\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-5}\\ =2\sqrt{5}-\dfrac{8\left(1+\sqrt{5}\right)}{4}\\ =2\sqrt{5}-2\left(1+\sqrt{5}\right)\\ =2\sqrt{5}-2-2\sqrt{5}\\ =-2\)

`n_{BaCl_2}={20,8}/{208}=0,1(mol)`

`n_{H_2SO_4}={20.19,6\%}/{98}=0,04(mol)`

`H_2SO_4+BaCl_2->BaSO_4+2HCl`

`0,04->0,04->0,04->0,08(mol)`

Do `0,1>0,04->BaCl_2` dư.

`C\%_{BaCl_2\ du}={208(0,1-0,04)}/{20,8+20-0,04.233}.100\%\approx 39,64\%`

`C\%_{HCl}={0,08.36,5}/{20,8+20-0,04.233}.100\%\approx 9,28\%`

Điều chế `FeCl_2:`

`2NaCl+2H_2O`  $\xrightarrow{đpcmn}$  `2NaOH+Cl_2+H_2`

`2H_2O`  $\xrightarrow{đp}$  `2H_2+O_2`

`Cl_2+H_2`  $\xrightarrow{askt}$  `2HCl`

`4FeS_2+11O_2`  $\xrightarrow{t^o}$  `2Fe_2O_3+8SO_2`

`Fe_2O_3+3H_2`  $\xrightarrow{t^o}$  `2Fe+3H_2O`

`Fe+2HCl->FeCl_2+H_2`

Điều chế `FeCl_3:`

`Fe_2O_3+6HCl->2FeCl_3+3H_2O`

Điều chế `FeSO_4:`

`2SO_2+O_2+2H_2O->2H_2SO_4`

`Fe+H_2SO_4->FeSO_4+H_2`

Điều chế `Fe_2(SO_4)_3`

`Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O`

Điều chế `CuCl_2:`

`Cu+Cl_2`  $\xrightarrow{t^o}$  `CuCl_2`

Điều chế `CuSO_4:`

`2Cu+O_2`  $\xrightarrow{t^o}$  `2CuO`

`CuO+H_2SO_4->CuSO_4+H_2O`

Điều chế `Na_2SO_4:`

`2NaOH+H_2SO_4->Na_2SO_4+2H_2O`

Điều chế nước Gia-ven:

`Cl_2+2NaOH->NaCl+NaClO+H_2O`

Điều chế clorua vôi:

`CaCO_3`  $\xrightarrow{t^o}$  `CaO+CO_2`

`CaO+H_2O->Ca(OH)_2`

`Cl_2+Ca(OH)_2->CaOCl_2+H_2O`

Đr b.

`n_{Fe}={11,2}/{56}=0,2(mol)`

`Fe+2HCl->FeCl_2+H_2`

`0,2`  `->0,4`    `->0,2`  `->0,2(mol)`

`a)`

`200ml=0,2l`

`V_{H_2(dkc)}=0,2.24,79=4,958(l)`

`x=C_{M\ HCl}={0,4}/{0,2}=2M`

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

`a)\ CH_4+2O_2`  $\xrightarrow{t^o}$  `CO_2+2H_2O`

`b)`

`M_X=14,4.2=28,8(g//mol)`

`->{n_{CH_4}}/{n_{O_2}}={32-28,8}/{28,8-16}=1/4`

Cho `n_{CH_4}=1(mol)->n_{O_2}=4(mol)`

`CH_4+2O_2`  $\xrightarrow{t^o}$  `CO_2+2H_2O`

`1->2->1->2(mol)`

Do `1<4/2->O_2` dư `4-2=2(mol)`

`->Y` gồm `CO_2:1(mol);O_2:2(mol)`

`->M_Y={1.44+2.32}/{1+2}=36(g//mol)`

`->d_{X //Y}={28,8}/{36}=0,8`

Hh: `Zn:x(mol);Al:y(mol)`

`->65x+27y=16,24(1)`

`Zn+2HCl->ZnCl_2+H_2`

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

Theo PT: `n_{H_2}=x+1,5y={9,4202}/{24,79}=0,38(2)`

`(1)(2)->x=0,2;y=0,12`

`m_{Zn}=0,2.65=13(g)`

`m_{Al}=16,24-13=3,24(g)`

`->{0,5b}/{a-0,5b}=1/3`

\(1,\\ \text{Với }x>0;x\ne4\\ A=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\\ A=\dfrac{5\sqrt{x}\left(2\sqrt{x}+3\right)}{\left(2\sqrt{x}+3\right)\left(2\sqrt{x}+1\right)}\\ A=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\\ 2,\\ A=\dfrac{\dfrac{5}{2}\left(2\sqrt{x}+1\right)-\dfrac{5}{2}}{2\sqrt{x}+1}=\dfrac{5}{2}-\dfrac{5}{4\sqrt{x}+2}< \dfrac{5}{2}\\ \text{Mà }x>0\Leftrightarrow4\sqrt{x}+2>2\\ \Leftrightarrow\dfrac{5}{4\sqrt{x}+2}< \dfrac{5}{2}\\ \Leftrightarrow A=\dfrac{5}{2}-\dfrac{5}{4\sqrt{x}+2}>\dfrac{5}{2}-\dfrac{5}{2}=0\\ \to0< A< \dfrac{5}{2}\\ \to A\in\left\{1;2\right\}\\ A=1\Leftrightarrow5\sqrt{x}=2\sqrt{x}+1\\ \Leftrightarrow x=\dfrac{1}{9}\left(n\right)\\ A=2\Leftrightarrow5\sqrt{x}=4\sqrt{x}+2\\ \Leftrightarrow x=4\left(l\right)\\ \text{Vậy }x=\dfrac{1}{9}\text{ thỏa.}\)

`a)`

`n_{Fe}=0,1(mol);n_S=0,05(mol)`

`S+Fe`  $\xrightarrow{t^o}$  `FeS`

`0,05->0,05->0,05(mol)`

`0,1>0,05->Fe` dư.

`->X` gồm `Fe:0,1-0,05=0,05(mol);FeS:0,05(mol)`

`Fe+2HCl->FeCl_2+H_2`

`FeS+2HCl->FeCl_2+H_2S`

Theo PT: `n_{H_2}=n_{Fe}=0,05(mol);n_{H_2S}=n_{FeS}=0,05(mol)`

`->\%V_{H_2}=\%V_{H_2S}={0,05}/{0,05+0,05}.100\%=50\%`

`b)`

`n_{NaOH}=0,125.0,1=0,0125(mol)`

`NaOH+HCl->NaCl+H_2O`

Theo PT: `\sum n_{HCl}=n_{NaOH}+2n_{H_2}+2n_{H_2S}=0,2125(mol)`

`->C_{M\ HCl}={0,2125}/{0,5}=0,425M`