PTHH: Mg + 2HCl ---> MgCl2 + H2↑ (1)
MgO + 2HCl ---> MgCl2 + H2O (2)
Ta có: \(n_{H_2}=\dfrac{672:1000}{22,4}=0,03\left(mol\right)\)
Theo PT(1): \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)
=> \(m_{Mg}=0,03.24=0,72\left(g\right)\)
=> \(m_{MgO}=3-0,72=2,28\left(g\right)\)
=> \(\%_{m_{Mg}}=\dfrac{0,72}{3}.100\%=24\%\)
=> \(\%_{m_{MgO}}=100\%-24\%=76\%\)