`a.`
`Al_2O_3+6HCl->2AlCl_3+3H_2O`
`b.`
`n_{Al_2O_3}=(20,4)/102=0,2(mol)`
`n_{HCl}=0,1.0,1=0,01(mol)`
`(0,2)/1<(0,01)/6=>Al` dư
`n_{AlCl_3}=1/3n_{HCl}=(0,01)/3(mol)`
`V_{ddspư}=V_{ddHCl}=0,1(l)`
`C_{M_{ddspư}}=((0,01)/3)/0,1=1/30(M)`
`\text{___________________________________}`
Đáp án:
`a.`
`Al_2O_3+6HCl->2AlCl_3+3H_2O`
`b.`
`C_{M_{ddspư}}=1/30(M)`