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:

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

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

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)

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

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

Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\Rightarrow n_{O_2\left(2\right)}=0,075\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(1\right)}=1,425\left(mol\right)\) \(\Rightarrow n_{Mg}=2,85\left(mol\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{2,85\cdot24}{2,85\cdot24+2,7}\cdot100\%\approx96,2\%\)

\(\Rightarrow\%m_{Al}=3,8\%\) 

\(n_{O_2} =\dfrac{33,6}{22,4} = 1,5(mol)\\ n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ n_{O_2} = \dfrac{1}{2}n_{Mg} + \dfrac{3}{4}n_{Al}\\ \Rightarrow n_{Mg} = 2,85(mol)\)

Vậy :

\(\%m_{Mg} = \dfrac{2,85.24}{2,85.24 + 2,7}.100\% = 96,2\%\\ \%m_{Al} = 100\% - 96,2\% = 3,8\%\)

no2 = 3,36 /22,4 = 0,15mol 

gọi x,y lần lượt là mol của Mg và Cu 

lập pthh của pu 

2Mg + O2 ----------->2MgO  (có nhiệt độ ) (1)

2mol    1mol              2mol 

x mol            1/2x mol

2Cu + O2 ------------>2CuO (có nhiệt độ ) (2) 

2mol    1mol               2mol 

y mol   1/2y mol    

m hỗn hợp =  mMg  + mCu 

11,2              24x       64y  (3)

tổng số mol O2 = nO2 (1) + nO2 (2)   

                 0,15  =    1/2x   + 1/2 y   (4)

từ 3,4 ta có hệ pt  11,2 = 24x  +     64y

                               0.15  = 1/2 x  + 1/2 y

   giải hệ pt ta được : x =0,2  y= 0,1

mMg = 0,2 . 24 =4,8 g

%mMg = (4,8 /11,2).100% =42,85% 

%mCu = 100% - 42,85% =57,15% 

Theo ĐLBT KL, có: m_KL + m_O2 = m _oxit

⇒ m_O2 = 28,4 - 15,6 = 12,8 (g)

\(\Rightarrow n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}27x+56y=22,2\\\dfrac{1}{2}x\cdot102+\dfrac{1}{3}y\cdot232=33,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

a)\(\%m_{Al}=\dfrac{0,2\cdot27}{22,2}\cdot100\%=24,32\%\)

\(\%m_{Fe}=100\%-24,32\%=75,68\%\)

b)Theo hai pt trên:

\(\Rightarrow n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{2}{3}n_{Fe}=\dfrac{3}{4}\cdot0,2+\dfrac{2}{3}\cdot0,3=0,35mol\)

\(H=80\%\Rightarrow n_{O_2}=80\%\cdot0,35=0,28mol\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)

\(\dfrac{14}{75}\)                         0,28

\(m_{KClO_3}=\dfrac{14}{75}\cdot122,5=22,87g\)

a)

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,1<-0,05<-------0,1

            2CO + O₂ --t^o--> 2CO₂

            0,2<--0,1-------->0,2

=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\V_{CO}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)

b) \(m_{CO_2}=0,2.44=8,8\left(g\right)\)

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(n_{H_2O}=\dfrac{1,8}{18}=0,1mol\)

\(2CO+O_2\rightarrow2CO_2\)

  a         0,5a     a

\(2H_2+O_2\rightarrow2H_2O\)

 0,1   0,05 \(\leftarrow\)    0,1

\(\Sigma n_{O_2}=0,5a+0,05=0,15\)

\(\Rightarrow a=n_{O_2\left(CO\right)}=0,2mol\)

\(V_{CO}=2\cdot0,2\cdot22,4=8,96l\)

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{CO_2}=0,2\cdot44=8,8g\)

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

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

a/ \(2CO\left(0,2\right)+O_2\left(0,1\right)\rightarrow2CO_2\left(0,2\right)\)

\(2H_2\left(0,1\right)+O_2\left(0,05\right)\rightarrow2H_2O\left(0,1\right)\)

\(n_{H_2O}=\frac{1,8}{18}=0,1\)

\(n_{O_2}=\frac{3,36}{22,4}=0,15\)

Số mol O₂ phản ứng ở phản ứng đầu là: \(0,15-0,05=0,1\)

\(\Rightarrow m_{CO_2}=0,2.44=8,8\)

b/ \(m_{CO}=0,2.28=5,6\)

\(m_{H_2}=0,1.2=0,2\)

c/ \(\%CO=\frac{0,2}{0,3}.100\%=66,67\%\)

\(\Rightarrow\%H_2=100\%-66,67\%=33,33\%\)