$\rm PTHH : MgO + 2HCl \xrightarrow{} MgCl_2 +H_2$ 

`n_{HCl}=\frac{30,5.10%}{36,5}≈0,084(mol)`

`n_{MgO}=\frac{12}{40}=0,3(mol)`

Tỉ lệ $\dfrac{0,3}{1}>\dfrac{0,084}{2}$ suy ra $HCl$ hết , `MgO` dư 

`c)`

theo pthh $n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,042(mol)$ 

`->m_{MgCl_2}=0,042.95=3,99(g)`

`d)` 

theo pthh 

`n_{H_2}=\frac{1}{2}n_{HCl}=0,042(mol)`

`->V_{H_2}=0,042.24,79=1,04118(l)`

`n_{MgO}=12/40=0,3(mol)`

`n_{HCl}={30,5.10}/{100.36,5}~~0,08(mol)`

`a.PTHH:MgO+2HCl->MgCl_2+H_2O`

`b.`ta có: `{0,3}/1 >{0,08}/2=>MgO` dư

`n_{MgO pư}=1/2 n_{HCl}=0,04(mol)`

`n_{MgO dư}=0,3-0,04=0,26(mol)`

`c.n_{MgCl_2}=1/2 n_{HCl}=0,04(mol)`

`m_{MgCl_2}=0,04.95=3,8(g)`

`d,` không có khí `H_2` thoát ra