`a.`
`Fe+2HCl->FeCl_2+H_2↑`
`b.`
`n_{H_2}=(3,36)/(22,4)=0,15(mol)`
`n_{Fe}=n_{H_2}=0,15(mol)`
`m_{Fe}=0,15.56=8,4(g)`
`c.`
`n_{HCl}=2n_{H_2}=0,3(mol)`
`C_{M_{ddHCl}}=(0,3)/(0,05)=6(M)`
`\text{___________________________________}`
Đáp án:
`a.`
`Fe+2HCl->FeCl_2+H_2↑`
`b.`
`m_{Fe}=8,4(g)`
`c.`
`C_{M_{ddHCl}}=6(M)`