4 M + 3 O2 -to-> 2 M2O3
nO2= 3,36/22,4=0,15(mol)
=> nM2O3= 2/3 . 0,15=0,1(mol)
=> M(M2O3)= 10,2/0,1=102(g/mol)
Ta lại có: M(M2O3)=2.M(M)+3.16
=> 2.M(M)+3.16=102
=>M(M)=27(g/mol)
Vậy M là nhôm (Al=27)
4M + 3O2 -> 2M2O3
nO2 = 3,36 : 22,4 = 0,15 mol
nM2O3 = 0,15x2 : 3 = o,1 mol
MM2O3 = 10,2 : 0,1 = 102
M = Mx2 + 16x3 = 102 => M = 27
Vậy M là Al
\(n_{O_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ 4M + 3O_2 \xrightarrow{t^o} 2M_2O_3\\ n_{oxit} = \dfrac{2}{3}n_{O_2} = 0,1(mol)\\ \Rightarrow 2M + 16.3 = \dfrac{10,2}{0,1} = 102\\ \Rightarrow M = 27(Al)\)
\(30_2+4M\text{→}2M_2O_3\)
\(nO_2=\dfrac{3,36}{22,4}=0,15mol\)
\(\text{⇒}nM_2O_3=0,1mol\)
\(MM_2O_3=\dfrac{10,2}{0,1}=102g\)/\(mol\)
\(\text{⇒}2MM+16.3=102\)
\(\text{⇒}MM=27g\)/\(mol\)
\(\text{⇒}M:Al\)
