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

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

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

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)

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

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

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

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

a) Gọi số mol Mg, Al, Fe trong m gam hỗn hợp là a, b, c (mol)

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

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

_______a--------------------->a------->a_______(mol)

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

_b-------------------->b------->1,5b___________(mol)

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

_c------------------>c------->c_______________(mol)

=> \(\left\{{}\begin{matrix}a+1,5b+c=0,35\left(1\right)\\95a+133,5b+127c=35,55\left(2\right)\end{matrix}\right.\)

Mặt khác:

PTHH: \(Mg+Cl_2\underrightarrow{t^o}MgCl_2\)

_______a--------------->a_________(mol)

\(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

_b----------------->b______________(mol)

\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

_c------------------>c______________(mol)

=> 95a + 133,5b + 162,5 = 39,1 (3)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\) 

=> m = 24.0,1 + 27.01 + 56.0,1 = 10,7(g)

b) \(\left\{{}\begin{matrix}m_{Mg}=24.0,1=2,4\left(g\right)\\m_{Al}=27.0,1=2,7\left(g\right)\\m_{Fe}=56.0,1=5,6\left(g\right)\end{matrix}\right.\)

PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)

            \(K_2SO_3+2HCl\rightarrow2KCl+SO_2\uparrow+H_2O\)

a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200\cdot14,6\%}{36,5}=0,8\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_3}=0,3\left(mol\right)\\n_{K_2O}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{K_2O}=\dfrac{0,1\cdot94}{0,1\cdot94+0,3\cdot158}\cdot100\%\approx16,55\%\\\%m_{K_2SO_3}=83,45\%\end{matrix}\right.\)

b) Theo các PTHH: \(n_{KCl}=0,8\left(mol\right)\) \(\Rightarrow m_{KCl}=74,5\cdot0,8=59,6\left(g\right)\)

Mặt khác: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)

Khối lượng sắt:mFe=0,46298*60,5=28 g=>mZn=32,5g

=>nFe=0,5;nZn=0,5

ta có:nCl- =2nFe+2nZn=2 mol =>nHCl=2mol

=>nH2 =1mol.=>V H2=44,8 lit

m muối =m kim loại +m Cl- =60,5 +2*35,5=131,5

131,5 tích nha

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1                           0,05           0,15

\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)

\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)

a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)

\(\%m_{Ag}=100\%-64,28\%=35,72\%\)

b)\(m_{muối}=0,05\cdot342=17,1g\)

a. PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

Cu + H₂SO₄ ---x--->

b. Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.\dfrac{6,72}{22,4}=0,2\left(mol\right)\)

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

=> \(m_{Cu}=10-5,4=4,6\left(g\right)\)

c. \(\%_{m_{Al}}=\dfrac{5,4}{10}.100\%=54\%\)

\(\%_{m_{Cu}}=100\%-54\%=46\%\)

d. Theo PT: \(n_{H_2SO_4}=n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{29,4}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=147\left(g\right)\)

a) Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow27a+24b=1,26\)  (1)

Ta có: \(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

Bảo toàn electron: \(3a+2b=0,12\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{Al}=0,02\left(mol\right)\\b=n_{Mg}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02\cdot27}{1,26}\cdot100\%\approx42,86\%\\\%m_{Mg}=57,14\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,02\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,02\cdot133,5=2,67\left(g\right)\\m_{MgCl_2}=0,03\cdot95=2,85\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{ddHCl}=40\cdot1,25=50\left(g\right)\\m_{H_2}=0,06\cdot2=0,12\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{H_2}=51,14\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{2,67}{51,14}\cdot100\%\approx5,22\%\\C\%_{MgCl_2}=\dfrac{2,85}{51,14}\cdot100\%\approx5,57\%\end{matrix}\right.\)

\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

        1          2               1         1

       0,05    0,1           0,05      0,05 

    \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

       1           2              1            1

      0,2       0,4            0,2

a) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(m_{Mg}=0,05.24=1,2\left(g\right)\)

\(m_{MgO}=9,2-1,2=8\left(g\right)\)

⁰/₀Mg = \(\dfrac{1,2.100}{9,2}=13,04\)⁰/₀

⁰/₀MgO = \(\dfrac{8.100}{9,2}=86,96\)⁰/₀

b) Có : \(m_{MgO}=8\left(g\right)\)

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

c) \(n_{MgCl2\left(tổng\right)}=0,05+0,2=0,25\left(mol\right)\)

⇒ \(m_{MgCl2}=0,25.95=23,75\left(g\right)\)

\(m_{ddspu}=9,2+125-\left(0,05.2\right)=134,1\left(g\right)\)

\(C_{MgCl2}=\dfrac{23,75.100}{134,1}=17,71\)⁰/₀

 Chúc bạn học tốt

a)\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:    0,05    0,1           0,05        0,05

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:     0,2         0,4         0,2

\(\%m_{Mg}=\dfrac{0,05.24.100\%}{9,2}=13,04\%;\%m_{MgO}=100-13,04=86,96\%\)

\(n_{MgO}=\dfrac{9,2-0,05.24}{40}=0,2\left(mol\right)\)

b,\(m_{HCl}=\left(0,1+0,4\right).36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

c,m_{dd sau pứ} = 9,2+125-0,05.2 = 134,1 (g)

 \(C\%_{ddMgCl_2}=\dfrac{\left(0,05+0,2\right).95.100\%}{134,1}=17,71\%\)

\(Mg + 2HCl \rightarrow MgCl_2 + H_2\)     (1)

\(MgO + 2HCl \rightarrow MgCl_2 + H_2O\)      (2)

Khí thu được là H₂

\(n_{H_2}= \dfrac{1,12}{22,4}=0,05 mol\)

Theo PTHH (1):

\(n_{Mg}= n_{H_2}= 0,05 mol\)

\(\Rightarrow m_{Mg}= 0,05 . 24= 1,2 g\)

\(\Rightarrow m_{MgO}= 9,2 - 1,2= 8g\)

C%_\(Mg\)= \(\dfrac{1,2}{9,2} .100\)%=13,04%

C%_\(MgO\)= 100% - 13,04%=86,96%

b)

\(n_{MgO}= \dfrac{8}{40}=0,2 mol\)

Theo PTHH (1) và (2):

\(n_{HCl(1)}= 2n_{Mg}= 0,1 mol\)

\(n_{HCl(2)}= 2n_{MgO}= 0,4 mol\)

Suy ra: \(n_{HCl}= n_{HCl(1)} + n_{HCl(2)}\)

                     \(= 0,1 + 0,4= 0,5 mol\)

\(\Rightarrow m_{HCl}= 0,5 . 36,5= 18,25g\)

\(\Rightarrow m_{dd HCl} = \dfrac{18,25 . 100%}{14,6%}=125 g\)

_c)

\(\)Dung dịch sau pư: MgCl₂

Theo PTHH:

\(n_{MgCl_2}= \dfrac{1}{2} n_{HCl}= 0,25 mol\)

\(\Rightarrow m_{MgCl_2}= 0,25 . 95=23,75g\)

\(m_{dd sau pư} = m_{Mg} + m_{MgO} + m_{dd HCl}- m_{H_2}\)

               \(= 9,2 + 125 + 2 . 0,05\)

             \(=134,1 g\)

C%_{\(MgCl_2\)}=\(\dfrac{23,75}{134,1}. 100\)%=17,71%

nH2= 0,35(mol)

a) PTHH: Mg +  2 HCl -> MgCl2 + H2

x_________2x_______x______x(mol)

PTHH: Fe + 2 HCl -> FeCl2 + H2

y________2y________y_____y(mol)

Ta có hpt: \(\left\{{}\begin{matrix}24x+56y=13,2\\x+y=0,35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,15\end{matrix}\right.\)

b) m=m(muối khan)= mMgCl2 + mFeCl2= 95.x+127y=95.0,2+127.0,15= 38,05(g)

a)

Gọi 

\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 13,2(1)\)

\(Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2\)

Theo PTHH : \(n_{H_2} = a + b = 0,35(mol)\)(2)

Từ (1)(2) suy ra a = 0,15 ;b = 0,2

Vậy : 

\(\%m_{Fe} = \dfrac{0,15.56}{13,2}.100\% = 63,64\%\\ \Rightarrow m_{Mg} = 100\% - 63,64\% = 36,36\%\)

b)

Ta có :\(n_{HCl} = 2n_{H_2} = 0,7(mol)\)

Bảo toàn khối lượng :

\(m_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13,2 + 0,7.36,5 - 0,35.2=38,05(gam)\)