`a.`
`Fe+2HCl->FeCl_2+H_2↑`
`Fe_2O_3+6HCl->2FeCl_3+3H_2O`
`b.`
Gọi `n_{Fe}=n_{Fe_2O_3}=a(mol)`
`n_{FeCl_2}=n_{Fe}=a(mol)`
`n_{FeCl_3}=2n_{Fe_2O_3}=2a(mol)`
`m_{\text{muối}}=127a+162,5.2a=4,52(g)`
`=>a=0,01`
`=>n_{Fe}=a=0,01(mol)`
`=>n_{H_2}=n_{Fe}=0,01(mol)`
`V_{H_2}=0,01.22,4=0,224(l)`
`c.`
`m=m_A=56.0,01+160.0,01=2,16(g)`