\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)

Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)

Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)

\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)

\(n_{MgCl_2}=0,15.0,2=0,03(mol)\\ PTHH:MgCl_2+2NaOH\to Mg(OH)_2\downarrow +2NaCl\\ a,n_{Mg(OH)_2}=n_{MgCl_2}=0,03(mol)\\ \Rightarrow m_{\downarrow}=m_{Mg(OH)_2}=0,03.58=1,74(g)\\ b,n_{NaOH}=2n_{MgCl_2}=0,06(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{0,06}{0,3}=0,2M\\ c,PTHH:Mg(OH)_2\xrightarrow{t^o}MgO+H_2O\\ \Rightarrow n_{MgO}=n_{Mg(OH)_2}=0,03(mol)\\ \Rightarrow m_{A}=m_{MgO}=0,03.40=1,2(g)\)

\(n_{BaCl_2}=\dfrac{150.16,64\%}{137+35.2}=0,12\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.14,7\%}{98}=0,15\left(mol\right)\)

Phương trình hóa học : 

BaCl₂ + H₂SO₄ -----> BaSO₄ + 2HCl 

Dễ thấy \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{1}\Rightarrow H_2SO_4\text{ dư }0,15-0,12=0,03\left(mol\right)\)

c) Khối lượng kết tủa : 

\(m_{BaSO_4}=0,12.233=27,96\) (g) 

Khối lượng chất tan : \(m_{HCl}=0,24.36,5=8,76\left(g\right)\) ; 

\(m_{H_2SO_4\left(\text{dư}\right)}=0,03.98=2,94\left(g\right)\) 

c) \(C\%_{H_2SO_4}\)= \(\dfrac{2,94}{150+100}.100\%=1,176\%\)

_{\(C\%_{HCl}=\dfrac{8,76}{150+100}.100\%=3.504\%\)}

d) NaOH + HCl ---> NaCl + H₂O 

      0,24 <-- 0,24

       mol       mol

    2NaOH + H₂SO₄  ---> Na₂SO₄ + 2H₂O

  0,06 mol <-- 0,03 mol  

\(\Rightarrow n_{NaOH}=0,24+0,06=0,3\left(mol\right)\)

\(V_{NaOH}=0,3.2=0,6\left(l\right)\)

a. PTHH    BaCl2 + H2SO4 → BaSO4↓ + 2HCl

b. _ mBaCl2 = 16,61.150:100 = 24,96 g

nBaCl2 = 24.96:208 = 0,12 mol

_ mH2SO4 = 14,7.100:100 = 14,7 g

nH2SO4 = 14,7:98 = 0,15 mol

Ta thấy: nBaCl2 < nH2SO4 ⇒ BaCl2 hết, H2SO4 dư

`n_{Ba(OH)_2}=0,135.2=0,27(mol)`

`n_{Al_2(SO_4)_3}=0,08.1=0,08(mol)`

`Al_2(SO_4)_3+3Ba(OH)_2->2Al(OH)_3+3BaSO_4`

`0,08->0,24->0,16->0,24(mol)`

`->n_{Ba(OH)_2\ du}=0,27-0,24=0,03(mol)`

`Ba(OH)_2+2Al(OH)_3->Ba(AlO_2)_2+4H_2O`

`0,03->0,06(mol)`

`->n_{Al(OH)_3\ du}=0,16-0,06=0,1(mol)`

`2Al(OH)_3`  $\xrightarrow{t^o}$  `Al_2O_3+3H_2O`

`0,1->0,05(mol)`

`->a=m_{Al_2O_3}+m_{BaSO_4}=0,05.102+0,24.233=61,02(g)`