a) Gọi số mol Mg, Ca là a, b

=> 24a + 40b = 8,8 

PTHH: Mg + 2HCl --> MgCl₂ + H₂

______a---->2a------>a------->a

Ca + 2HCl --> CaCl₂ + H₂

b---->2b------->b------->b

=> a + b = \(\dfrac{6,72}{22,4}=0,3\)

=> a = 0,2 ; b = 0,1

=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,2.24}{8,8}.100\%=54,55\%\\\%Ca=\dfrac{0,1.40}{8,8}.100\%=45,45\%\end{matrix}\right.\)

b) n_HCl = 2a + 2b = 0,6 (mol)

=> \(V_{ddHCl}=\dfrac{0,6}{2}=0,3\left(l\right)\)

c) m_MgCl2 = 0,2.95 = 19 (g)

m_CaCl2 = 0,1.111 = 11,1 (g)

=> Tổng khối lượng muối = 19 + 11,1 = 30,1(g)

\(n_{H2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

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

        2           6              2           3

        a          0,6                       1,5a

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

      1         2                       1b

      b       0,8

a) Gọi a là số mol của Al

           b là số mol của Fe

\(m_{Al}+m_{Fe}=27,8\left(g\right)\)

⇒ \(n_{Al}.M_{Al}+n_{Fe}.M_{Fe}=27,8g\)

  ⇒ 27a + 56b = 27,8g(1)

Theo phương trình : 1,5a + 1b = 0,7(2)

Từ(1),(2), ta có hệ phương trình : 

          27a + 56b = 27,8

           1,5a + 1b = 0,7

           ⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,4\end{matrix}\right.\)

    \(m_{Al}=0,2.27=5,4\left(g\right)\)

    \(m_{Fe}=0,4.56=22,4\left(g\right)\)

b) ⁰/₀Al = \(\dfrac{5,4.100}{27,8}=19,42\)⁰/₀

    ⁰/₀Fe = \(\dfrac{22,4.100}{27,8}=80,58\)⁰/₀

c) \(n_{HCl\left(tổng\right)}=0,6+0,8=1,4\left(g\right)\)

⇒ \(m_{HCl}=1,4.36,5=51,1\left(g\right)\)

 Chúc bạn học tốt

\(n_{Al}=a\left(mol\right)\)

\(n_{Fe}=b\left(mol\right)\)

\(m=27a+56b=19.3\left(g\right)\left(1\right)\)

\(n_{H^+}=0.2\cdot2+0.2\cdot2.25\cdot2=1.3\left(mol\right)\)

\(2Al+6H^+\rightarrow2Al^{3+}+3H_2\)

\(Fe+2H^+\rightarrow Fe^{2+}+H_2\)

\(n_{H^+}=3a+2b=1.3\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.3,b=0.2\)

\(\%Al=\dfrac{0.3\cdot27}{19.3}\cdot100\%=41.96\%\)

\(\%Fe=58.04\%\)

\(b.\)

\(n_{H_2}=\dfrac{1}{2}n_{H^+}=0.65\left(mol\right)\)

Bảo toàn khối lượng : 

\(m_{Muối}=19.3+0.4\cdot36.5+0.45\cdot98-0.65\cdot2=76.7\left(g\right)\)

em cảm ơn ạ!

Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\end{matrix}\right.\) => 65a + 56b + 27c = 10,65 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            Fe + 2HCl --> FeCl₂ + H₂

            2Al + 6HCl --> 2AlCl₃ + 3H₂

=> \(n_{H_2}=a+b+1,5c=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) (2)

PTHH: Zn + Cl₂ --t^o--> ZnCl₂

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

            2Al + 3Cl₂ --t^o--> 2AlCl₃

=> \(n_{Cl_2}=a+1,5b+1,5c=\dfrac{5,6}{22,4}=0,25\left(mol\right)\) (3)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,05\left(mol\right)\\c=0,05\left(mol\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{Fe}=0,05.56=2,8\left(g\right)\\m_{Al}=0,05.27=1,35\left(g\right)\end{matrix}\right.\)

a) \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{6,5}{10,65}.100\%=61,033\%\\\%m_{Fe}=\dfrac{2,8}{10,65}.100\%=26,291\%\\\%m_{Al}=\dfrac{1,35}{10,65}.100\%=12,676\%\end{matrix}\right.\)

b) n_HCl = 2a + 2b + 3c = 0,45 (mol)

=> m_HCl = 0,45.36,5 = 16,425 (g)

=> \(a\%=C\%=\dfrac{16,425}{200}.100\%=8,2125\%\)

c) m_{dd sau pư} = 10,65 + 200 - 0,225.2 = 210,2 (g)

=> \(\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{210,2}.100\%=6,47\%\\C\%_{FeCl_2}=\dfrac{0,05.127}{210,2}.100\%=3,02\%\\C\%_{AlCl_3}=\dfrac{0,05.133,5}{210,2}.100\%=3,176\%\end{matrix}\right.\)

Gọi số mol Mg, Fe, Al là a, b, c

=> 24a + 56b + 27c = 23,8

PTHH: Mg + 2HCl --> MgCl₂ + H₂ 

            a------------------------->a

            Fe + 2HCl --> FeCl₂ + H₂

            b------------------------->b

            2Al + 6HCl --> 2AlCl₃ + 3H₂ 

            c------------------------->1,5c

=> a + b + 1,5c = \(\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

PTHH: Mg + Cl₂ --t^o--> MgCl₂

             a-->a

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

             b--->1,5b

             2Al + 3Cl₂ --t^o--> 2AlCl₃

             c--->1,5c

=> \(a+1,5b+1,5c=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

=> a = 0,3; b = 0,2; c = 0,2

=> \(\left\{{}\begin{matrix}m_{Mg}=0,3.24=7,2\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

giúp em vs ạ

Bài 1

\(a)n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,15      0,3          0,15         0,15

\(\%m_{Fe}=\dfrac{0,15.56}{12}\cdot100\%=70\%\\ \%m_{FeO}=100\%-70\%=30\%\\ b)n_{FeO}=\dfrac{12-0,15.56}{72}=0,05mol\\ FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05        0,1

\(V_{ddHCl}=\dfrac{0,1+0,3}{2}=0,2l\)

Bài 2

\(a)Na_2O+H_2O\rightarrow2NaOH\\ b)BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ c)BaSO_4?\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Ba\left(OH\right)_2+CuSO_4\rightarrow BaSO_4+Cu\left(OH\right)_2\\ d)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ e)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2+2NaCl\)

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ 2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)

Cho hỗn hợp tác dụng với NaOH, chất rắn không tan là Fe

=> m_Fe= 1,12 (g) \(\Rightarrow n_{Fe}=0,02\left(mol\right)\)

Ta có: \(n_{H_2\left(2\right)}=n_{Fe}=0,02\left(mol\right)\)

=> \(n_{H_2\left(1\right)}=\Sigma n_{H_2}-n_{H_2\left(2\right)}=0,065-0,02=0,045\left(mol\right)\)

\(\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2\left(1\right)}=0,03\left(mol\right)\)

\(\Rightarrow m_{Al}=0,03.27=0,81\left(g\right)\)

\(\Rightarrow\%m_{Al}=41,97\%,\%m_{Fe}=58,03\%\)

b) \(m_{FeCl_2}=0,02.127=2,54\left(g\right)\\ m_{AlCl_3}=0,03.133,5=4,005\left(g\right)\)

\(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)

\(m_X=27a+56b=11\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

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

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

\(\Rightarrow a+1.5b=0.4\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.2\)

\(n_{HCl}=2n_{H_2}=2\cdot0.4=0.8\left(mol\right)\)

\(m_{dd_{HCl}}=\dfrac{0.8\cdot36.5\cdot100}{14.6}=200\left(g\right)\)

\(\%Fe=\dfrac{0.1\cdot56}{11}\cdot100\%=50.91\%\)

\(\%Al=49.09\%\)