\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
Theo PTHH ta có tỉ lệ: \(\dfrac{0,5}{2}=0,25>\dfrac{0,2}{1}\)
=> Mg dư. O2 hết => tính theo \(n_{O_2}\)
Theo PT ta có: \(n_{MgO}=2.n_{O_2}=2.0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)