\(2Mg+O_2\xrightarrow{t^o}2MgO\\ 4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ \Rightarrow \begin{cases} 24.n_{Mg}+27.n_{Al}=5,1\\ 0,5.n_{Mg}+0,75.n_{Al}=n_{O_2}=\dfrac{2,8}{22,4}=0,125 \end{cases}\\ \Rightarrow \begin{cases} n_{Mg}=0,1(mol)\\ n_{Al_2O_3}=0,1(mol) \end{cases}\\ \Rightarrow \%m_{Mg}=\dfrac{0,1.24}{5,1}.100\%\approx 47,06\%\)

a) Gọi số mol Al, Zn là 2a, a (mol)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          2a-->1,5a---------->a

             2Zn + O₂ --t^o--> 2ZnO

            a---->0,5a------->a

=> \(102a+81a=18,3\)

=> a = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{0,2.27+0,1.65}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{0,2.27+0,1.65}.100\%=54,622\%\end{matrix}\right.\)

b) \(n_{O_2}=1,5a+0,5a=0,2\left(mol\right)\)

=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)

Gọi số mol Al, Mg là a, b (mol)

=> 27a + 24b = 12,6 (1)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

              a-->0,75a

            2Mg + O₂ --t^o--> 2MgO

              b-->0,5b

=> 0,75a + 0,5b = 0,3 (2)

(1)(2) => a = 0,2 (mol); b = 0,3 (mol)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{12,6}.100\%=42,86\%\\\%m_{Mg}=\dfrac{0,3.24}{12,6}.100\%=57,14\%\end{matrix}\right.\)

4Al (x mol) + 3O₂ (0,75x mol) \(\underrightarrow{t^o}\) 2Al₂O₃.

2Mg (y mol) + O₂ (0,5y mol) \(\underrightarrow{t^o}\) 2MgO.

Gọi x (mol) và y (mol) lần lượt là số mol của Al và Mg.

Số mol khí oxi cần dùng là 6,72:22,4=0,3 (mol)

Ta có: 27x+24y=12,6 (1).

Lại có: 0,75x+0,5y=0,3 (2).

Giải hệ phương trình gồm (1) và (2), ta suy ra x=0,2 (mol) và y=0,3 (mol).

%m_Al=0,2.27/12,6\(\approx\)42,86%, %m_Mg=0,3.24/12,6\(\approx\)57,14%.

a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)

\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)

\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)

b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)

\(\rightarrow V_{kk}=4,48.5=11,2l\)

c. Có \(n_{Mg}=2n_{O_2}=0,2l\)

\(\rightarrow m_{Mg}=0,2.24=4,8g\)

\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)

\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)

\(n_{O_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(a.......\dfrac{2a}{3}\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

\(b.......\dfrac{3b}{4}\)

\(n_{O_2}=\dfrac{2a}{3}+\dfrac{3b}{4}=0.25\left(mol\right)\left(1\right)\)

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

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

\(n_{H_2}=a+1.5b=0.45\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.15,b=0.2\)

\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)

\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)

\(\%m_{Fe}=\dfrac{8.4}{8.4+5.4}\cdot100\%=60.8\%\)

\(\%m_{Al}=100-60.8=39.2\%\)

PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)  (1)

            \(2Mg+O_2\underrightarrow{t^o}2MgO\)  (2)

Ta có: \(\left\{{}\begin{matrix}n_{O_2\left(1\right)}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}\cdot\dfrac{13,5}{27}=0,375\left(mol\right)\\n_{O_2\left(1\right)}+n_{O_2\left(2\right)}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(2\right)}=0,375\left(mol\right)\) \(\Rightarrow n_{Mg}=0,75\left(mol\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{0,75\cdot24}{0,75\cdot24+13,5}\cdot100\%\approx57,14\%\)