a) \(n_{Al2O3}=\frac{20,4}{102}=0,2\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^O}2Al_2O_3\)
0,4____0,3____0,2(mol)
\(m_{Al}=0,4.27=10,8\left(g\right)\)
\(V_{O2}=0,3.22,4=6,72\left(l\right)\)
b) \(n_{Al}=0,4\left(mol\right)\)
\(n_{O2}=\frac{11,2}{32}=0,35\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^O}2Al_2O_3\)
0,4___0,35___0,2(mol)
So sanhs 0,4/4 < 0,35/3 ->,Dư
->0,4__0,3____0,2(mol)
\(m_{Al2O3}=0,2.102=20,4\left(g\right)\)
ý ý b đọc nhầm :v
b) nO2=0,3(mol)
nFe=11,2/56=0,2(mol)
_______\(3Fe+2O_2\underrightarrow{t^O}Fe_3O_2\)
Ban đầu:0,2____0,3
PỨ: 0,2___2/15___1/15(mol)
\(m_{Fe3O4}=\frac{1}{15}.232=15,4\left(g\right)\)