\(n_{H2}=\dfrac{4,2}{22,4}=0,1875\left(mol\right)\)

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

            2          6                2          3

            a                                        1,5a

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

              1         2               1         1

              b                                     1b

b) Gọi a là số mol của Al

           b là số mol của Fe

\(m_{Al}+m_{Fe}=6,225\left(g\right)\)

⇒ \(n_{Al}.M_{Al}+n_{Fe}.M_{Fe}=6,225g\)

 ⇒ 27a + 56b = 6,225g

Theo phương trình : 1,5a + 1b = 01875(2)

Từ(1),(2) ta có hệ phương trình : 

        27a + 56b = 6,225g

         1,5a + 1b = 0,1875

        ⇒ \(\left\{{}\begin{matrix}a=0,075\\b=0,075\end{matrix}\right.\)

\(m_{Al}=0,075.27=2,025\left(g\right)\)

\(m_{Fe}=0,075.56=4,2\left(g\right)\)

⁰/₀Al = \(\dfrac{2,025.100}{6,225}=32,53\)⁰/₀

⁰/₀Fe = \(\dfrac{4,2.100}{6,225}=67,47\)⁰/₀

 Chúc bạn học tốt

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,1<--0,2<--------------0,1

=> m_Fe = 0,1.56 = 5,6 (g)

=> m_Ag = 32-5,6 = 26,4 (g)

c) \(\left\{{}\begin{matrix}\%Fe=\dfrac{5,6}{32}.100\%=17,5\%\\\%Ag=\dfrac{26,4}{32}100\%=82,5\%\end{matrix}\right.\)

d) \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)

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

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Theo PTHH: \(n_{Mg}=n_{H_2}=0,6\left(mol\right)\)

=> \(m_{Mg}=0,6.24=14,4\left(g\right)\)

=> \(m_{Cu}=50-14,4=35,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%Mg=\dfrac{14,4}{50}.100=28,8\%\\\%Cu=\dfrac{35,6}{50}.100=71,2\%\end{matrix}\right.\)

\(n\)_H2 =\(\dfrac{13,44}{22,4}\) =0,6(mol)
PTHH:
Mg +HCl →MgCl₂ + H₂
0,6 mol                 ←0,6 mol
a) \(m\)_Mg =0,6. 24 =14,4(g)
    \(m\)_Cu= 50- 14,4= 35,6(g)
b)\(m\)%_Mg= \(\dfrac{14,4}{50}\).100%= 28,8%
   \(m\)%_Cu=100%- 28,8%= 71,2%
 

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

Ta có: \(n_{Zn}=n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

Bạn xem lại xem đề cho bao nhiêu gam hỗn hợp nhé, vì m_Zn đã bằng 13 (g) rồi.

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

\(\Rightarrow m_{Ag}=20-13=7\left(g\right)\)

b, \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{20}.100\%=65\%\\\%m_{Ag}=100-65=35\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{Al}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ \Rightarrow \%_{Al}=\dfrac{5,4}{10}.100\%=54\%\\ \Rightarrow \%_{Ag}=100\%-54\%=46\%\\ n_{HCl}=0,6(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,6.36,5}{10\%}=219(g)\)

\(a,\) Đặt \(\begin{cases} n_{Fe}=x(mol)\\ n_{Al}=y(mol) \end{cases}\Rightarrow 56x+27y=22(1)\)

\(n_{H_2}=\dfrac{17,92}{22,4}=0,8(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow x+1,5y=0,8(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,2(mol)\\ y=0,4(mol) \end{cases} \Rightarrow \begin{cases} \%_{Fe}=\dfrac{0,2.56}{22}.100\%=50,91\%\\ \%_{Al}=100\%-50,91\%=49,09\% \end{cases} \)

\(b,\Sigma n_{HCl}=2n_{Fe}+3n_{Al}=0,4+1,2=1,6(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{1,6.36,5}{3,7\%}=1578,38\%\)

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

         0,15<--------------------0,15

=> \(\%Fe=\dfrac{0,15.56}{14,8}.100\%=56,757\%\)

=> \(\%Cu=100\%-56,757\%=43,243\%\)

nH2 = 3,36/22,4 = 0,15 (mol)

PTHH: Fe + 2HCl -> FeCl2 + H2

nFe = 0,15 (mol)

mFe = 0,15 . 56 = 8,4 (g)

%mFe = 8,4/14,8 = 56,75%

%mCu = 100% - 56,75% = 43,25%

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ a,\%m_{Fe}=\dfrac{0,02.56}{4,36}.100\approx25,688\%\\ \Rightarrow\%m_{Ag}\approx74,312\%\\ b,Ta.thấy:2,18=\dfrac{1}{2}.4,36\\ \Rightarrow m_{hh\left(câuB\right)}=\dfrac{1}{2}.m_{hh\left(câuA\right)}\\ n_{Fe}=\dfrac{0,02}{2}=0,01\left(mol\right)\\ n_{Ag}=\dfrac{2,18-0,01.56}{108}=0,015\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ 2Ag+Cl_2\rightarrow\left(t^o\right)2AgCl\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+\dfrac{1}{2}.n_{Ag}=\dfrac{3}{2}.0,01+\dfrac{1}{2}.0,015=0,0225\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc\right)}=0,0225.22,4=0,504\left(l\right)\)