PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)  (1)

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

a) Gọi số mol của Mg là a (mol) \(\Rightarrow n_{Al}=\dfrac{2}{3}a\left(mol\right)\)

\(\Rightarrow24a+27\cdot\dfrac{2}{3}a=6,3\) \(\Rightarrow a=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{MgO}=0,15\left(mol\right)\\n_{Al_2O_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgO}=0,15\cdot40=6\left(g\right)\\m_{Al_2O_3}=0,05\cdot102=5,1\left(g\right)\end{matrix}\right.\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,075\left(mol\right)\\n_{O_2\left(2\right)}=0,075\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{O_2}=0,15\left(mol\right)\) \(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)

Gọi số mol Fe là x

số mol Mg là y

Số mol oxi là:

\(n_{O_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

\(3Fe+2O_2\rightarrow Fe_3O_4\\ x.....\dfrac{2}{3}x\)

\(2Mg+O_2\rightarrow2MgO\\ y.....\dfrac{y}{2}\)

Ta có:

\(\left\{{}\begin{matrix}56x+24y=1,92\\\dfrac{2}{3}x+\dfrac{y}{2}=0,025\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}56x+24y=1,92\\84\left(\dfrac{2}{3}x+\dfrac{y}{2}\right)=2,1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}56x+24y=1,92\\56x+42y=2,1\end{matrix}\right.\Leftrightarrow18y=0.18\Leftrightarrow y=0,01\left(mol\right)\)

Khối lượng magie trong hỗn hợp là:

\(m_{Mg}=0,01.24=0,24\left(g\right)\)

\(\%m_{Mg}=\dfrac{0,24}{1,92}.100=12.5\%\Rightarrow\%m_{Fe}=87,5\%\)

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Pt : \(2Zn+O_2\underrightarrow{t^o}2ZnO|\)

         2         1        2

        a        0,5b    0,2

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

          2        1           2

         b       0,5b       0,1

a) Gọi a là số mol của Zn

           b là số mol của Mg

\(m_{Zn}+m_{Mg}=15,4\left(g\right)\)

⇒ \(n_{Zn}.M_{Zn}+n_{Mg}.M_{Mg}=15,4g\)

  ⇒ 65a + 24b = 15,4g (1)

Theo phương trình : 0,5a + 0,5b = 0,15 (2)

            65a + 24b = 15,4g

            0,5a + 0,5b = 0,15

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

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

         \(m_{Mg}=01.24=2,4\left(g\right)\)

b) Có : \(n_{ZnO}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

            ⇒ \(m_{ZnO}=0,2.81=16,2\left(g\right)\)

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

           ⇒ \(m_{MgO}=0,1.40=4\left(g\right)\)

 Chúc bạn học tốt

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ 2Zn+O_2\underrightarrow{^{to}}2ZnO\\a.........0,5a..........a\left(mol\right)\\ 2Mg+O_2\underrightarrow{^{to}}2MgO\\b..........0,5b...........b\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}65a+24b=15,4\\0,5a+0,5b=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow m_{Zn}=0,2.65=13\left(g\right);m_{Mg}=0,1.24=2,4\left(g\right)\\ b.m_{ZnO}=81.0,2=16,2\left(g\right)\\ m_{MgO}=0,1.40=4\left(g\right)\)

Gọi số mol Al, Na trong a gam hỗn hợp là x, y (mol)

=> 27x + 23y = a (1)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

             x---------------->0,5x

            4Na + O₂ --t^o--> 2Na₂O

              y---------------->0,5y

=> 102.0,5x + 62.0,5y = 1,64.a

=> 51x + 31y = 1,64a (2)

(1)(2) => 51x + 31y = 1,64(27x + 23y)

=> 6,72x = 6,72y

=> x = y

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27x}{27x+23y}.100\%=54\%\\\%m_{Na}=\dfrac{23y}{27x+23y}.100\%=46\%\end{matrix}\right.\)

3Fe + 2O₂ --to> Fe₃O₄                            4Al + 3O₂    -to-> 2Al₂O₃

x ---------------> x/3                           y------------------> y/2

Theo đề bài\(\dfrac{\dfrac{x.232}{3}+\dfrac{y.102}{2}}{56x+27y}=\dfrac{283}{195}\)

Giải pt => x = 3y 

=> %mFe =\(\dfrac{3y.56}{3y.56+27y}100=\) 86,15%

<=> %mAl = 100 - 86,15 = 13,85%