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)
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)
a)PT: 4Al+3O2\(\underrightarrow{t^o}\)2Al2O3
b)nAl=\(\dfrac{3,5}{27}\)=0,1296(mol)
Theo PT ta có: n\(_{Al_2O_3}\)=\(\dfrac{1}{2}\)nAl=0,0648(mol)⇒m\(_{Al_2O_3}\)=0,0648.102=6,6096(g)
c)Theo PT ta có:n\(_{O_2}\)=\(\dfrac{3}{4}\)nAl=0,0972(mol)⇒V\(_{O_2}\)=0,0972.22,4=2,17728(l)
a , AL + O2 = Al2O3
PT 4Al + 3O2=2AL2O3
\(a,4Al+3O_2\rightarrow^{t^0}2Al_2O_3\)
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 :
\(n_{O_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,13=0,195\left(mol\right)\)
\(\Rightarrow V_{O_2}=n.22,4=0,195.22,4=4,368\left(l\right)\)
