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\(a,n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}95a+133,5b=10,475\\a+1,5b=0,115\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\\ \%m_{Mg}=\dfrac{0,04.24}{0,04.24+0,05.27}.100\approx41,558\%\Rightarrow\%m_{Al}\approx58,442\%\\ b,n_{HCl}=2.n_{H_2}=2.0,115=0,23\left(mol\right)\\ \Rightarrow x=C\%_{ddHCl}=\dfrac{0,23.36,5}{100}.100=8,395\%\)
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) Gọi số mol Zn, Fe là a, b (mol)
=> 65a + 56b = 8,56 (1)
\(n_{H_2}=\dfrac{3,136}{22,4}=0,14\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a--->2a-------->a----->a
Fe + 2HCl --> FeCl2 + H2
b----->2b------->b------>b
=> a + b = 0,14 (2)
(1)(2) => a = 0,08; b = 0,06
=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,08.65}{8,56}.100\%=60,748\%\\\%m_{Fe}=\dfrac{0,06.56}{8,56}.100\%=39,252\%\end{matrix}\right.\)
b)
nKOH = 0,2.0,1 = 0,02 (mol)
PTHH: KOH + HCl --> KCl + H2O
0,02-->0,02
=> nHCl = 0,02 + 2a + 2b = 0,3 (mol)
=> \(C_{M\left(HCl\right)}=xM=\dfrac{0,3}{0,15}=2M\)
c) m = 0,08.136 + 0,06.127 = 18,5(g)
Coi hỗn hợp Y gồm :
Kim loại : 14,3(gam)
O :(x mol)
\(2H^+ + O^{2-}\to H_2O\\ 2H^+ + 2e \to H_2\)
Ta có : \(n_{Cl^-} = n_{HCl} = n_{H^+} = 2n_O + 2n_{H_2} = 2x + 0,4(mol)\)
Mà :
\(m_{muối} = m_{kim\ loại} + m_{Cl^-} = 14,3 + (2x + 0,4).35,5 = 49,8(gam)\\ \Rightarrow x = 0,3\)
Vậy : \(a = m_{kim\ loại} + m_O = 14,3 + 0,3.16 = 19,1(gam)\)
\(19,1gam\) \(:\left\{{}\begin{matrix}Al\\Mg\\Zn\end{matrix}\right.\)\(\underrightarrow{+O_2}\)\(Y:25,5gam\)\(\underrightarrow{+HCl}\left\{{}\begin{matrix}AgCl_3\\MgCl_2\\ZnCl_2\end{matrix}\right.\) + H2 : 0,3 mol
H2O
Áp dụng định luật bảo toàn khối lượng:
\(mO_2=25,5-19,1=6,4gam\) \(\Rightarrow nO_2=0,2\left(mol\right)\)
BTNT O : nH2O = 0,4mol
\(\rightarrow nHCl^-\left(tdOxi\right)=0,8\left(mol\right)\)
\(nH_2=0,3\left(mol\right)\rightarrow nCl^-\left(tdKl\right)=0,6\left(mol\right)\)
\(m_{muối}=19,1+\left(0,8+0,6\right).35,5=68,8\left(g\right)\)
\(n_{HCl}=0,12.2=0,24\left(mol\right)\)
=> \(n_{H_2O}=0,12\left(mol\right)\)
=> \(n_{O\left(oxit\right)}=0,12\left(mol\right)\)
=> \(n_{O_2}=0,06\left(mol\right)\)
\(n_{Mg}=\dfrac{1,68}{24}=0,07\left(mol\right)\); \(n_{Al}=\dfrac{2,16}{27}=0,08\left(mol\right)\)
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) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)
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
=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)