Đặt `n_(Mg)=x(mol), n_(MgO)=y(mol)`
`→24x+40y=9.2`
`Mg+2HCl→MgCl_2+H_2`
` x.........2x............x.............x(mol)`
`MgO+2HCl→MgCl_2+H_2O`
` y...........2y...........y..............y(mol)`
`n_(Mg)=x=n_(H_2)=1.12/22.4=0.05(mol)`
`→24·0.05+40·y=9.2`
`→y=0.2(mol)`
`a, %m_(Mg)=(0.05·24)/9.2·100%≈13%`
`%m_(MgO)=(0.2·40)/9.2·100%≈87%`
`b, n_(HCl)=2·0.05+2·0.2=0.5(mol)`
`m=m_(ddHCl)=(0.5·36.5·100%)/(14.6%)=125(g)`
`c,` Dd sau pư chứa `MgCl_2`
`∑n_(MgCl_2)=x+y=0.05+0.2=0.25(mol)`
Theo ĐLBTKL:
`m_(dd)=m_(hh)+m_(ddHCl)-m_(H_2)`
`=9.2+125-0.05·2=134.1(g)`
`C%(ddMgCl_2)=(0.25·95)/134.1·100%≈17.71%`