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

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

ta có :

2Mg+O2-to>2MgO

x--------0,5x

2Zn+O2-to>2ZnO

y-------0,5y

=>\(\left\{{}\begin{matrix}24x+65y=14,58\\0,5x+0,5y=0,15\end{matrix}\right.\)

=>x=0,12 mol ,y=0,18 mol

=>%mMg=\(\dfrac{0,12.24}{14,58}100\)=19,753%

=>%mZn=80,247%

\(n_{O_2}=\dfrac{4,8}{32}=0,15mol\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Zn}=y\end{matrix}\right.\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

  x        1/2 x                       ( mol )

\(2Zn+O_2\rightarrow\left(t^o\right)2ZnO\)

  y       1/2 y                         ( mol )

Ta có:

\(\left\{{}\begin{matrix}24x+65y=14,58\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,12\\y=0,18\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,12.24=2,88g\\m_{Zn}=0,18.65=11,7g\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,88}{14,58}.100=19,75\%\\\%m_{Zn}=100\%-19,75\%=80,25\%\end{matrix}\right.\)

\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\) 
gọi số mol Mg : a , số mol Zn : b (a,b>0) 
pthh : \(2Mg+O_2\underrightarrow{t^o}2MgO\) 
         a          \(\dfrac{1}{2}a\) 
         \(2Zn+O_2\underrightarrow{t^o}2ZnO\) 
         b         \(\dfrac{1}{2}b\) 
mà 24a + 65b=14,58  
có : \(\dfrac{1}{2}a+\dfrac{1}{2}b=0,15\) 
=> a = 0,12 ( mol ) , b=0,18 (mol) 
=> \(\%m_{Zn}=\dfrac{0,18.65}{14,68}.100\%=79,7\%\\ \%m_{Mg}=100\%-79,7\%=20,3\%\)

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

a, Giả sử: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 27x + 56y = 13,8 (1)

BTNT Al và Fe, có: \(\left\{{}\begin{matrix}n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}x\left(mol\right)\\n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{2}x.102+\dfrac{1}{3}y.232=21,8\left(g\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{13,8}.100\%\approx39,1\%\\\%m_{Fe}\approx60,9\%\end{matrix}\right.\)

b, BTNT O, có: \(n_{O_2}=\dfrac{3n_{Al_2O_3}+4n_{Fe_3O_4}}{2}=0,6\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)

Bạn tham khảo nhé!

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

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

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

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

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

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

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

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 :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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%.

Theo đề bài, ta có: \(\dfrac{m_{Mg}}{3}=\dfrac{m_{Al}}{2}\) và \(m_{Mg}+m_{Al}=6,3\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{m_{Mg}}{3}=\dfrac{m_{Al}}{2}=\dfrac{m_{Mg}+m_{Al}}{3+2}=\dfrac{6,3}{5}=1,26\\ \Rightarrow\left\{{}\begin{matrix}m_{Mg}=3.1,26=3,78\left(g\right)\\m_{Al}=2.1,26=2,52\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}n_{Mg}=\dfrac{3,78}{24}=0,1575\left(mol\right)\\n_{Al}=\dfrac{2,52}{27}=\dfrac{7}{75}\left(mol\right)\end{matrix}\right.\)

\(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ \left(mol\right)0,1575\rightarrow.......0,1575\\ PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ \left(mol\right)....\dfrac{7}{75}\rightarrow.........\dfrac{7}{150}\)

Bạn dựa theo đề làm tiếp nhé, tại không thấy cái đề :v

Rồi đề là gì :v