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a)Gọi x,y lần lượt là số mol của Al, Fe trong hỗn hợp ban đầu (x,y>0)
Sau phản ứng hỗn hợp muối khan gồm: \(\left\{{}\begin{matrix}AlCl_3:x\left(mol\right)\\FeCl_2:y\left(mol\right)\end{matrix}\right.\)
Ta có hệ phương trình: \(\left\{{}\begin{matrix}27x+56y=13,9\\133,5x+127y=38\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\approx0,0896\\y\approx0,205\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,0896\cdot27\cdot100\%}{13,9}\approx17,4\%\\\%m_{Fe}=\dfrac{0,205\cdot56\cdot100\%}{13,9}\approx82,6\%\end{matrix}\right.\)
Theo Bảo toàn nguyên tố Cl, H ta có:\(n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{3n_{AlCl_3}+2n_{FeCl_2}}{2}\\ =\dfrac{3\cdot0,0896+2\cdot0,205}{2}=0,3394mol\\ \Rightarrow V_{H_2}=0,3394\cdot22,4\approx7,6l\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x 3x x 1,5x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y 2y y y
\(\left\{{}\begin{matrix}27x+56y=22\\1,5x+y=\dfrac{17,92}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)
\(m_{Al}=0,4\cdot27=10,8g\)
\(m_{Fe}=22-10,8=11,2g\)
\(m_{HCl}=36,5\cdot\left(3x+2y\right)=36,5\cdot\left(3\cdot0,4+2\cdot0,2\right)=58,4g\)
\(m_{ddHCl}=\dfrac{m_{HCl}}{C\%}\cdot100\%=\dfrac{58,4}{25\%}\cdot100\%=233,6g\)
\(Đặt:n_{Al}=u\left(mol\right);n_{Fe}=v\left(mol\right)\left(u,v>0\right)\\ n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56u=22\\1,5a+u=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,4\\u=0,2\end{matrix}\right.\\ \Rightarrow m_{Al}=0,4.27=10,8\left(g\right);m_{Fe}=56.0,2=11,2\left(g\right)\\ n_{HCl}=2.0,8=1,6\left(mol\right)\\ m_{HCl}=1,6.36,5=58,4\left(g\right)\\ m_{ddHCl}=\dfrac{58,4.100}{25}=233,6\left(g\right)\)
\(n_{H_2}=\dfrac{2,464}{22,4}=0,11mol\)
\(\left\{{}\begin{matrix}Al:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\Rightarrow Muối\left\{{}\begin{matrix}Al_2\left(SO_4\right)_3\\FeSO_4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}BTe:3x+2y=2n_{H_2}=0,22\\\dfrac{x}{2}\cdot342+y\cdot152=14,44\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,04mol\\y=0,05mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,04\cdot27=1,08g\\m_{Fe}=0,05\cdot56=2,8g\end{matrix}\right.\)
\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\downarrow\)
0,02 0,06
\(FeSO_4+BaCl_2\rightarrow BaSO_4\downarrow+FeCl_2\)
0,05 0,05
\(\Rightarrow\Sigma n_{\downarrow}=0,06+0,05=0,11\Rightarrow m_{BaSO_4}=x=25,63g\)
Gọi n Fe = a (mol )
n Mg = b (mol ) (a,b > 0)
--> 56a+24b = 13,2
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a 2a a a
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b 2b b b
----> a+b=0,35
Ta có hệ Pt :
\(\left\{{}\begin{matrix}56a+24b=13,2\\a+b=0,35\end{matrix}\right.\)
Giải hệ PT , ta có :
a= 0,15
b = 0,2 (mol )
\(V_{HClđủ}=\left(0,15.2+0,2.2\right):0,5=1,4\left(l\right)\)
\(a,m_{Fe}=0,15.56=8,4\left(g\right)\)
\(m_{Mg}=0,2.24=4,8\left(g\right)\)
\(\%m_{Fe}=\dfrac{8,4}{13,2}.100\%\approx63,64\%\)
\(\%m_{Mg}=\dfrac{4,8}{13,2}.100\%\approx36,36\%\)
\(b,m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
\(m_{MgCl_2}=0,2.95=19\left(g\right)\)
\(c,HCl+NaOH\rightarrow NaCl+H_2O\)
0,2 0,2
\(m_{NaOH}=\dfrac{100.8}{100}=8\left(g\right)\)
\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(V_{HCldư}=\dfrac{n}{C_M}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
\(V_{HCl}=V_{HClđủ}+V_{HCldư}=1,4+0,4=1,8\left(l\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=0,2\cdot4=0,8mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(x\) \(\rightarrow\) \(3x\) \(x\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(y\) \(\rightarrow\) \(2y\) \(y\)
\(\Rightarrow\left\{{}\begin{matrix}27x+65y=11,9\\3x+2y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
a)\(\%m_{Al}=\dfrac{0,2\cdot27}{11,9}\cdot100\%=45,38\%\)
\(\%m_{Zn}=100\%-45,38\%=54,62\%\)
b)\(\Sigma n_{H_2}=\dfrac{3}{2}x+y=\dfrac{3}{2}\cdot0,2+0,1=0,4mol\)
\(V_{H_2}=0,4\cdot22.4=8,96l\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)
\(n_{SO_2}=\dfrac{12,32}{22,4}=0,55mol\)
\(2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O+3SO_2\uparrow\)
x 3x 0,5x 3x 1,5x
\(2Ag+2H_2SO_4\rightarrow2H_2O+SO_2\uparrow+Ag_2SO_4\)
y y y 0,5y 0,5y
\(\Rightarrow\left\{{}\begin{matrix}1,5x+0,5y=0,55\\56x+108y=38,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
a)\(\%m_{Fe}=\dfrac{0,3\cdot56}{38,4}\cdot100\%=43,75\%\)
\(\%m_{Ag}=100\%-43,75\%=56,25\%\)
b)\(m_{muối}=m_{Fe_2\left(SO_4\right)_3}+m_{Ag_2SO_4}\)
\(\Rightarrow muối=0,5\cdot0,3\cdot400+0,5\cdot0,2\cdot312=91,2g\)
c)Cho hỗn hợp trên tác dụng \(H_2SO_4\) loãng chỉ có Fe tác dụng.
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,3 0,3 0,3
\(C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5M\)
\(V_{H_2}=0,3\cdot22,4=6,72l\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:a\left(mol\right)\\n_{Zn}:b\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}56a+65b=18,6\\127a+136b=39,9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,1.56}{18,6}.100\%=30,11\%\)
\(\%m_{Fe}=100\%-30,11\%=69,89\%\)
b, \(n_{HCl}=0,1.2+0,2.2=0,6\left(mol\right)\)
\(V_{dd\left(HCl\right)}=\frac{0,6}{4}=0,15\left(l\right)=150\left(ml\right)\)
\(m_{dd\left(HCl\right)}=150.1,1=165\left(g\right)\)
c,\(ZnCl_2+2AgNO_3\rightarrow Zn\left(NO_3\right)_2+2AgCl\)
\(FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_3+2AgCl\)
\(n_{AgCl}=0,1.2+0,2.2=0,6\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,6.143,5=86,1\left(g\right)\)