nH2= 1,344 / 22,4=0,06 mol

Khoi luong Fe va Al la    2,52 - 0,3= 2,22 g

Fe +     2HCl --> FeCl2 + H2

x mol     2xmol                x mol

2Al +     6HCl --> 2AlCl3 + 3H2

y mol     3ymol                  3/2 mol

56x + 27y = 2,22     (1)

x + 3/2y = 0,06       (2)

Tu (1) va (2) ta duoc => x= 0,03

                             => y=0,02

a/ mFe=0,03.56=1,68g

mAl=0,02.27=0,54g

%mCu=(0,3/2,52).100=11,9%

%mFe=(1,68/2,52).100=66,6%

%mAl=21,5%

b/ m_ct=0,12.36,5=4,38g

%C_HCl= (4,38/200).100=2,19%

nH2= 1,12/22,4=0,05(mol)

PTHH: Fe +2 HCl -> FeCl2 + H2

0,05_______0,1__0,05___0,05(mol)

a) mFe=0,05.56=2,8(g)

=>%mFe=(2,8/10).100=28% => %mCu=100%- 28%= 72%

b) mHCl=0,1.36,5=3,65(g)

=> C%ddHCl= (3,65/200).100= 1,825%

nH2= 1,12/22,4=0,05(mol)

PTHH: Fe +2 HCl -> FeCl2 + H2

0,05_______0,1__0,05___0,05(mol)

a) mFe=0,05.56=2,8(g)

=>%mFe=(2,8/10).100=28% => %mCu=100%- 28%= 72%

b) mHCl=0,1.36,5=3,65(g)

=> C%ddHCl= (3,65/200).100= 1,825%

\(n_{Fe}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)

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

       1          2             1           1

     0,01     0,02                     0,01

a) \(n_{Fe}=\dfrac{0,01.1}{1}=0,01\left(mol\right)\)

   \(m_{Fe}=0,01.56=0,56\left(g\right)\)

  \(m_{Cu}=1,2-0,56=0,64\left(g\right)\)

⁰/₀Fe = \(\dfrac{0,56.100}{1,2}=46,67\)⁰/₀

 ⁰/₀Cu = \(\dfrac{0,64.100}{1,2}=53,33\)⁰/₀

b) \(n_{HCl}=\dfrac{0,01.2}{1}=0,02\left(mol\right)\)

⇒ \(m_{HCl}=0,02.36,5=0,73\left(g\right)\)

\(C_{ddHCl}=\dfrac{0,73.100}{10}=7,3\)⁰/₀

 Chúc bạn học tốt

Vì Cu không tác dụng với HCl 

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

       1         2              1           1

      0,1     0,2                           0,1

a) \(n_{Fe}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{Fe}=0,1.56=5,6\left(g\right)\)

\(m_{Cu}=12-5,6=6,4\left(g\right)\)

b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddHCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

c) ⁰/₀Fe = \(\dfrac{5,6.100}{12}=46,67\)⁰/₀

    ⁰/₀Cu = \(\dfrac{6,4.100}{12}=53,33\)⁰/₀

 Chúc bạn học tốt

a)

Chất rắn sau phản ứng là Cu

=> m_Cu = 0,3 (g)

Gọi số mol Al, Fe là x,y (mol)

=> \(m_{Al}+m_{Fe}=2,22\left(g\right)\)

=> 27x + 56y = 2,22 (1)

\(n_{H_2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)

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

_______x------->3x------------------>1,5x____(mol)

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

_y----->2y------------------->y______(mol)

=> 1,5x + y = 0,06 (2)

(1)(2) => \(\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,03\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\frac{0,02.27}{2,52}.100\%=21,43\%\\\%m_{Fe}=\frac{0,03.56}{2,52}.100\%=66,67\%\\\%m_{Cu}=\frac{0,3}{2,52}.100\%=11,9\%\end{matrix}\right.\)

b)

\(n_{HCl}=3x+2y=0,12\left(mol\right)\)

=> m dd HCl = \(\frac{4,38.100}{25}=17,52\left(g\right)\)

=> V dd HCl = \(\frac{17,52}{1,19}=14,72\left(ml\right)\)

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

                a_____3a______________\(\dfrac{3}{2}\)a                   (mol)

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

                b_____2b_____________b                        (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}27a+24b=7,8\\\dfrac{3}{2}a+b=\dfrac{8,96}{22,4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{7,8}\cdot100\%\approx69,23\%\\\%m_{Mg}=30,77\%\\C_{M_{HCl}}=\dfrac{0,2\cdot3+0,1\cdot2}{0,2}=4\left(M\right)\end{matrix}\right.\) 

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

\(n_{Mg}=n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(\Rightarrow m_{Mg}=0.25\cdot24=6\left(g\right)\)

\(m_{Mg}>m_{hh}=5.6\left(g\right)\)

=> Đề sai