\(PTHH:2Mg+O_2\overset{t^o}{--->}2MgO\)
Ta có: \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{Mg}=0,3\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,3.40=12\left(g\right)\)
Chọn D
\(2Mg+O_2\rightarrow2MgO\)
\(nMg=\dfrac{7,2}{24}=0,3mol\)
\(\rightarrow mMgO=0,3.40=12g\)
D