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\(Fe+2HCl-->FeCl2+H2\)
\(2Al+6HCl-->2AlCl4+3H2\)
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow m_{H2}=0,4\left(g\right)\)
\(n_{HCl}=2n_{H2}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{muối}=m_{hh}+m_{HCl}-m_{H2}-m_{cr}=11,9+14,6-0,4-6,4=19,7\left(g\right)\)
\(m_{không.tan}=m_{Cu}=2\left(g\right)\\ \Rightarrow m_{Al,Fe}=10,3-2=8,3\left(g\right)\\ Đặt:a=n_{Al}\left(mol\right);b=n_{Fe}\left(mol\right)\left(a,b>0\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=8,3\\1,5.22,4a+22,4b=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{2}{10,3}.100\approx19,417\%\\ \%m_{Fe}=\dfrac{56.0,1}{10,3}.100\approx54,369\%\\ \Rightarrow\%m_{Al}\approx26,214\%\)
1) Ptpư:
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
Cu + HCl \(\rightarrow\) không phản ứng
=> 0,6 gam chất rắn còn lại chính là Cu:
Gọi x, y lần lượt là số mol Al, Fe
Ta có:
3x + 2y = 2.0,06 = 0,12
27x + 56 y = 2,25 – 0,6 = 1,65
=> x = 0,03 (mol) ; y = 0,015 (mol)
=> \(\%Cu=\frac{0,6}{2,25}.100\%=26,67\%\); \(\%Fe=\frac{56.0,015}{2,25}.100\%=37,33\%\); %Al = 36%
2) \(n_{SO_2}=\frac{1,344}{22,4}=0,06mol\); m (dd KOH) = 13,95.1,147 = 16 (gam)
=> mKOH = 0,28.16 = 4,48 (gam)=> nKOH = 0,08 (mol)=> \(1<\)\(\frac{n_{KOH}}{n_{SO_2}}<2\)
=> tạo ra hỗn hợp 2 muối: KHSO3: 0,04 (mol) và K2SO3: 0,02 (mol)
Khối lượng dung dịch sau pu = 16 + 0,06.64 = 19,84 gam
=> \(C\%\left(KHSO_3\right)=\frac{0,04.120}{19,84}.100\%\)\(=24,19\%\)
\(C\%\left(K_2SO_3\right)=\frac{0,02.158}{19,84}.100\%\)\(=15,93\%\)
Gọi số mol Al, Fe là a, b
\(m_{Cu}=m_B=6,4\left(g\right)\)
=> \(m_{Al}+m_{Fe}=17,4-6,4=11\left(g\right)\)
=> 27a + 56b = 11
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
b----------------------->b
2Al + 6HCl --> 2AlCl3 + 3H2
a------------------------>1,5a
=> 1,5a + b = 0,4
=> a = 0,2; b = 0,1
=> \(\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)
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
=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)
mCu = 2 (g)
Gọi số mol Al, Fe là a, b (mol)
=> 27a + 56b = 10,3 - 2 = 8,3 (1)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a--------------------->1,5a
Fe + 2HCl --> FeCl2 + H2
b---------------------->b
=> 1,5a + b = 0,25 (2)
(1)(2) => a = 0,1 (mol); b = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,1.27}{10,3}.100\%=26,21\%\\\%Fe=\dfrac{0,1.56}{10,3}.100\%=54,37\%\\\%Cu=\dfrac{2}{10,3}.100\%=19,42\%\end{matrix}\right.\)
\(n_{HCl}=2.n_{H_2}=0,5\left(mol\right)\)
=> \(V_{dd.HCl}=\dfrac{0,5}{2}=0,25\left(l\right)\)
\(n_{H_2}=\dfrac{0,784}{22,4}=0,035\left(mol\right)\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=1,39\\1,5a+b=0,035\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,01\\b=0,02\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,01.27}{1,39}.100=37,53\%\\ \Rightarrow\%m_{Fe}=100\%-37,53\%=62,47\%\)
làm rồi mà??
dạ tại em ko thấy tb bài kia nên nảy em up lại , em cảm ơn ạ