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

b, \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)

c, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)

nAl = 2,7/27 = 0,1 (mol)

PTHH: 4Al + 3O2 -> (t°) 2Al2O3

Mol: 0,1 ---> 0,075 ---> 0,05

mAl2O3 = 0,05 . 102 = 5,1 (g)

VO2 = 0,075 . 22,4 = 1,68 (l)

Vkk = 1,68 . 5 = 8,4 (l)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

0,1     0,075           0,05          ( mol )

\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,05.102=5,1g\)

\(V_{kk}=V_{O_2}.5=\left(0,075.22,4\right).5=8,4l\)

Giúp vs

4Al+3O2-to>2Al2O3

0,2-----0,15-----0,1 mol

n Al=0,2 mol

=>m O2=0,15.32=4,8g

=>Vkk=0,15.22,4.5=16,8l

=>m Al2O3=0,1.102=10,2g

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

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

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

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

c, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

\(a,\) Nhôm + Oxi \(\xrightarrow{t^o}\) Nhôm Oxit

\(b,4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ c,\text{Bảo toàn KL: }m_{O_2}=m_{Al_2O_3}-m_{Al}=40,8-21,6=19,2(g)\)

\(a.Nhôm+Oxi\rightarrow NhômOxit\\ b.4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\\ c.BTKL\Rightarrow m_{O_2}=m_{Al_2O_3}-m_{Al}=40,8-21,6=19,2\left(g\right)\)

a.\(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

 0,6     0,45           0,3                ( mol )

\(m_{Al}=0,6.27=16,2g\)

\(V_{O_2}=0,45.22,4=10,08l\)

\(V_{kk}=10,08.5=50,4l\)

b.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

       0,3                                           0,45 ( mol )

\(m_{KClO_3}=0,3.122,5=36,75g\)

c.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

       0,3                                           0,45  ( mol )

\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)

\(m_{KClO_3}=0,4.122,5=49g\)

\(a.4Al+3O_2\rightarrow2Al_2O_3\\ b.n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=0,05.102=5,1\left(g\right)\\ c.n_{O_2}=\dfrac{3}{4}n_{Al}=0,075\left(mol\right)\\ \Rightarrow V_{O_2}=0,075.22,4=1,68\left(l\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1(mol)\\ a,PTHH:4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ b,n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,05(mol)\\ \Rightarrow m_{Al_2O_3}=0,05.102=5,1(g)\\ c,n_{O_2}=\dfrac{3}{4}n_{Al}=0,075(mol)\\ \Rightarrow V_{O_2}=0,075.22,4=1,68(l)\)

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

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=16,2/27= 0,6(mol)

a) PTHH: 4 Al +3 O2 -to-> 2 Al2O3

nO2= 3/4 . nAl=3/4 . 0,6= 0,45(mol)

=> V(O2,đktc)=0,45 x 22,4=10,08(l)

b) nAl2O3= nAl/2=0,6/2=0,3(mol)

=>mAl2O3=102. 0,3= 30,6(g)

c) 2KMnO4 -to-> K2MnO4 + MnO2 + O2

nKMnO4= 2.nO2=2. 0,45=0,9(mol)

=>mKMnO4= 158 x 0,9= 142,2(g)