\(5.MgO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+2H_2O\\ MgCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\\ n_{CO_2}=0,15\left(mol\right)\\ n_{MgCO_3}=n_{CO_2}=0,15\left(mol\right)\\ \Rightarrow\%m_{MgCO_3}=\dfrac{0,15.84}{20,6}.100=61,17\%\\ \%m_{MgO}=100-61,17=38,83\%\)










