Sửa đề: 36 (g) → 3,6 (g)
Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
Theo PT: \(n_{MgO\left(LT\right)}=n_{Mg}=0,1\left(mol\right)\Rightarrow m_{MgO\left(LT\right)}=0,1.40=4\left(g\right)\)
\(\Rightarrow H=\dfrac{3,6}{4}.100\%=90\%\)