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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)
=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,12.95}{157,28}.100\%=7,25\%\\C\%_{FeCl_2}=\dfrac{0,16.127}{157,28}.100\%=12,92\%\end{matrix}\right.\)
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) 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)
\(a,n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}95a+133,5b=10,475\\a+1,5b=0,115\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\\ \%m_{Mg}=\dfrac{0,04.24}{0,04.24+0,05.27}.100\approx41,558\%\Rightarrow\%m_{Al}\approx58,442\%\\ b,n_{HCl}=2.n_{H_2}=2.0,115=0,23\left(mol\right)\\ \Rightarrow x=C\%_{ddHCl}=\dfrac{0,23.36,5}{100}.100=8,395\%\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5(mol)\\ n_{Mg}=x(mol);n_{Fe}=y(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow \begin{cases} 24x+56y=20\\ x+y=0,5\\ \end{cases} \Rightarrow x=y=0,25(mol)\\ \Rightarrow \begin{cases} m_{Fe}=56.0,25=14(g)\\ m_{Mg}=24.0,25=6(g)\\ \end{cases}\)
Gọi số mol của Al, Mg, CuO là a, b, c
=> 27a + 24b + 80c = 23,8
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a---->3a----------------->1,5a
Mg + 2HCl --> MgCl2 + H2
b----->2b---------------->b
CuO + 2HCl --> CuCl2 + H2O
c------->2c
=> \(\left\{{}\begin{matrix}1,5a+b=\dfrac{8,96}{22,4}=0,4\\3a+2b+2c=0,6.2=1,2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a=0,2\\b=0,1\\c=0,2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\\m_{CuO}=0,2.80=16\left(g\right)\end{matrix}\right.\)
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
=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)
a) Gọi số mol Zn, Fe là a, b (mol)
=> 65a + 56b = 7,35 (1)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a---->2a------->a------>a
Fe + 2HCl --> FeCl2 + H2
b------>2b----->b------>b
=> \(a+b=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)
=> a + b = 0,12 (2)
(1)(2) => a = 0,07; b = 0,05
=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,07.65}{7,35}.100\%=61,9\%\\\%m_{Fe}=\dfrac{0,05.56}{7,35}.100\%=38,1\%\end{matrix}\right.\)
b) nHCl(dư) = 0,3.1 - 0,07.2 - 0,05.2 = 0,06 (mol)
PTHH: Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
0,03<-----0,06
=> \(x=C_{M\left(ddCa\left(OH\right)_2\right)}=\dfrac{0,03}{0,1}=0,3M\)
c) Chất rắn thu được là Fe2O3
Bảo toàn Fe: \(n_{Fe_2O_3}=0,025\left(mol\right)\)
=> \(a=m_{Fe_2O_3}=0,025.160=4\left(g\right)\)
Kết tủa thu được là Fe(OH)2
Bảo toàn Fe: \(n_{Fe\left(OH\right)_2}=0,05\left(mol\right)\)
=> \(m=m_{Fe\left(OH\right)_2}=0,05.90=4,5\left(g\right)\)
Gọi $n_{Mg} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 24a + 65b = 15,3(1)$
$Mg + 2HCl \to MgCl_2 + H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH : $n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)$
Từ (1)(2) suy ra : $a = \dfrac{81}{410} ; b = \dfrac{21}{205}$
$m_{Mg} = \dfrac{81}{410}.24 = 4,74(gam)$
$m_{Zn} = 15,3 - 4,74 = 10,56(gam)$