\(n_{BaCl_2}=\dfrac{208.15\%}{208}=0,15\left(mol\right)\\ n_{H_2SO_4}=\dfrac{150.19,6\%}{98}=0,3\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ Vì:\dfrac{0,15}{1}< \dfrac{0,3}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{BaSO_4}=n_{BaCl_2}=0,15\left(mol\right)\\ n_{HCl}=2.0,15=0,3\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,13-0,15=0,15\left(mol\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\\ m_{BaSO_4}=233.0,15=34,95\left(g\right)\\ m_{H_2SO_4\left(dư\right)}=0,15.98=14,7\left(g\right)\\ m_{ddsau}=208+150-34,95=323,05\left(g\right)\\ C\%_{ddHCl}=\dfrac{10,95}{323,05}.100\approx3,39\%\)

\(C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{14,7}{323,05}.100\approx4,55\%\)

a,\(m_{BaCl_2}=208.15\%=31,2\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{31,2}{208}=0,15\left(mol\right)\)

\(m_{H_2SO_4}=150.19,6\%=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

PTHH: BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Mol:      0,15       0,15          0,15        0,3

Ta có: \(\dfrac{0,15}{1}< \dfrac{0,3}{1}\) ⇒ BaCl₂ hết, H₂SO₄ dư

\(m_{H_2SO_4dư}=\left(0,3-0,15\right).98=14,7\left(g\right)\)

b, \(m_{BaSO_4}=0,15.233=34,95\left(g\right)\)

  \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

a, \(n_{NaOH}=0,2.0,3=0,06\left(mol\right)\)

\(m_{MgCl_2}=400.30\%=120\left(g\right)\Rightarrow n_{MgCl_2}=\dfrac{120}{95}=\dfrac{24}{19}\left(mol\right)\)

PT: \(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2+2NaCl\)

Xét tỉ lệ: \(\dfrac{0,06}{2}< \dfrac{\dfrac{24}{19}}{1}\), ta được MgCl₂ dư.

Theo PT: \(n_{MgCl_2\left(pư\right)}=\dfrac{1}{2}n_{NaOH}=0,03\left(mol\right)\Rightarrow n_{MgCl_2\left(dư\right)}=\dfrac{24}{19}-0,03=\dfrac{2343}{1900}\left(mol\right)\)

\(\Rightarrow m_{MgCl_2\left(dư\right)}=\dfrac{2343}{1900}.95=117,15\left(g\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{NaCl}=n_{NaOH}=0,06\left(mol\right)\\n_{Mg\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,06.58,5=3,51\left(g\right)\\m_{Mg\left(OH\right)_2}=0,03.58=1,74\left(g\right)\end{matrix}\right.\)

nCuSO4=0,01 mol
Fe+CuSO4=> FeSO4+Cu
        0,01 mol          =>0,01 mol
mCu=0,01.64=0,64gam
FeSO4+2NaOH=>Fe(OH)2 +Na2SO4
0,01 mol=>0,02 mol
Vdd NaOH=0,02/1=0,02 lit

a)b)c)d) mBaCl2=150.16,64%=24,96g

=>nBaCl2=0,12 mol

mH2SO4=100.14,7%=14,7g=>nH2SO4=0,15mol

     BaCl2       + H2SO4 =>BaSO4    +2HCl

Bđ: 0,12 mol;    0,15 mol

Pứ: 0,12 mol=>0,12 mol=>0,12 mol=>0,24 mol

Dư:                   0,03 mol

Dd ban đầu chứa BaCl2 0,12 mol và H2SO4 0,15 mol

Dd A sau phản ứng chứa HCl 0,24 mol và H2SO4 dư 0,03 mol

mHCl=0,24.36,5=8,76g

mH2SO4=0,03.98=2,94g

Kết tủa B là BaSO4 0,12 mol=>mBaSO4=0,12.233=27,96g

mddA=mddBaCl2+mddH2SO4-mBaSO4

=150+100-27,96=222,04g

C%dd HCl=8,76/222,04.100%=3,945%

C% dd H2SO4=2,94/222,04.100%=1,324%

e) HCl     +NaOH =>NaCl +H2O

0,24 mol=>0,24 mol

H2SO4 +2NaOH =>Na2SO4 + 2H2O

0,03 mol=>0,06 mol

TÔNG nNaOH=0,3 mol

=>V dd NaOH=0,3/2=0,15 lit

nNa= 4,6/23=0,2(mol)

PTHH: Na + H2O ->NaOH + 1/2 H2

nNaOH=nNa=0,2(mol)

VddNaOH=V(H2O)= 100(ml)=0,1(l)

=> CMddNaOH= 0,2/0,1= 2(M)