g\)
\(m_{ddFeCl_2}=8,4+270-0,15\cdot64=268,8g\)
\(C\%=\dfrac{19,05}{268,8}\cdot100\%=7,09\%\)
đây ạ
nFe = 8.4/56=0.15 mol
nCu = 6.4/64=0.1 mol
nAgNO3 = 0.35*2=0.7 mol
Fe + 2AgNO3 --> Fe(NO3)2 + 2Ag
0.15___0.3________0.15_____0.3
Cu + 2AgNO3 --> Cu(NO3)2 + 2Ag
0.1____0.2________0.1_______0.2
nAgNO3( còn lại ) = 0.7 - 0.3 - 0.2 = 0.2 mol
Vì : AgNO3 còn dư nên tiếp tục phản ứng với Fe(NO3)2
Fe(NO3)2 + AgNO3 --> Fe(NO3)3 + Ag
Bđ: 0.15________0.2
Pư: 0.15________0.15_______________0.15
Kt: 0___________0.05_______________0.15
Chất rắn : 0.65 (mol) Ag
mAg = 0.65*108 = 70.2g
B gồm 3 kim loại là Fe, Cu, Ag
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,03<----------------0,03
Gọi số mol Cu, Ag là a, b (mol)
=> 64a + 108b = 8,12 - 0,03.56 = 6,44 (g) (1)
\(\left\{{}\begin{matrix}n_{Al^{3+}}=\dfrac{0,81}{27}=0,03\left(mol\right)\\n_{Fe^{2+}}=\dfrac{2,8}{56}-0,03=0,02\left(mol\right)\end{matrix}\right.\)
=> \(n_{NO_3^-}=0,03.3+0,02.2=0,13\left(mol\right)\)
\(\left\{{}\begin{matrix}n_{Cu\left(NO_3\right)_2}=a\left(mol\right)\\n_{AgNO_3}=b\left(mol\right)\end{matrix}\right.\)
=> 2a + b = 0,13 (2)
(1)(2) => a = 0,05 (mol); b = 0,03 (mol)
=> \(C_{M\left(Cu\left(NO_3\right)_2\right)}=\dfrac{0,05}{0,1}=0,5M\)
