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

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          2<---1,5---------->1

=> \(m_{Al_2O_3}=1.102=102\left(g\right)\)

\(m_{Al}=2.27=54\left(g\right)\)

Ta có: n_{\(O_2\)}=\(\dfrac{6.72}{22.4}\)=0.3 (mol)

PTHH:    4Al   +    3O₂    ^______>    2Al₂O₃     (1)

Ta có: theo (1): n_Al =\(\dfrac{4}{3}n_{O_2}\)=\(\dfrac{4}{3}0.3=0.4\left(mol\right)\)

=> m_Al = 0.4 .  27=10.8(g)

PTHH: 4Al+3O₂->^to 2Al₂O₃

            4        3           2        (mol)

                      0,3                  (mol)

n_O2= V/22,4=6,72/22,4=0,3 (mol)

n_Al= n_O2.4/3=0,3.4/3=0.4 (mol)

m_Al=n.M=0,4.27=10,8 (g)

Em chú ý lần sau môn Hóa đăng ở box hóa nha!

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

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

a, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)

\(\Rightarrow V_{kk}=5V_{O_2}=33,6\left(l\right)\)

b, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

a) số mol của 10,8 gam Al:

\(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

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

 tỉ lệ      4    :    3     :  2 

          0,4 ->   0,3    :  0,2

Thể tích của 0,3 mol \(O_2\) :

\(V_{O_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)

Khối lượng của 0,2 mol \(Al_2O_3\) :

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

nAl = 10,8/27 = 0,4 (mol)

nO2 = 13,44/22,4 = 0,6 (mol)

PTHH: 4Al + 3O2 -> (t°) 2Al2O3

LTL: 0,4/4 < 0,6/3 => O2 dư

mAl2O3 = 0,4/2 = 0,2 (mol)

mAl2O3 = 0,2. 102 = 20,4 (g)

a ) PTHH : 4Al + 3O2 -t-> 2Al2O3
b) nAl = 10,8 : 27= 0,4(mol) 
theo pthh : nAl2O3 = 1/2 nAl = 0,2 (mol) 
=> m = mAl2O3 = 0,2.102=20,4(g)  

a/ PTHH của PƯ: 4Al + 3O₂ ^{\(\rightarrow\)} 2A
nhớ thêm t^o trên dấu mũi tên

Theo ĐLBTKL thì

m _Al + m _O2 = m _Al2O3

10,8 + 9,6 = a

=> a = 20,4 g

\(n_{KMnO_4}=\dfrac{18.96}{158}=0.12\left(mol\right)\)

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

\(0.12...........................................0.06\)

\(V_{O_2}=0.06\cdot22.4=1.344\left(l\right)\)

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

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

\(0.08.....0.06.......0.04\)

\(m_{Al\left(dư\right)}=\left(0.2-0.08\right)\cdot27=3.24\left(g\right)\)

\(m_{Al_2O_3}=0.04\cdot102=4.08\left(g\right)\)

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

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

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

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

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

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!

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:4Al+3O_2\rightarrow^{t^o}2Al_2O_3\\ \Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}V=V_{O_2}=0,3\cdot22,4=6,72\left(l\right)\\a=m_{Al_2O_3}=0,2\cdot102=20,4\left(g\right)\end{matrix}\right.\)