\(a,PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ b,n_{Al_2O_3}=\dfrac{30,6}{102}=0,3\left(mol\right)\\ n_{Al}=\dfrac{4}{2}.n_{Al_2O_3}=2.0,3=0,6\left(mol\right)\\ \Rightarrow m_{Al}=0,6.27=16,2\left(g\right)\\ c,n_{O_2}=\dfrac{3}{2}.n_{Al_2O_3}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\\ \Rightarrow V_{O_2\left(đkc\right)}=0,45.24,79=11,1555\left(l\right)\)

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

b, \(nAl_2O_3=\dfrac{30,6}{102}=0,3\left(mol\right)\)

\(nAl=\dfrac{4}{2}.0,3=0,6\left(mol\right)\)

\(mAl=0,6.27=16,2\left(g\right)\)

c, \(nO_2=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)

\(VO_{2\left(đkc\right)}=0,45.24,79=11,1555\left(l\right)\)

a) PTHH: \(4Al+3O_2\rightarrow^{t^0}2Al_2O_3\).

b) \(n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{30,6}{102}=0,3\left(mol\right)\)

-Theo PTHH trên, ta có:

\(n_{Al}=\dfrac{n_{Al_2O_3}}{2}.4=\dfrac{0,3}{2}.4=0,6\left(mol\right)\)

\(\Rightarrow m_{Al}=0,6.27=16,2\left(mol\right)\)

c) -Theo PTHH trên, ta có:

\(n_{O_2}=\dfrac{n_{Al_2O_3}}{2}.3=\dfrac{0,3}{2}.3=0,45\left(mol\right)\).

\(\Rightarrow V_{O_2\left(đktc\right)}=n.24,79=0,45.24,79=11,1555\left(l\right)\)

\(a)n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 4Al+3O_2\xrightarrow[]{t^0}2Al_2O_3\\ n_{Al_2O_3}=\dfrac{0,2.2}{4}=0,1mol\\ m_{Al_2O_3}=0,1.102=10,2g\\ b)n_{H_2}=\dfrac{0,2.3}{4}=0,15mol\\ V_{O_2}=0,15.24,79=3,7185l\)

1. \(n_{O_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

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

Theo PT: \(n_{Al}=\dfrac{4}{3}n_{O_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)

2. \(n_{KCl\left(25\%\right)}=300.25\%=75\left(g\right)\)

Gọi: m _{dd KCl 10%} = a (g) ⇒ m_{KCl (10%)} = 10%a (g)

\(\Rightarrow\dfrac{75+10\%a}{a+300}=0,15\Rightarrow a=600\left(g\right)\)

\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)

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

     0,02-->0,015-->0,01

a) \(m_{Al2O3}=0,01.102=1,02\left(g\right)\)

b) \(V_{O2\left(dktc\right)}=0,015.24,79=0,37185\left(l\right)\)

`#\text{id07 - 25th8}.`

a.

Số mol của Al có trong pứ là:

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{0,54}{27}=0,02\left(\text{mol}\right)\)

PTHH: \(\text{4Al + 3O}_2\text{ }\text{ }\underrightarrow{t^0}\text{ }\text{ 2Al}_2\text{O}_3\) 

Theo pt:   4     :     2  (mol)

`=> 0,02`  :  `0,01` (mol)

`=>` n của Al₂O₃ thu được sau pứ là `0,01` mol

m của Al₂O₃ thu được sau pứ là:

\(m_{Al_2O_3}=n_{Al_2O_3}\cdot M_{Al_2O_3}=0,01\cdot\left(27\cdot2+16\cdot3\right)=0,01\cdot102=1,02\left(g\right)\)

b.

Theo pt: 4   :   3   (mol)

`=>` 0,02  :  0,015  (mol)

`=>` n của O₂ có trong pứ là `0,015` mol

V của khí O₂ tham gia pứ ở đkc là:

\(V_{O_2}=n_{O_2}\cdot24,79=0,015\cdot24,79=0,37185\left(l\right).\)

Vậy: 

a. m Al₂O₃ `=` `1,02` g

b. V O₂ `= 0,37185` l.

a) PTHH:

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

b) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Theo PTHH: \(n_{Al_2O_3}=\dfrac{0,1\cdot2}{4}=0,05\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,05\cdot102=5,1\left(g\right)\)

c) Theo PTHH: \(n_{O_2}=\dfrac{0,1\cdot3}{4}=0,075\left(mol\right)\)

\(\Rightarrow V_{O_2\left(dktc\right)}=0,075\cdot22,4=0,168\left(l\right)\)

Phương trình hoá học: 4Al + 3O₂ → 2Al₂O₃.

Số mol Al tham gia phản ứng:

n Al = mAl : M Al = 0,54 : 27 = 0,02 mol

a)     Từ phương trình hóa học ta có:

n Al₂O₃ = ½ n Al = 0,02 : 2 = 0,01 mol

n Al₂O₃ = 0,01 x 102 = 10,2 gam

b)    theo phương trình hóa học ta có:

n O₂ = ¾ n Al = ¾ x 0,02 = 0,015 mol

V O₂ (đkc) = 0,015 x 24,79 = 0,37185 (lít)

\(PTHH:4Al+3O_2\left(t^o\right)\rightarrow2Al_2O_3\\ n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\Rightarrow n_{Al}=\dfrac{2}{4}.0,02=0,01\left(mol\right);n_{O_2}=\dfrac{3}{4}.0,02=0,15\left(mol\right)\\ a,m_{Al_2O_3}=0,01.27=0,27\left(g\right)\\ b,V_{O_2\left(đkc\right)}=0,15.24,79=3,7185\left(l\right)\)

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

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

            0,2-->0,15

=> V = 0,15.22,4 = 3,36 (l)

b) 

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

              0,3<----------------------------0,15

=> m_{KMnO4(lý thuyết)} = 0,3.158 = 47,4 (g)

=> \(m_{KMnO_4\left(tt\right)}=\dfrac{47,4.110}{100}=52,14\left(g\right)\)

1.\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01mol\)

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

0,3                             0,1    ( mol )

\(m_{Fe}=0,3.56=16,8g\)

2.\(n_{Cu}=\dfrac{3,2}{64}=0,05mol\)

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)

0,05                       0,05     ( mol )

\(m_{CuO}=0,05.80=4g\)

3.\(n_{Na}=\dfrac{4,6}{23}=0,2mol\)

\(4Na+O_2\rightarrow\left(t^o\right)2Na_2O\)

0,2      0,05                         ( mol )

\(V_{O_2}=0,05.24,79=1,2395l\)

4.\(n_{Cu}=\dfrac{1,6}{64}=0,025mol\)

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)

0,025 0,0125                     ( mol )

\(V_{O_2}=0,0125.24,79=0,309875l\)

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