\(PTHH:2Al+\frac{3}{2}O_2\rightarrow Al_2O_3\)
_________0,1___0,075___0,05
Ta có :
\(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\)
\(\rightarrow V_{O2}=0,075.22,4=1,68\left(l\right)\)
\(\rightarrow m_{Al2O3}=0,05.\left(27.2+16.3\right)=5,1\left(g\right)\)
a) 4Al+3O2--->2Al2O3
n Al=2,7/27=0,1(mol)
n O2=3/4n Al=0,075(mol)
V O2=0,075.22,4=1,68%
b) n Al2O3=1/2n Al=0,05(mol)
m Al2O3=0,05.102=5,1(g)