Làm gộp cả phần a và b

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

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,15mol\\n_{Al_2O_3}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{Al_2O_3}=0,1\cdot102=10,2\left(g\right)\end{matrix}\right.\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ n_{O_2}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ V_{kk\left(đktc\right)}=5.3,36=16,8\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

a) $4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$

b) $n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)$
$n_{Al\ pư} = \dfrac{4}{3}n_{O_2} = 0,4(mol)$
$m_{Al\ pư} = 0,4.27 = 10,8(gam)$

c) 

Cách 1 : 

$m_{Al_2O_3} = m_{Al} + m_{O_2} = 10,8 + 0,3.32 = 20,4(gam)$

Cách 2 : 

Theo PTHH, $n_{Al_2O_3} = \dfrac{1}{2}n_{Al\ pư} = 0,2(mol)$
$m_{Al_2O_3} = 0,2.102 = 20,4(gam)$

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

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

$n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
\(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

0,2         0,15            0,1                (mol)

$V_{O_2} = 0,15.22,4 = 3,36(lít)$
$m_{Al_2O_3} =0,1.102 = 10,2(gam)$

tách nhỏ nhé

8

4Al+3O2-to>2al2O3

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

n O2=0,15 mol

=>m AL2o3=0,1.102=10,2g

10

4H2+Fe3O4->3Fe+4H2O

0,8-----0,2--------0,6

n Fe3O4=0,2 mol

=>m Fe=0,6.56=33,6g

=>VH2=0,8.22,4=17,92l

a/ PTHH

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

b/

Ta có: \(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)

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

4                   2

x                    0.1

\(=>x=\dfrac{0.1\cdot4}{2}=0.2=n_{Al}\)  

\(=>m_{Al}=0.2\cdot27=5.4\left(g\right)\)

7. Ta có: n_Zn = \(\dfrac{13}{65}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\)

PTHH: 2Zn + O₂ ---t^o---> 2ZnO

Ta thấy: \(\dfrac{0,2}{2}< \dfrac{1}{1}\)

=> Oxi dư

Theo PT: n_ZnO = n_Zn = 0,2(mol)

=> m_ZnO = 81.0,2 = 16,2(g)

8. Ta có: n_Al = \(\dfrac{21,6}{27}=0,8\left(mol\right)\)

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 4Al + 3O₂ ---t^o---> 2Al₂O₃.

Ta thấy: \(\dfrac{0,8}{4}=\dfrac{0,6}{3}\)

Vậy không có chất dư.

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,8=0,4\left(mol\right)\)

=> \(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)

Em xem bài này là kiểu tính theo PT mà, bài này dễ lắm, bài 8 là bài toán dư

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)