`n_(MgO)=\frac{50}{40}=1,25(mol)`
`a,`
`MgO+2HCl->MgCl_2+H_2O`
`b,`
Theo PT
`n_(MgCl_2)=n_(MgO)=1,25(mol)`
`=>m_(MgCl_2)=1,25.95=118,75(g)`
`c,`
Theo PT
`n_(HCl)=2n_(Mg)=2,5(mol)`
`=>m_(HCl)=2,5.36,5=91,25(g)`
`C%_(HCl)=\frac{91,25}{200}.100=45,625%`