`2)`

`Fe+2HCl->FeCl_2+H_2`

BTKL: `m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}`

`->a=m_{Fe}=12,7+0,2-7,3=5,6(g)`

`3)`

`n_P={15,5}/{31}=0,5(mol)`

`4P+5O_2` $\xrightarrow{t^o}$ `2P_2O_5`

`0,5->0,625->0,25(mol)`

`m_{P_2O_5}=0,25.142=35,5(g)`

`4)`

`H_2+FeO` $\xrightarrow{t^o}$ `Fe+H_2O`

`n_{Fe\ l t}=n_{FeO\ l t}={14,4}/{72}=0,2(mol)`

`H={8,4}/{0,2.56}.100\%=75\%`

Câu `2:`

BTKL : `m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}`

`=>a+7,3=12,7+0,2`

`<=>a=5,6(gam)`

Câu `3:`

`n_{P}=(15,5)/31=0,5(mol)`

`4P+5O_2overset(t^o)(->)2P_2O_5`

`n_{P_2O_5}=1/(2).n_{P}=0,25(mol)`

`->m_{P_2O_5}=0,25.142=35,5(g)`

Câu `4:`

`n_{FeO}=(14,4)/72=0,2(mol)`

`n_{Fe}=(8,4)/56=0,15(mol)`

`H_2+FeOoverset(t^o)(->)Fe+H_2O`

`n_{FeO}` thực tế `=n_{Fe}=0,15(mol)`

`->H%=(0,15)/(0,2).100%=75%`