1a.\(nMg=\dfrac{2.4}{24}=0.1mol\Rightarrow nO2=\dfrac{0.1}{2}=0.05mol\)
\(\Rightarrow V_{O2}=0.05\times22.4=11.2l\)
b.nMg = nMgO = 0.1 mol
\(\Rightarrow mMgO=0.1\times40=4g\)
2.4Al + 3O2 -> 2Al2O3
a.\(nAl=\dfrac{1.35}{27}=0.05mol\Rightarrow nO2=\dfrac{0.05\times3}{4}=0.0375mol\)
\(V_{O2}=0.0375\times22.4=0.84l\)
b.\(nAl2O3=\dfrac{0.05}{2}=0.025mol\)
\(mAl2O3=0.025\times102=2.55g\)