\(n_{C_3H_8}=\dfrac{8,8}{44}=0,2mol\)

\(C_3H_8+5O_2\rightarrow\left(t^o\right)3CO_2+4H_2O\)

 0,2         1                                           ( mol )

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

     2                                                          1   ( mol )

\(m_{KMnO_4}=2.158=316g\)

nO2 = 4.48/22.4 = 0.2 (mol) 

4P + 5O2 -to-> 2P2O5 

0.16__0.2______0.08

mP2O5 = 0.08*142= 11.36(g) 

mP=0.16*31=4.96 (g) 

2KMnO4 -to-> K2MnO4 + MnO2 + O2 

0.4_________________________0.2 

mKMnO4 = 0.4*158=63.2 (g) 

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

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

\(0.2..........0.15\)

\(V_{O_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)

\(0.3...............................................0.15\)

\(m_{KMnO_4}=0.3\cdot158=47.4\left(g\right)\)

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) 
=> \(\%m_{Zn}=\dfrac{0,18.65}{14,68}.100\%=79,7\%\\ \%m_{Mg}=100\%-79,7\%=20,3\%\)

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 )

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

Vậy ...

Có lẽ đề cho "Đốt cháy hoàn toàn 5,4 g Al" bạn nhỉ?

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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

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

\(\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)

b, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)

Câu 6.

\(n_{O_2}=\dfrac{16,8}{22,4}=0,75mol\)

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

1,5                                                 0,75

\(m_{KMnO_4}=1,5\cdot158=237g\)

Câu 7.

\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02mol\)

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

           0,04      0,02

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

\(\dfrac{2}{75}\)                           0,04

\(m_{KClO_3}=\dfrac{2}{75}\cdot122,5=\dfrac{49}{15}\approx3,27g\)

Bài 6 :

\(n_{O_2}=\dfrac{16.8}{22,4}=0,75\left(mol\right)\)

PTHH : 2KMnO₄ ----t⁰-----> K₂MnO₄ + MnO₂ + O₂

               1,5                                                        0,75

\(m_{KMnO_4}=1,5.158=237\left(g\right)\)

PTHH: \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

           \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,15\left(mol\right)\\n_{KMnO_4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{KMnO_4}=0,3\cdot158=47,4\left(g\right)\end{matrix}\right.\)

con cc