a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+27b=5,1\)  (1)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Bảo toàn electron: \(2a+3b=0,5\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%\approx47,06\%\\\%m_{Al}=52,94\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{7,3\%}=250\left(g\right)\)

\(\Rightarrow V_{HCl}=\dfrac{250}{1,2}\approx208,33\left(ml\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ FeO+H_2SO_4\rightarrow FeSO_4+H_2O\\ n_{Zn}=n_{H_2}=0,2mol\\ \%m_{Zn}=\dfrac{0,2.65}{20}\cdot100\%=65\%\\ \%m_{Fe}=100\%-65\%=35\%\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=0,15(mol)\\ \Rightarrow \%_{Ag}=\dfrac{20-0,15.65}{20}.100\%=51,25\%\)

\(a)n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ n_{Fe}=a;n_{Al}=b\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}56a+27b=11\\a+1,5b=0,4\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,2\)

\(\%m_{Fe}=\dfrac{0,1.56}{11}\cdot100=50,91\%\\ \%m_{Al}=100-50,91=49,09\%\)

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

0,1            0,2          0,1          0,1

\(2Al+6HCl\rightarrow2AlCl_2+3H_2\)

0,2          0,6           0,2            0,3

\(m_{HCl}=\dfrac{\left(0,2+0,6\right).36,5}{9,125}\cdot100=320g\)       

\(c)m_{dd}=320+11-0,1.2-0,3.2=308,2g\)

\(C_{\%FeCl_2}=\dfrac{0,1.127}{308,2}\cdot100=4,12\%\\ C_{\%AlCl_3}=\dfrac{0,2.133,5}{308,2}\cdot100=8,66\%\)

Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 27y = 7,8 (1)

Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

BT e, có: 2x + 3y = 0,8 (2)

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

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{7,8}.100\%\approx30,77\%\\\%m_{Al}\approx69,23\%\end{matrix}\right.\)

b, BTNT Mg và Al, có:

n_MgCl2 = n_Mg = 0,1 (mol)

 n_AlCl3 = n_Al = 0,2 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCl_2}=\dfrac{0,1.95}{0,1.95+0,2.133,5}.100\%\approx26,24\%\\\%m_{AlCl_3}\approx73,76\%\end{matrix}\right.\)

Bạn tham khảo nhé!

^{trả lời nhanh cho em ạ}

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

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

\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

 x                2x                            x                     x    ( mol )

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

 y                2y                           y                      y      ( mol )

Ta có:

\(\left\{{}\begin{matrix}24x+65y=11,3\\x+y=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8g\\m_{Zn}=0,1.65=6,5g\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{4,8}{11,3}.100=42,47\%\\\%m_{Zn}=100\%-42,47\%=57,53\%\end{matrix}\right.\)

\(m_{CH_3COOH}=60.\left(0,2+0,1\right)=18g\)

\(C\%_{CH_3COOH}=\dfrac{18}{200}.100=9\%\)

\(\left\{{}\begin{matrix}m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4g\\m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3g\end{matrix}\right.\)

\(m_{ddspứ}=11,3+200-0,3.2=210,7g\)

\(\rightarrow\left\{{}\begin{matrix}C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{28,4}{210,7}.100=13,47\%\\C\%_{\left(CH_3COO\right)_2Zn}=\dfrac{18,3}{210,7}.100=8,68\%\end{matrix}\right.\)

Giải thích: 

Zn + 2HCl -> ZnCl₂ + H₂

ZnO + 2HCl -> ZnCl₂ + H₂O

n_H2 = n_Zn = 0,4 mol

n_HCl = 1,2 mol = 2n_Zn + 2n_ZnO => n_ZnO = 0,2 mol

=> %m_ZnO = 38,4%

Đáp án B