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\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ a,\%m_{Fe}=\dfrac{0,02.56}{4,36}.100\approx25,688\%\\ \Rightarrow\%m_{Ag}\approx74,312\%\\ b,Ta.thấy:2,18=\dfrac{1}{2}.4,36\\ \Rightarrow m_{hh\left(câuB\right)}=\dfrac{1}{2}.m_{hh\left(câuA\right)}\\ n_{Fe}=\dfrac{0,02}{2}=0,01\left(mol\right)\\ n_{Ag}=\dfrac{2,18-0,01.56}{108}=0,015\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ 2Ag+Cl_2\rightarrow\left(t^o\right)2AgCl\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+\dfrac{1}{2}.n_{Ag}=\dfrac{3}{2}.0,01+\dfrac{1}{2}.0,015=0,0225\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc\right)}=0,0225.22,4=0,504\left(l\right)\)
a) Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow27a+56b=11\) (1)
Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Bảo toàn electron: \(3a+2b=0,4\cdot2=0,8\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{11}\cdot100\%\approx49,09\%\\\%m_{Fe}=50,91\%\end{matrix}\right.\)
b) Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{muối}=m_{AlCl_3}+m_{FeCl_2}=0,2\cdot133,5+0,1\cdot127=39,4\left(g\right)\)
c) Bảo toàn electron: \(3\cdot0,2+3\cdot0,1=2n_{SO_2}\)
\(\Rightarrow n_{SO_2}=0,45\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,45\cdot22,4=10,08\left(l\right)\)
a) Gọi nAl = x, nFe = y
Có 27x + 56y = 11 (1)
Bảo toàn e
3x + 2y = 2.0,4 (2)
Từ 1 và 2 => x = 0,2, y = 0,1
\(\%mAl=\dfrac{0,2.27}{11}.100\%=49,09\%\)
\(\%mFe=100-49,09=50,91\%\)
b) BTKL:
m muối = mkim loại + mHCl - mH2
= 11 + 0,4.2.36,5 - 0,4.2 = 39,4g
c)
Bảo toàn e
Al => Al+3 + 3e S+6 + 2e => S+4
0,2 0,6 2x x
Fe => Fe+3 + 3e
0,1 0,3
=> 2x = 0,6 + 0,3 => x = 0,45 mol
=> VSO2 = 0,45.22,4 = 10,08 lít
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)
=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,12.95}{157,28}.100\%=7,25\%\\C\%_{FeCl_2}=\dfrac{0,16.127}{157,28}.100\%=12,92\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15<--------------------0,15
=> \(\%Fe=\dfrac{0,15.56}{14,8}.100\%=56,757\%\)
=> \(\%Cu=100\%-56,757\%=43,243\%\)
nH2 = 3,36/22,4 = 0,15 (mol)
PTHH: Fe + 2HCl -> FeCl2 + H2
nFe = 0,15 (mol)
mFe = 0,15 . 56 = 8,4 (g)
%mFe = 8,4/14,8 = 56,75%
%mCu = 100% - 56,75% = 43,25%
\(a)n_{Mg} = a(mol) ; n_{Fe} = b(mol)\Rightarrow 24a + 56b = 10,4(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} =a +b = \dfrac{6,72}{22,4} = 0,3(2)\\ (1)(2)\Rightarrow a = 0,2 ; b = 0,1\\ n_{MgCl_2} = a = 0,2(mol) \Rightarrow m_{MgCl_2} = 0,2.95 = 19(gam)\\ n_{FeCl_2} = b = 0,1(mol) \Rightarrow m_{FeCl_2} = 0,1.127 = 12,7(gam)\\ b)\\ \%m_{Mg} = \dfrac{0,2.24}{10,4}.100\% = 46,15\%\\ \%m_{Fe} = 100\% -46,15\% = 53,85\%\)
chỉ vs ạ
\(n_{Mg}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)
\(m_{hh}=24a+56b=18.8\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\Rightarrow a+b=0.45\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.2,b=0.25\)
\(\%Mg=\dfrac{0.2\cdot24}{18.8}\cdot100\%=25.53\%\)
\(\%Fe=100-25.53=77.47\%\)