a) 2Mg + O₂ --t^o--> 2MgO

4Al + 3O₂ --t^o--> 2Al₂O₃

b) Gọi số mol Mg, Al là a, b

=> 24a + 27b = 7,8 

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2Mg + O₂ --t^o--> 2MgO

______a--->0,5a-------->a

4Al + 3O₂ --t^o--> 2Al₂O₃

b-->0,75b------->0,5b

=> 0,5a + 0,75b = 0,2

=> a = 0,1 ; b = 0,2

=> m_Mg = 0,1.24 = 2,4 (g); m_Al = 0,2.27 = 5,4 (g)

=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{2,4}{7,8}.100\%=30,769\%\\\%Al=\dfrac{5,4}{7,8}.100\%=69,231\%\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}n_{MgO}=0,1\left(mol\right)\\n_{Al_2O_3}=0,1\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{MgO}=0,1.40=4\left(g\right)\\m_{Al_2O_3}=0,1.102=10,2\left(g\right)\end{matrix}\right.\)

=> m = 4 + 10,2 = 14,2 (g)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{50,4}{2.22,4}=0,45\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

LTL: \(\dfrac{0,2}{4}< \dfrac{0,45}{5}\rightarrow\) O₂ dư

Theo pthh: \(\left\{{}\begin{matrix}n_{Al_2O_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\n_{O_2\left(pư\right)}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(0,45-0,15\right).32=9,6\left(g\right)\\m_{Al_2O_3}=0,1.102=10,2\left(g\right)\end{matrix}\right.\)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(n_{O_2}=\dfrac{9,6}{32}=0,3mol\)

\(4Al+3O_2\rightarrow2Al_2O_3\)

0,2  <  0,3                       ( mol )

0,2       0,15       0,1           ( mol )

\(m_{O_2\left(dư\right)}=\left(0,3-0,15\right).32=4,8g\)

\(n_{Al_2O_3}=0,1.80\%=0,08mol\)

\(m_{Al_2O_3}=0,08.102=8,16g\)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(n_{O_2}=\dfrac{9,6}{32}=0,3mol\)

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

0,2      0,3      0

0,2      0,15    0,1

0         0,15    0,1

Chất dư: \(O_2\) và có khối lượng \(m_{O_2dư}=0,15\cdot32=4,8g\)

\(m_{Al_2O_3}=0,1\cdot102\cdot80\%=8,16g\)

a, Ta có : \(\dfrac{n_{Al}}{n_{Mg}}=\dfrac{2}{1}\)

Mà \(m_{hh}=m_{Al}+m_{Mg}=27n_{Al}+24n_{Mg}=7,8\)

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

b, Ta có : \(\left\{{}\begin{matrix}m_{Al}=n.M=5,4\\m_{Mg}=n.M=2,4\end{matrix}\right.\) g

Vậy ...

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%

3Fe + 2O₂ --> Fe₃O₄                            4Al + 3O₂    --> 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 = \(\dfrac{mFe}{mFe+mAl}.100\%\)= \(\dfrac{3y.56}{3y.56+27y}.100\%\) = 86,15%

<=> %mAl = 100 - 86,15 = 13,85%

4Al+3O2-to>2Al2O3

0,2-----0,15-----0,1 mol

n Al=0,2 mol

=>m O2=0,15.32=4,8g

=>Vkk=0,15.22,4.5=16,8l

=>m Al2O3=0,1.102=10,2g