Question

Đốt cháy 3,5g Al trong bình chứa khí ôxi thu được AL2O3

a/Viết PTHH

b/Tính khối lượng của AL2O3

c/Tính V ôxi tham gia phản ứng

Answer 1

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

b, nAl=\(\dfrac{3,5}{27}\approx0,13\) mol

Theo pt: nAl2O3=\(\dfrac{1}{2}.nAl=\dfrac{1}{2}.0,13=0,065\)mol

=> mAl2O3= 0,065.102= 6,63 (g)

c, Theo pt: nO2=\(\dfrac{3}{4}.nAl=\dfrac{3}{4}.0,13=0,0975\) mol

=> VO2= 0,0975.22,4= 2,184 (l)

Answer 2

a)ta có PTHH : 4Al+3O2--->2Al2O3

b)nAl = 3,5:27=0,13mol

theo PTHH cứ 4 mol Al tạo thành 2 mol Al2O3

0,13 mol Al tạo thành 0,065 mol Al2O3

mAl2O3=0,065.102=6,63g

c)theo PTHH cứ 4 mol Al cần 3mol O2

0,13 mol Al cần 0,0975 mol O2

VO2=0,0975.22,4=2,184(l)

Answer 3

a)PT: 4Al+3O₂\(\underrightarrow{t^o}\)2Al₂O₃

b)n_Al=\(\dfrac{3,5}{27}\)=0,1296(mol)

Theo PT ta có: n\(_{Al_2O_3}\)=\(\dfrac{1}{2}\)n_Al=0,0648(mol)⇒m\(_{Al_2O_3}\)=0,0648.102=6,6096(g)

c)Theo PT ta có:n\(_{O_2}\)=\(\dfrac{3}{4}\)n_Al=0,0972(mol)⇒V\(_{O_2}\)=0,0972.22,4=2,17728(l)

Answer 4

a , AL + O₂ = Al₂O₃

PT 4Al + 3O₂=2AL₂O₃

Answer 5

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

b, \(n_{Al}=\dfrac{m}{M}=\dfrac{3,5}{27}\sim0,13\left(mol\right)\)

Theo PTHH :

\(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,13=0,065\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=n.M=0,065.102=6,63\left(g\right)\)

c, Theo PTHH :

\(n_{O_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,13=0,195\left(mol\right)\)

\(\Rightarrow V_{O_2}=n.22,4=0,195.22,4=4,368\left(l\right)\)