Đặt `n_(Mg)=x(mol)`, `n_(MgO)=y(mol)`
`→24x+40y=11.6 (*)`
`PTHH: Mg+2HCl→MgCl_2+H_2 (1)`
`MgO+2HCl→MgCl_2+H_2O (2)`
`NaOH+HCl→NaCl+H_2O (3)`
`n_(HCl)=(400·7.3%)/(100%·36.5)=0.8(mol)`
`m_(NaOH)=(400·4%)/(100%·40)=0.1(mol)`
`Theopt: n_(HCl)(3)=n_(NaOH)=0.1(mol)`
`→∑n_(HCl)(1)+(2)=0.8-0.1=0.7(mol)(**)`
Từ `(*)(**)→x=0.15(mol), y=0.2(mol)`
`→m_(Mg)=0.15·24=3.6(g)`
`→%m_(Mg)=3.6/11.6·100%=31.03%`
`→%m_(MgO)=100%-31.03%=68.97%`
`b,` Dd thu đc chứa `MgCl_2`
Ta có:
`∑n_(MgCl_2)=n_(Mg)+n_(MgO)=0.15+0.2=0.35(mol)`
`→m_(MgCl_2)=0.35·95=33.25`
`n_(H_2)=n_(Mg)=0.15(mol)`
`m_(dd)=m_(hh)+m_(HCl)-m_(H_2)`
`=11.6+400-0.15·2=411.3(g)`
`C%(ddMgCl_2)=33.25/411.3·100%=8.08%`