\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(\Rightarrow n_{HCl}=2\cdot n_{H_2}=2\cdot0.3=0.6\left(mol\right)\)

BTKL : 

\(m_{Muối}=m_{hh}+m_{HCl}-m_{H_2}=18.6+0.6\cdot36.5-0.3\cdot2=39.9\left(g\right)\)

a.\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{8,96}{22,4}=0,4mol\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)

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

 x                                      x ( mol )

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 y                                        y   ( mol )

Ta có:

\(\left\{{}\begin{matrix}56x+65y=25,55\\x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,35\end{matrix}\right.\)

\(\Rightarrow m_{Fe}=0,05.56=2,8g\)

\(\Rightarrow m_{Zn}=0,35.65=22,75g\)

\(\%m_{Fe}=\dfrac{2,8}{25,55}.100=10,95\%\)

\(\%m_{Zn}=100\%-10,95\%=89,05\%\)

b.\(n_{HCl}=2.0,05+2.0,35=0,8mol\)

\(C_M=\dfrac{n}{V}\Rightarrow V=\dfrac{n}{C_M}=\dfrac{0,8}{2}=0,4l\)

Đáp án B.

Số mol của Hiđro bằng:

n H 2 = 6 , 72 22 , 4 = 0 , 3 ( m o l )

lại có  n C l - t r o n g   m u ố i = 2 n H 2 = 2 . 0 , 3 = 0 , 6

Khối lượng muối tạo thành bằng: 12,45 + 0,6. 35,5 = 33,75 (gam).

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\)

\(BTe:\) \(2n_{Fe}+2n_{Zn}=2n_{H_2}\)

\(BTKL:\) \(56n_{Fe}+65n_{Zn}=18,6\)

\(\Rightarrow\left\{{}\begin{matrix}Fe:0,1mol\\Zn:0,2mol\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,1\cdot56=5,6g\\m_{Zn}=65\cdot0,2=13g\end{matrix}\right.\)

\(n_{Cl}=n_{HCl}=2n_{H_2}=2\cdot\dfrac{8.96}{22.4}=0.8\left(mol\right)\)

\(m_{muối}=m_{kl}+m_{Cl}=11+0.8\cdot35.5=39.4\left(g\right)\)

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

            \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

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

\(\Rightarrow m_{Fe}=0,2\cdot56=11,2\left(g\right)\) \(\Rightarrow m_{Fe_2O_3}=16\left(g\right)\)

b+c) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=0,2\left(mol\right)\\n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\end{matrix}\right.\) 

\(\Rightarrow n_{HCl}=2n_{Fe}+6n_{Fe_2O_3}=1\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{36,5}{20\%}=182,5\left(g\right)\)

Mặt khác: \(n_{FeCl_2}=0,2\left(mol\right)=n_{H_2}=n_{FeCl_3}\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\m_{FeCl_3}=0,2\cdot162,5=32,5\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hhA}+m_{ddHCl}-m_{H_2}=209,3\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{25,4}{209,3}\cdot100\%\approx12,14\%\\C\%_{FeCl_3}=\dfrac{32,5}{209,3}\cdot100\%\approx15,53\%\end{matrix}\right.\)

\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

        1         2             1           1

       0,2      0,4          0,2         0,2

     \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)

         1             6               2            3

        0,1         0,6             0,2

a) \(n_{Fe}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{Fe}=0,2.56=11,2\left(g\right)\)

\(m_{Fe2O3}=27,2-11,2=16\left(g\right)\)

b) Có : \(m_{Fe2O3}=16\left(g\right)\)

 \(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)

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

⇒ \(m_{HCl}=1.36,5=36,5\left(g\right)\)

\(m_{ddHCl}=\dfrac{36,5.100}{20}=182,5\left(g\right)\)

c) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{FeCl2}=0,2.127=25,4\left(g\right)\)

\(n_{FeCl3}=\dfrac{0,6.2}{6}=0,2\left(mol\right)\)

⇒ \(m_{FeCl3}=0,2.162,5=32,5\left(g\right)\)

\(m_{ddspu}=27,2+182,5-\left(0,2.2\right)=209,3\left(g\right)\)

\(C_{FeCl2}=\dfrac{25,4.100}{209,3}=12,14\)⁰/₀

\(C_{FeCl3}=\dfrac{32,5.100}{209,3}=15,53\)⁰/₀

 Chúc bạn học tốt

a, Ta có: 65n_Zn + 27n_Al = 11,9 (1)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)

⇒ m_Zn = 0,1.65 = 6,5 (g)

m_Al = 0,2.27 = 5,4 (g)

b, Theo PT: n_ZnCl2 = n_Zn = 0,1 (mol)

n_AlCl3 = n_Al = 0,2 (mol)

⇒ m _muối = 0,1.136 + 0,2.133,5 = 40,3 (g)

c, Theo PT: n_HCl = 2n_H2 = 0,8 (mol)

\(\Rightarrow m_{ddHCl}=\dfrac{0,8.36,5}{10\%}=292\left(g\right)\)

1. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

2. Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 9,2 (1)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(2\right)\)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%\approx70,65\%\\\%m_{Al}\approx29,35\%\end{matrix}\right.\)

3. Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnSO_4}=0,1.160=16\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\end{matrix}\right.\)