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- Từ giả thiết tính được : n Cl 2 = 0,035 mol; n O 2 = 0,025 mol
Theo ĐLBT khối lượng :
Từ (3)(4) ⇒ x = 0,04; y = 0,03
⇒ m Mg = 0,04.24 = 0,96g; m Al = 0,03.27 = 0,81g
Phản ứng không oxi hoá - khử
\(n_{H^+} = n_{HCl} = 0,12.2 = 0,24(mol)\\ 2H^+ + O^{2-} \to H_2O\\ n_{O(oxit)} = \dfrac{1}{2}n_{H^+} = 0,12(mol)\\ \Rightarrow n_{O_2} = \dfrac{n_{O(oxit)}}{2} = 0,06(mol)\\ n_{Mg} = \dfrac{1,68}{24} = 0,07(mol) ; n_{Al} = \dfrac{2,16}{27} = 0,08(mol)\)
Bảo toàn electron :
\(2n_{Mg} + 3n_{Al} = 4n_{O_2} + 2n_{Cl_2}\\ \Rightarrow n_{Cl_2} = \dfrac{0,07.2 + 0,08.3-0,06.4}{2} = 0,07(mol)\\ \Rightarrow \%V_{Cl_2} = \dfrac{0,07}{0,07+0,06}.100\% = 53,85\%\)
\(a,n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}95a+133,5b=10,475\\a+1,5b=0,115\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\\ \%m_{Mg}=\dfrac{0,04.24}{0,04.24+0,05.27}.100\approx41,558\%\Rightarrow\%m_{Al}\approx58,442\%\\ b,n_{HCl}=2.n_{H_2}=2.0,115=0,23\left(mol\right)\\ \Rightarrow x=C\%_{ddHCl}=\dfrac{0,23.36,5}{100}.100=8,395\%\)
\(n_{HCl}=0,12.2=0,24\left(mol\right)\)
=> \(n_{H_2O}=0,12\left(mol\right)\)
=> \(n_{O\left(oxit\right)}=0,12\left(mol\right)\)
=> \(n_{O_2}=0,06\left(mol\right)\)
\(n_{Mg}=\dfrac{1,68}{24}=0,07\left(mol\right)\); \(n_{Al}=\dfrac{2,16}{27}=0,08\left(mol\right)\)
Gọi số mol Cl2 là x (mol)
Mg0 - 2e --> Mg+2
0,07->0,14
Al0 - 3e --> Al+3
0,08->0,24
O20 + 4e --> 2O-2
0,06->0,24
Cl20 + 2e --> 2Cl-
x--->2x
Bảo toàn e: 2x + 0,24 = 0,24 + 0,14
=> x = 0,07 (mol)
=> \(\%V_{Cl_2}=\dfrac{0,07}{0,06+0,07}.100\%=53,846\%\)
a) Gọi số mol Mg, Fe là a, b (mol)
=> 24a + 56b = 11,84
\(n_{HCl}=\dfrac{146.14\%}{36,5}=0,56\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
a--->2a--------->a----->a
Fe + 2HCl --> FeCl2 + H2
b-->2b-------->b------>b
=> 2a + 2b = 0,56
=> a = 0,12; b = 0,16
=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,12.24}{11,84}.100\%=24,324\%\\\%Fe=\dfrac{0,16.56}{11,84}.100\%=75,676\%\end{matrix}\right.\)
b) \(n_{H_2}=a+b=0,28\left(mol\right)\)
=> \(V_{H_2}=0,28.22,4=6,272\left(l\right)\)
c) mdd sau pư = 11,84 + 146 - 0,28.2 = 157,28 (g)
=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,12.95}{157,28}.100\%=7,25\%\\C\%_{FeCl_2}=\dfrac{0,16.127}{157,28}.100\%=12,92\%\end{matrix}\right.\)
a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,05<-----------0,05---->0,075
=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)
=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)
b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)
c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
0,05->0,0375
2Cu + O2 --to--> 2CuO
0,2-->0,1
=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)
a) Gọi số mol Zn, Fe là a, b (mol)
=> 65a + 56b = 7,35 (1)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a---->2a------->a------>a
Fe + 2HCl --> FeCl2 + H2
b------>2b----->b------>b
=> \(a+b=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)
=> a + b = 0,12 (2)
(1)(2) => a = 0,07; b = 0,05
=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,07.65}{7,35}.100\%=61,9\%\\\%m_{Fe}=\dfrac{0,05.56}{7,35}.100\%=38,1\%\end{matrix}\right.\)
b) nHCl(dư) = 0,3.1 - 0,07.2 - 0,05.2 = 0,06 (mol)
PTHH: Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
0,03<-----0,06
=> \(x=C_{M\left(ddCa\left(OH\right)_2\right)}=\dfrac{0,03}{0,1}=0,3M\)
c) Chất rắn thu được là Fe2O3
Bảo toàn Fe: \(n_{Fe_2O_3}=0,025\left(mol\right)\)
=> \(a=m_{Fe_2O_3}=0,025.160=4\left(g\right)\)
Kết tủa thu được là Fe(OH)2
Bảo toàn Fe: \(n_{Fe\left(OH\right)_2}=0,05\left(mol\right)\)
=> \(m=m_{Fe\left(OH\right)_2}=0,05.90=4,5\left(g\right)\)
