a)

\(n_{H_2} = \dfrac{3,808}{22,4} = 0,17(mol)\\ \Rightarrow n_{HCl} = 2n_{H_2} = 0,17.2 = 0,34(mol)\)

Bảo toàn khối lượng :

\(m_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 4 + 0,34.36,5 - 0,17.2 = 16,07(gam)\)

b)

\(n_A = a(mol) \Rightarrow n_{Al} =5a(mol)\\ A + 2HCl \to 2ACl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + H_2\\ n_{H_2} = n_A + \dfrac{3}{2}n_{Al} = a + \dfrac{3}{2}.5a = 0,17\\ \Rightarrow a = 0,02\\ m_{hỗn\ hợp} = 0,02A + 0,02.5.27 = 4\\ \Rightarrow A = 65(Zn)\)

Vậy kim loại hóa trị II cần tìm là Kẽm.

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\(a.n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ a............3a.......a.........1,5a\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.........2b........b.........b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}27a+56b=5,5\\1,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,5}.100\approx49,091\%\\\%m_{Fe}=\dfrac{0,05.56}{5,5}.100\approx50,909\%\end{matrix}\right.\\ b.C_{MddHCl}=\dfrac{3a+2b}{0,5}=\dfrac{3.0,1+2.0,05}{0,5}=0,8\left(M\right)\)

a, PTHH:

\(A+2HCl\rightarrow ACl_2+H_2\left(1\right)\)

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

\(AlCl_3+4NaOH\rightarrow NaAlO_2+3NaCl+2H_2O\)

b, Ta có \(n_{AlCl_3}=n_{NaAlO_2}=\dfrac{2,7}{82}=0,03\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=n_{AlCl_3}=0,03\left(mol\right)\\n_{H_2\left(2\right)}=\dfrac{3}{2}n_{AlCl_3}=0,045\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=27.0,03=0,81\left(g\right)\\n_A=n_{H_2\left(1\right)}=\dfrac{1,68}{22,4}-n_{H_2\left(2\right)}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_A=2,49-0,81=1,68\left(g\right)\\n_A=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow M_A=\dfrac{1,68}{0,03}=56\left(g/mol\right)\Rightarrow A\) là \(Fe\)

c, \(m_{\text{muối}}=m_{FeCl_2}+m_{AlCl_3}\)

\(=127.n_{Fe}+133,5.n_{Al}\)

\(=127.0,03+133,5.0,03=7,815\left(g\right)\)

Bn nào giúp mình với ạk

a)

PTHH: 2A + 2nHCl --> 2ACl_n + nH₂

            2B + 2mHCl --> 2BCl_m + mH₂

Gọi số mol H₂ là a (mol)

=> n_HCl = 2a (mol)

Theo ĐLBTKL: m_{kim loại} + m_HCl = m_muối + m_H2

=> 8,9 + 36,5.2a = 23,1 + 2a

=> a = 0,2 (mol)

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) 

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

=> \(n_{H_2\left(tăng\right)}=0,25-0,2=0,05\left(mol\right)\)

PTHH: 2B + 2mHCl --> 2BCl_m + mH₂

           \(\dfrac{0,1}{m}\)<------------\(\dfrac{0,1}{m}\)<---0,05

Khối lượng rắn sau pư tăng lên do có thêm BCl_m sinh ra

=> \(m_{BCl_m}=\dfrac{0,1}{m}\left(M_B+35,5m\right)=27,85-23,1=4,75\left(g\right)\)

=> M_B = 12m (g/mol)

Xét m = 2 thỏa mãn => M_B = 24 (g/mol) => B là Mg

\(n_{Mg\left(thêm\right)}=\dfrac{0,1}{m}=\dfrac{0,1}{2}=0,05\left(mol\right)\)

=> \(n_{Mg\left(bđ\right)}=0,1\left(mol\right)\)

=> \(m_A=8,9-0,1.24=6,5\left(g\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             0,1-------------------->0,1

             2A + 2nHCl --> 2ACl_n + nH₂

            \(\dfrac{0,2}{n}\)<-------------------0,1

=> \(M_A=\dfrac{6,5}{\dfrac{0,2}{n}}=32,5n\left(g/mol\right)\)

 Xét n = 2 thỏa mãn => M_A = 65 (g/mol)

=> A là Zn

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Tương tự nhau chỉ thay

340 ml = 0,34l

Số mol HCl là: 0,34 x 1= 0,34 (mol)

\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)

\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)

\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

Fe + 2HCl --> FeCl₂ + H₂

b) Gọi số mol Al, Fe lần lượt là a,b 

=> 27a + 56b = 13,9

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

2Al + 6HCl --> 2AlCl₃ + 3H₂

a----->3a--------->a------->1,5a______(mol)

Fe + 2HCl --> FeCl₂ + H₂

b------>2b-------->b----->b__________(mol)

=> 1,5a + b = 0,35

=> \(\left\{{}\begin{matrix}a=0,1=>m_{Al}=0,1.27=2,7\left(g\right)\\b=0,2=>m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

c) n_HCl = 3a + 2b = 0,7 (mol)

=> m_HCl = 0,7.36,5 = 25,55(g)

=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)

\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)

\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)