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\(a,n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}95a+133,5b=10,475\\a+1,5b=0,115\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\\ \%m_{Mg}=\dfrac{0,04.24}{0,04.24+0,05.27}.100\approx41,558\%\Rightarrow\%m_{Al}\approx58,442\%\\ b,n_{HCl}=2.n_{H_2}=2.0,115=0,23\left(mol\right)\\ \Rightarrow x=C\%_{ddHCl}=\dfrac{0,23.36,5}{100}.100=8,395\%\)
\(n_{Fe}=a\left(mol\right),n_{Mg}=b\left(mol\right)\)
\(m_{hh}=56a+24b=10.16\left(g\right)\)
\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{H_2}=a+b=0.25\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.13,b=0.12\)
\(m_{Fe}=0.13\cdot56=7.28\left(g\right)\)
\(m_{Mg}=0.12\cdot24=2.88\left(g\right)\)
\(n_{HCl}=2\cdot n_{H_2}=2\cdot0.25=0.5\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0.5}{0.5}=1\left(M\right)\)
Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)
⇒ 24x + 27y = 7,8 (1)
Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
BT e, có: 2x + 3y = 0,8 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{7,8}.100\%\approx30,77\%\\\%m_{Al}\approx69,23\%\end{matrix}\right.\)
b, BTNT Mg và Al, có:
nMgCl2 = nMg = 0,1 (mol)
nAlCl3 = nAl = 0,2 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCl_2}=\dfrac{0,1.95}{0,1.95+0,2.133,5}.100\%\approx26,24\%\\\%m_{AlCl_3}\approx73,76\%\end{matrix}\right.\)
Bạn tham khảo nhé!
a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+27b=5,1\) (1)
Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Bảo toàn electron: \(2a+3b=0,5\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%\approx47,06\%\\\%m_{Al}=52,94\%\end{matrix}\right.\)
b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{7,3\%}=250\left(g\right)\)
\(\Rightarrow V_{HCl}=\dfrac{250}{1,2}\approx208,33\left(ml\right)\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{20,16}{22,4}=0,9mol\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Mg}=y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=27x\\m_{Mg}=24y\end{matrix}\right.\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x \(\dfrac{3}{2}x\) ( mol )
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}27x+24y=19,8\\\dfrac{3}{2}x+y=0,9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,6\end{matrix}\right.\)
\(\Rightarrow m_{Al}=0,2.27=5,4g\)
\(\Rightarrow m_{Mg}=0,6.24=14,4g\)
\(\%m_{Al}=\dfrac{5,4}{19,8}.100=27,27\%\)
\(\%m_{Mg}=100\%-27,27\%=72,73\%\)
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)
=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,12.95}{157,28}.100\%=7,25\%\\C\%_{FeCl_2}=\dfrac{0,16.127}{157,28}.100\%=12,92\%\end{matrix}\right.\)
\(a) n_{Mg}= a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b =2,55(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{2,8}{22,4}=0,125(2)\\ (1)(2) \Rightarrow a = b = 0,05\\ \%m_{Mg} = \dfrac{0,05.24}{2,55}.100\% = 47,06\%\ ;\ \%m_{Al} =100\% -47,06\% = 52,94\%\\ b) n_{HCl} = 2n_{H_2} = 0,125.2 = 0,25(mol)\\ m_{dd\ HCl} = \dfrac{0,25.36,5}{7,3\%} = 125(gam)\\ V_{dd\ HCl} = \dfrac{125}{1,2} = 104,17(ml)\)
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
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