\(n_{Al}=0,1\left(mol\right)\\ n_{CO_2}=1,5\left(mol\right)\\ 4Al+3CO_2\underrightarrow{t^o}2Al_2O_3+3C\)

n(bd) 0,1 1,5

n(spu) 0 1,425

\(2Mg+CO_2\underrightarrow{t^o}2MgO+C\)

2,85 1,425

\(m_{hh}=2,7+24\cdot2,85=71,1\left(g\right)\\ \Rightarrow\%m_{Al}=\dfrac{2,7\cdot100\%}{71,1}\approx3,797\%\\ \Rightarrow\%m_{Mg}=100\%-3,797\%=96,203\%\)

bạn viết đề bài hơi rồi nên mk làm theo cách mk hiểu, ban kiểm tra lại thể tích CO2 xem có phải là 3,36 l khong vi mk thay so mol CO2 hoi lon

nhấn vào đây

PTHH:

2Mg + O2 =(nhiệt)=> 2MgO (1)

4Al + 3O2 =(nhiệt=> 2Al2O3 (2)

Ta có: n_Al = \(\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(\Sigma n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

+) n_O2 (2) = \(\dfrac{0,1.3}{4}=0,075\left(mol\right)\)

=> nO2(1) = 1,5 - 0,075 = 1,425 (mol)

=> nMg = 1,425 x 2 = 2,85 (mol)

=> mMg = 2,85 x 24 = 68,4 (gam)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{68,4}{68,4+2,7}.100\%=96,2\%\\\%m_{Al}=100\%-96,2\%=3,8\%\end{matrix}\right.\)

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

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

Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Mg}\) \(\Rightarrow n_{Mg}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{2,7}{2,7+0,15.24}.100\%\approx42,86\%\\\%m_{Mg}\approx57,14\%\end{matrix}\right.\)

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

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

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

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

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

Bài 1 :

\(n_{Na}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(4Na+O_2\rightarrow2Na_2O\)

..0,1....0,025....0,05.......

a, \(V_{O_2}=n.22,4=0,56\left(l\right)\)

b, \(m=m_{Na_2o}=n.M=3,1\left(g\right)\)

Bài 2 :

\(n_{Al}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(4Al+3O_2\rightarrow2Al_2O_3\)

..0,1...0,075...

\(\Rightarrow n_{O_2}=0,075\left(mol\right)\)

Mà : \(\Sigma n_{O_2}=\dfrac{V}{22,4}=0,4\left(mol\right)\)

\(\Rightarrow n_{O_2\left(Mg\right)}=0,4-0,075=0,325\left(mol\right)\)

\(2Mg+O_2\rightarrow2MgO\)

.0,65.....0,325........

\(\Rightarrow m_{Mg}=15,6\left(g\right)\)

\(\Rightarrow m_{hh}=2,7+15,6=18,3\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%Al=~14,75\\\%Mg=~85,25\end{matrix}\right.\) %

Bài 3 :

- Gọi số mol Al và Mg lần lượt là x , y

\(4Al+3O_2\rightarrow2Al_2O_3\)

..x....0,75x

\(2Mg+O_2\rightarrow2MgO\)

..y........0,5y...........

Có : \(n_{O_2}=0,75x+0,5y=\dfrac{V}{22,4}=0,1\left(mol\right)\left(I\right)\)

Lại có : \(m_{hh}=m_{Al}+m_{Mg}=27x+24y=3,9\left(II\right)\)

- Giair ( i ) và ( ii ) ta được : \(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\) ( mol )

\(\Rightarrow\left\{{}\begin{matrix}\%Al=~69,23\\\%Mg=~30,77\end{matrix}\right.\) %

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