Đáp án:
`%m_{MgO}=60%`
`V_{dd\ HCl}=63,045(ml)`
Giải thích các bước giải:
`a)`
`Pt: Mg+2HCl->MgCl_2+H_2`
`MgO+2HCl->MgCl_2+H_2O`
`n_{H_2}={2,24}/{22,4}=0,1(mol)`
`n_{Mg}=n_{H_2}=0,1(mol)`
`%m_{MgO}=100%-({0,1.24}/{6}.100%)=60%`
`b)`
`n_{MgO}={6-(0,1.24)}/{40}=0,09(mol)`
`\Sigman_{HCl}=2(n_{Mg}+n_{MgO})=0,38(mol)`
`m_{dd\ HCl}={0,38.36,5}/{20%}=69,35(g)`
`V_{dd\ HCl}={69,35}/{1,1}=63,045(ml)`