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Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=18,6(1)\)
\(n_{HCl}=3.0,2=0,6(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow 2x+2y=0,6(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Zn}=0,2.65=13(g)\)
Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=9,3(1)\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow x+y=0,15(2)\\ (1)(2)\Rightarrow x=0,05(mol);y=0,1(mol)\\ \Rightarrow \%_{Zn}=\dfrac{0,1.65}{9,3}.100\%=69,89\%\)
Đặt \(n_{Fe}=x(mol);n_{Mg}=y(mol)\Rightarrow 56x+24y=8(1)\)
\(PTHH:Fe+2HCl\to FeCl_2+H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow 127x+95y=22,2(2)\\ (1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{8}.100\%=70\%\)
Đặt \(n_{FeO}=x(mol);n_{Zn}=y(mol)\Rightarrow 72x+65y=9,52(1)\)
\(n_{HCl}=0,28.1=0,28(mol)\\ PTHH:FeO+2HCl\to FeCl_2+H_2O\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow 2x+2y=0,28(2)\\ (1)(2)\Rightarrow x=0,06(mol);y=0,08(mol)\\ \Rightarrow m_{FeO}=0,06.72=4,32(g)\)
Gọi số mol FeO, Zn là a,b
=> 72a + 65b = 9,52 (1)
nHCl = 0,28.1 = 0,28(mol)
PTHH: FeO + 2HCl --> FeCl2 + H2O
_______a--->2a
Zn + 2HCl --> ZnCl2 + H2
b----->2b
=> 2a + 2b = 0,28
=> a + b = 0,14 (2)
(1)(2) => \(\left\{{}\begin{matrix}a=0,06\\b=0,08\end{matrix}\right.\)
=> mFeO = 0,06.72 = 4,32(g)
\(n_{Br_2}=0.3\cdot1=0.3\left(mol\right)\)
\(C_3H_6+Br_2\rightarrow C_3H_6Br_2\)
\(0.3.........0.3\)
\(n_{CO_2}=\dfrac{48.4}{44}=1.1\left(mol\right)\)
\(BảotoànC:\)
\(n_{C_2H_6}=\dfrac{1.1-0.3\cdot3}{2}=0.2\left(mol\right)\)
\(\%C_2H_6=\dfrac{0.2\cdot30}{0.2\cdot30+0.3\cdot42}\cdot100\%=32.25\%\)
\(\%C_3H_6=67.75\%\)
\(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)
\(n_{HCl}=0.35\cdot2=0.7\left(mol\right)\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(1.................6\)
\(0.1.............0.7\)
Lập tỉ lệ : \(\dfrac{0.1}{1}< \dfrac{0.7}{6}\Rightarrow HCldư\)
\(m_{AlCl_3}=0.1\cdot2\cdot133.5=26.7\left(g\right)\)
Gọi số mol Fe, Zn là a,b
=> 56a + 65b = 12,1 (1)
PTHH: Fe + 2HCl --> FeCl2 + H2
______a---->2a
Zn + 2HCl --> ZnCl2 + H2
_b---->2b
=> 2a + 2b = 0,4 (2)
(1)(2) => \(\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
=> \(\%Fe=\dfrac{56.0,1}{12,1}.100\%=46,28\%\)
Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=12,1(1)\)
\(n_{HCl}=2.0,2=0,4(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow 2x+2y=0,4(2)\\ (1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\)