a) 

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o-- 2Al₂O₃

_____0,5-->0,375--->0,25

=> m_Al2O3 = 0,25.102 = 25,5 (g)

b) V_O2 = 0,375.22,4 = 8,4 (l)

=> V_kk = 8,4.5 = 42 (l)

Bài 1:

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

Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,3mol\\n_{Al_2O_3}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0.3\cdot22.4=6,72\left(l\right)\\m_{Al_2O_3}=0,2\cdot102=20,4\left(g\right)\end{matrix}\right.\)

Bài 2 : 

\(n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)\)

4Na + O₂ \(\xrightarrow{t^o}\) 2Na₂O

0,4......0,1.........0,2..................(mol)

Vậy  :

\(V_{O_2} = 0,1.22,4 = 2,24(lít)\\ m_{Na} = 0,4.23 = 9,2(gam)\)

Bài 1:

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

Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

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

b, Ta có: \(V_{O_2}=0,3.22,4=6,72\left(l\right)\)

c, Ta có: \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

Bài 2:

PT: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)

Ta có: \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{Na}=2n_{Na_2O}=0,4\left(mol\right)\\n_{O_2}=\dfrac{1}{2}n_{Na_2O}=0,1\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(V_{O_2}=0,1.22,4=2,24\left(l\right)\)

b, Ta cóL \(m_{Na}=0,4.23=9,2\left(g\right)\)

Bạn tham khảo nhé!

\(n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

 0,03   0,02               0,01  ( mol )

\(m_{Fe_3O_4}=0,01.232=2,32\left(g\right)\)

\(V_{kk}=0,02.22,4.5=2,24\left(l\right)\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

   \(\dfrac{1}{75}\)                               0,02   ( mol )

\(m_{KClO_3}=\dfrac{1}{75}.122,5=1,63\left(g\right)\)

a,nFe=1,68/56=0,03 mol

Ta có PTHH : 3Fe + 2O2 --> Fe3O4 (1) ( ở trên dấu --> có to nha )

Theo PTHH ta có :

nFe3O4=1/3nFe=1/3.0,03=0,01 mol

nO2=2/3nFe=2/3.0,03=0,02 mol

=>mFe3O4= 0,01.232=2,32g

=>Vkk=5.(0,02.22,4)=2,24 l

b, Ta có PTHH: 2KClO3 --> 2KCl + 3O2 (2) ( trên dấu --> vẫn có to )

Gọi x là số mol KClO3 cần dùng ( x > 0 )

Theo PTHH (3) và theo bài ra ta có PTHH sau:

2/3x=0,02 

=> x=0,03 mol

=> mKClO3= 0,03.122,5= 3,675g

nAl = 10,8: 27=0,4 (mol) 
pthh : 4Al + 3O2 -t--->2 Al2O3
          0,4---> 0,3 (mol) 
=>VO2  = 0,3 .22,4 = 6,72 (l)
ta  có  : VO2 = 1/5 Vkk <=> Vkk = VO2 : 1/5= 33,6 (l)
pthh : 2KClO3 -t--> 2KCl + 3O2 
               0,2<---------------------0,3  (mol) 
=> mKClO3 = 0,2 . 122,5 (g) 
pthh : 2KMnO4-t--> K2MnO4 + MnO2+ O2 
                 0,6<--------------------------------  0,3(mol) 
=> mKMnO4 = 0,6.158 = 94,8 (g)
     

a) PTHH: \(4Al+3O_2\rightarrow2Al_2O_3\)

b) \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

Theo PTHH: \(n_{Al_2O_3}=\dfrac{0,5.2}{4}=0,25\left(mol\right)\)

 Khối lượng sản phẩm tạo thành: \(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,25.102=25,5\left(g\right)\)

c) Theo PTHH: \(n_{O_2}=\dfrac{0,5.3}{4}=0,375\left(mol\right)\)

Thể tích không khí cần dùng: \(V_{O_2}=n_{O_2}.22,4=0,375.22,4=8,4\left(l\right)\)

b, tính m của cái nào bn?

\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)

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

\(0.4......0.3..........0.2\)

\(m_{Al_2O_3}=0.2\cdot102=20.4\left(g\right)\)

\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)

Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

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

_____0,4____0,3___0,2 (mol)

b, \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

c, \(V_{O_2}=0,3.22,4=6,72\left(l\right)\)

Bạn tham khảo nhé!

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

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

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.24,79=3,7185\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=18,5925\left(l\right)\)

b, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

a)\(n_{Al}=\dfrac{5,4}{27}=0,2\left(m\right)\)

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

tỉ lệ       :4          3        2

số mol  :0,2       0,15   0,1

\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)

\(V_{kk}=\dfrac{2,24.20\%}{100\%}=11,2\left(l\right)\)