\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)
\(2Mg+O_2\underrightarrow{t^0}2MgO\)
\(0.2.......0.1........0.2\)
\(V_{O_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{MgO}=0.2\cdot40=8\left(g\right)\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(PTHH:2Mg+O_2\overset{t^o}{--->}2MgO\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(lít\right)\)