Đáp án:
`a.`
`%m_{Fe}=41,176%`
`%m_{Fe_2O_3}=58,824%`
`b.182,5(g)`
`c.195,9(g)`
`d.`
`C%_{FeCl_2}=6,483%`
`C%_{FeCl_3}=8,295%`
Giải thích các bước giải:
`a.`
`Fe+2HCl->FeCl_2+H_2↑`
`Fe_2O_3+6HCl->2FeCl_3+3H_2O`
`n_{H_2}=(2,24)/(22,4)=0,1(mol)`
`n_{Fe}=n_{H_2}=0,1(mol)`
`m_{Fe}=0,1.56=5,6(g)`
`m_{Fe_2O_3}=13,6-5,6=8(g)`
`%m_{Fe}=(5,6)/(13,6).100=41,176%`
`%m_{Fe_2O_3}=8/(13,6).100=58,824%`
`b.`
`n_{Fe_2O_3}=8/160=0,05(mol)`
`∑n_{HCl}=2n_{Fe}+6n_{Fe_2O_3}=0,5(mol)`
`m_{HCl}=0,5.36,5=18,25(g)`
`m_{ddHCl}=18,25:10%=182,5(g)`
`c.`
`m_X=13,6+182,5-0,1.2=195,9(g)`
`d.`
`n_{FeCl_2}=n_{Fe}=0,1(mol)`
`n_{FeCl_3}=2n_{Fe_2O_3}=0,1(mol)`
`C%_{FeCl_2}=(0,1.127)/(195,9).100=6,483%`
`C%_{FeCl_3}=(0,1.162,5)/(195,9).100=8,295%`
