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1-B

2: 1-B/2-B

3-D

4-A

5-B

6-C

\(n_{O_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

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

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

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

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

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

0,2                           0,1          ( mol )

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

nAl = 5,4 : 27 = 0,2 (mol) 
pthh : 4Al+ 3O2 -t--> 2Al2O3 
         0,2 --------------> 0,1 (mol) 
=> mAl2O3 = 0,1. 102 = 10,2 (g)

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

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 \(Al_2O_3\) :

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

PTHH: \(4Al+3O_2\xrightarrow[]{t^0}2Al_2O_3\)

Theo ĐLBTKL: \(m_{Al}+m_{O_2}=m_{Al_2O_3}\)

\(\Rightarrow m_{Al}=m_{Al_2O_3}-m_{O_2}=20,4-9,6=10,8\left(g\right)\Rightarrow A\)

Chọn A

A

a.\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{3,16}{158}=0,02mol\)

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

0,02                                                       0,01 ( mol )

\(V_{O_2}=n_{O_2}.22,4=0,01.22,4=0,224l\)

b.

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

1/75   0,01              1/150   ( mol )

\(m_{Al}=n_{Al}.M_{Al}=\dfrac{1}{75}.27=0,36g\)

\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=\dfrac{1}{150}.102=0,68g\)

2KMnO4-to>K2MnO4+MnO2+O2

0,02-------------------------------------0,01

4Al+3O2-to->2Al2O3

\(\dfrac{1}{75}\)---0,01---------\(\dfrac{1}{150}\)

n KMnO4=\(\dfrac{3,16}{158}\)=0,02 mol

=>VO2=0,01.22,4=0,224 l

b)m Al=\(\dfrac{1}{75}\).27=0,36g

=>m Al2O3=\(\dfrac{1}{150}\)102=0,68g

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

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

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          2<---1,5---------->1

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

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

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)