a, Ta có: 27n_Al + 56n_Fe = 22 (1)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)

b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)

\(n_{HCl}=0,3.2=0,6\left(mol\right)\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,6}{2}>\dfrac{0,25}{1}\Rightarrow HCldư\\ Đặt:n_{Al}=t\left(mol\right);n_{Fe}=r\left(mol\right)\\ \left(t,r>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27t+56r=8,3\\1,5t+r=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=0,1\\r=0,1\end{matrix}\right.\\ \Rightarrow m_{Al}=0,1.27=2,7\left(g\right);m_{Fe}=0,1.56=5,6\left(g\right)\\ b,n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\Rightarrow m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\ n_{Fe}=n_{FeCl_2}=0,1\left(mol\right)\Rightarrow m_{ddFeCl_2}=127.0,1=12,7\left(g\right)\\ m_{ddHCl}=300.1,15=345\left(g\right)\\ m_{ddsau}=8,3+345-0,25.2=352,8\left(g\right)\)

\(n_{HCl\left(dư\right)}=0,6-0,25.2=0,1\left(mol\right)\\ \Rightarrow m_{ddHCl}=0,1.36,5=3,65\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{3,65}{352,8}.100\approx1,035\%\\ C\%_{ddAlCl_3}=\dfrac{13,35}{352,8}.100\approx3,784\%\\ C\%_{ddFeCl_2}=\dfrac{12,7}{352,8}.100\approx3,6\%\)

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:

n_MgCl2 = n_Mg = 0,1 (mol)

 n_AlCl3 = n_Al = 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!

^{trả lời nhanh cho em ạ}

\(a)n_{Mg} = a ; n_{Al} = b \Rightarrow 24a +27b = 5,1(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{5,6}{22,4} = 0,25(2)\\ (1)(2) \Rightarrow a = 0,1 ; b = 0,1\\ \%m_{Mg} = \dfrac{0,1.24}{5,1}.100\% = 44,44\%\ ;\ \%m_{Al} = 100\% -44,44\% = 55,56\%\\ b) n_{MgCl_2} = n_{Mg} = 0,1 \Rightarrow m_{MgCl_2} = 0,1.95 = 9,5(gam)\\ n_{AlCl_3} = n_{Al} = 0,1 \Rightarrow m_{AlCl_3} = 0,1.133,5 = 13,35(gam)\\ c)n_{HCl} = 2n_{Mg} + 3n_{Al} = 0,5(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,5.36,5}{3,65\%} = 500(gam)\)

\(m_{dd\ sau\ pư} = 5,1 + 500 - 0,25.2 = 504,6(gam)\\ C\%_{MgCl_2} = \dfrac{9,5}{504,6}.100\% = 1,89\%\\ C\%_{AlCl_3} = \dfrac{13,35}{504,6}.100\% = 2,65\%\)

a) Gọi $n_{Al} =a (mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b = 13(1)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : $n_{H_2} = 1,5a + b = \dfrac{6,72}{22,4} = 0,3(2)$
Từ (1)(2) suy ra : $a = \dfrac{1}{15} ; b = 0,2$

$\%m_{Al} = \dfrac{ \dfrac{1}{15}.27}{13}.100\% = 13,8\%$

$\%m_{Fe} = 100\% - 13,8\% = 86,2\%$

b) $n_{HCl} = 2n_{H_2} = 0,3.2 = 0,6(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,6}{0,15} = 4M$

c) $n_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13 + 0,6.36,5 - 0,3.2 = 34,3(gam)$

\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}65x+24y=12,5\\x+y=0,35\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,25\end{matrix}\right.\\ \Rightarrow m_{Zn}=6,5\left(g\right);m_{Mg}=6\left(g\right)\\ b.Tacó:BTNT\left(H\right):n_{HCl}.1>n_{H_2}.2\\ \Rightarrow HCldưsauphảnứng\\ Dungdịchsauphảnứnggồm:\left\{{}\begin{matrix}ZnCl_2:0,1\left(mol\right)\\MgCl_2:0,25\left(mol\right)\\HCl_{dư}:0,8-0,7=0,1\left(mol\right)\end{matrix}\right.\\ m_{ddsaupu}=200+12,5-0,35.2=212,8\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,1.136}{212,8}.100=6,39\%;C\%_{MgCl_2}=\dfrac{0,25.95}{212,8}.100=11,16\%;C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{212,8}.100=1,72\%\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 17,05 (1)

Ta có: \(n_{H_2}=\dfrac{9,52}{22,4}=0,425\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=0,425\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,2\left(mol\right)\\n_{Al}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)

c, Ta có: m _{dd HCl} = 1,05.500 = 525 (g)

m _{dd sau pư} = m_hh + m _{dd HCl} - m_H2 = 541,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,2.136}{541,2}.100\%\approx5,03\%\\C\%_{AlCl_3}=\dfrac{0,15.133,5}{541,2}.100\%\approx3,7\%\end{matrix}\right.\)

$n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)$

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)$

$\Rightarrow \%m_{Al}=\dfrac{0,2.27}{24,6}.100\%=21,95\%$

$\Rightarrow \%m_{Cu}=100-21,95=78,05\%$

$b)n_{AlCl_3}=n_{Al}=0,2(mol)$

$\Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)$

$c)n_{HCl}=3n_{Al}=0,6(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M$