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)

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

1. Theo ĐLBT KL, có: m_Al + m_O2 = m_Al2O3

2. \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

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

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,15\left(mol\right)\)

\(\Rightarrow m_{O_2}=0,15.32=4,8\left(g\right)\)

\(\left[1\right]BTKL:m_{Al}+m_{O_2}=m_{Al_2O_3}\\ \left[2\right]n_{Al_2O_3}=\dfrac{10,2}{102}=0,1mol\\ 4Al+3O_2\xrightarrow[]{t^0}2Al_2O_3\\ n_{O_2}=\dfrac{0,1.4}{2}=0,2mol\\ m_{Al}=0,2.27=5,4g\)

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

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

\(1,PTHH:4Al+3O_2\rightarrow^{t^o}2Al_2O_3\\ \Rightarrow\text{Số nguyên tử Al }:\text{ số nguyên tử O}=4:3\\ 2,\text{Bảo toàn KL: }m_{O_2}=m_{Al_2O_3}-m_{Al}=9,6\left(g\right)\)

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

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

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

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

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

b, Gọi: m_O2 = x (g) ⇒ m_Al = 1,5x (g)

Theo ĐLBT KL, có: m_Al + m_O2 = m_Al2O3

⇒ 1,5x + x = 10

⇒ x = 4 (g) = m_O2

m_Al = 1,5.4 = 6 (g)