Giúp mình câu 34 với ạ iu iu

`a)`

Đặt `n_{Al}=n_{Mg}=x(mol);n_{Fe}=y(mol)`

`->27x+24x+56y=15,8(1)`

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

`Mg+2HCl->MgCl_2+H_2`

`Fe+2HCl->FeCl_2+H_2`

Theo pt: `1,5x+x+y=n_{H_2}={13,44}/{22,4}=0,6(2)`

`(1)(2)->x=0,2;y=0,1`

`\%m_{Al}={0,2.27}/{15,8}.100\%=34,18\%`

`\%m_{Mg}={0,2.24}/{15,8}.100\%=30,38\%`

`\%m_{Fe}=100-34,18-30,38=35,44\%`

`b)`

Theo pt:

`n_{AlCl_3}=n_{MgCl_2}=x=0,2(mol)`

`n_{FeCl_2}=y=0,1(mol)`

`n_{HCl\ pu}=2n_{H_2}=1,2(mol)`

`n_{HCl\ du}=0,5.2,5-1,2=0,05(mol)`

`m_{dd\ sau}=15,8+500.1,2-0,6.2=614,6(g)`

`C\%_{AlCl_3}={0,2.133,5}/{614,6}.100\%=4,34\%`

`C\%_{MgCl_2}={0,2.95}/{614,6}.100\%=3,09\%`

`C\%_{FeCl_2}={0,1.127}/{614,6}.100\%=2,07\%`

`C\%_{HCl\ du}={0,05.36,5}/{614,6}.100\%=0,3\%`

`c)`

Bảo toàn Mg: `n_{MgO}=x=0,2(mol)`

Bảo toàn Fe: `n_{Fe_2O_3}=1/2y=0,05(mol)`

`->m_{CR}=0,2.40+0,05.160=16(g)`

Câu 33:

`a)`

`n_{HCl} = 0,5.2,5 = 1,25(mol)`

`n_{H_2} = (13,44)/(22,4) = 0,6(mol)`

Gọi `n_{Al} = n_{Mg} = a(mol); n_{Fe} = b(mol)`

`=> 27a + 24a + 56b = 15,8`

`<=> 51a + 56b = 15,8(1)`

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

`a -> 3a -> a -> 1,5a`

`Mg + 2HCl -> MgCl_2 + H_2`

`a -> 2a -> a -> a`

`Fe + 2HCl -> FeCl_2 + H_2`

`b -> 2b -> b -> b`

Theo PT: `n_{HCl(pư)} = 2n_{H_2} = 1,2(mol) < 1,25`

`=> HCl` dư, kim loại tan hết

`=> 1,5a + a + b = 0,6`

`<=> 2,5a + b = 0,6(2)`

`(1),(2) => a = 0,2; b = 0,1`

Hay `n_{Al} = n_{Mg} = 0,2(mol); n_{Fe} = 0,1(mol)`

`\%m_{Al} = (0,2.27)/(15,8) .100\% = 34,18\%`

`\%m_{Mg} = (0,2.24)/(15,8) .100\% = 30,38\%`

`\%m_{Fe} = 100\% - 34,18\% - 30,38\% = 35,44\%`

`m_{\text{muối}} = 0,2.133,5 + 0,2.95 + 0,1.127 = 58,4(g)`

`b)`

`m_{ddHCl} = 500.1,2 = 600(g)`

`=> m_{dd} = 600 + 15,8 - 0,6.2 = 614,6(g)`

`C\%_{AlCl_3} = (0,2.133,5)/(614,6) .100\% = 4,34\%`

`C\%_{MgCl_2} = (0,2.95)/(614,6) .100\% = 3,09\%`

`C\%_{FeCl_2} = (0,1.127)/(614,6) .100\% = 2,07\%`

`C\%_{HCl} = ( (1,25 - 1,2).36,5)/(614,6) .100\% = 0,3\%`

`c)`

`NaOH + HCl -> NaCl + H_2O`

`AlCl_3 + 3NaOH -> Al(OH)_3 + 3NaCl`

`MgCl_2 + 2NaOH -> Mg(OH)_2 + 2NaCl`

`0,2 -> 0,4 -> 0,2`

`FeCl_2 + 2NaOH -> Fe(OH)_2 + 2NaCl`

`0,1 -> 0,2 _> 0,1`

`Al(OH)_3 + NaOH -> NaAlO_2 + 2H_2O`

$Mg(OH)_2 \xrightarrow{t^o} MgO + H_2O$

`0,2 -> 0,2`

$4Fe(OH)_2 + O_2 \xrightarrow{t^o} 2Fe_2O_3 + 4H_2O$

`0,1 -> 0,025 -> 0,05`

`=> m_{\text{chất rắn}} = 0,2.40 + 0,05.160 = 16(g)`