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Đáp án:

Giải thích các bước giải:

 `n_{H_2}={4,4622}/{24,79}=0,18(mol)`

`PTHH:FeO+2HCl->FeCl_2+H_2O`

`2Al+6HCl->2AlCl_3+3H_2`

`a.n_{Al}=2/3 n_{H_2}=0,12(mol)`

`m_{Al}=0,12.27=3,24(g)`

`%Al={3,24}/{10,44}.100%~~31,03%`

`%FeO=100-31,03=68,96%`

`b.n_{AlCl_3}=n_{Al}=0,12(mol)`

`C_M AlCl_3={0,12}/{0,2}=0,6(M)`

`n_{FeCl_2}=n_{FeO}={10,44-3,24}/72=0,1(mol)`

`C_M FeCl_2={0,1}/{0,2}=0,5(M)`

`a)`

`FeO+2HCl->FeCl_2+H_2O`

`2Al+6HCl->2AlCl_3+3H_2`

`n_{H_2}={4,4622}/{24,79}=0,18(mol)`

Theo PT: `n_{Al}=2/3n_{H_2}=0,12(mol)`

`->\%m_{Al}={0,12.27}/{10,44}.100\%\approx 31,03\%`

`->\%m_{FeO}=100-31,03=68,97\%`

`b)`

`n_{FeO}={10,44-0,12.27}/{72}=0,1(mol)`

Theo PT: `n_{AlCl_3}=n_{Al}=0,12(mol);n_{FeCl_2}=n_{FeO}=0,1(mol)`

`200ml=0,2l`

`C_{M\ AlCl_3}={0,12}/{0,2}=0,6M`

`C_{M\ FeCl_2}={0,1}/{0,2}=0,5M`