\(n_{Na_2CO_3}=\dfrac{200.0,159}{106}=0,3mol\\ n_{BaCl_2}=\dfrac{200.0,208}{208}=0,2mol\\ a.Na_2CO_3+BaCl_2->2NaCl+BaCO_3\\ n_{Na_2CO_3}:1>n_{BaCl_2}:1\\ m_B=197.0,2=39,4g\\ Na_2CO_3+2HCl->2NaCl+H_2O+CO_2\\ V=\dfrac{2.0,1}{1}=0,2\left(L\right)=200\left(mL\right)\\ b.m_A=200+200-39,4=360,6g\\ C\%_{Na_2CO_3du}=\dfrac{106.0,1}{360,6}.100\%=2,94\%\\ C\%_{NaCl}=\dfrac{58,5.0,4}{360,6}.100\%=6,49\%\)

a) 

\(Na_2CO_3+BaCl_2\rightarrow BaCO_3+2NaCl\)

0,2 <---------- 0,2 ------> 0,2 -----> 0,4

\(n_{Na_2CO_3}=\dfrac{200.15,9\%}{100\%}:106=0,3\left(mol\right)\)

\(n_{BaCl_2}=\dfrac{200.20,8\%}{100\%}:208=0,2\left(mol\right)\)

Do \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\) nên \(Na_2CO_3\) dư sau phản ứng.

Dung dịch A: \(n_{Na_2CO_3}=0,3-0,2=0,1\left(mol\right);n_{NaCl}:0,4\left(mol\right)\)

Kết tủa B: \(BaCO_3\)

\(m_B=m_{BaCO_3}=0,2.197=39,4\left(g\right)\)

Dung dịch A td với HCl:

\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

0,1 ---------> 0,2

\(V=V_{HCl}=\dfrac{0,2}{1}=0,2\left(l\right)\)

b)

\(m_{dd}=m_{dd.Na_2CO_3}+m_{dd.BaCl_2}-m_{BaCO_3}=200+200-39,4=360,6\left(g\right)\)

\(C\%_{Na_2CO_3}=\dfrac{0,1.106.100\%}{360,6}=2,94\%\)

\(C\%_{NaCl}=\dfrac{0,4.58,5.100\%}{360,6}=6,49\%\)

a, \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,2.84}{64,8}.100\%\approx25,93\%\\\%m_{MgSO_4}\approx74,07\%\end{matrix}\right.\)

b, - Dung dịch C gồm: MgCl₂, MgSO₄ và HCl dư.

Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{CO_2}=0,2\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{CO_2}=0,4\left(mol\right)\end{matrix}\right.\)

Ta có: \(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)

\(m_{MgSO_4}=64,8-0,2.84=48\left(g\right)\Rightarrow n_{MgSO_4}=\dfrac{48}{120}=0,4\left(mol\right)\)

Có: m _{dd sau pư} = 64,8 + 100 - 0,2.44 = 156 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{156}.100\%\approx12,18\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{156}.100\%\approx2,34\%\\C\%_{MgSO_4}=\dfrac{48}{156}.100\%\approx30,77\%\end{matrix}\right.\)

c, PT: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(HCl+NaOH\rightarrow NaCl+H_2O\)

\(MgSO_4+2NaOH\rightarrow Na_2SO_4+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}+n_{MgSO_4}=0,6\left(mol\right)\)

\(\Rightarrow m_{cr}=m_{MgO}=0,6.40=24\left(g\right)\)

\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(n_{NaCl}=n_{NaOH}=0,2.3=0,6\left(mol\right)\)

=> \(C_{M\left(NaCl\right)}=\dfrac{0,6}{0,2}=3M\)

\(n_{Fe\left(ỌH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,2\left(mol\right)\)

\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)

Ta có \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,1\left(mol\right)\)

=> m _Fe2O3 = 0,1 . 160=16(g)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)

a) CuSO4 + 2KOH -> K2SO4 + Cu(OH)2 (1)

0,13 -> 0,26 -> 0,13 (mol)

Cu(OH)2 -> CuO + H2O (2)

0,13 -> 0,13 (mol)

b) mCuSO4= \(\dfrac{5\cdot416}{100}\)=20,8(g)

nCuSO4= \(\dfrac{20,8}{160}\)=0,13(mol)

=>mCuO= 0,13.80=10,4(g)

c)250ml=0,25l

CmKOH= \(\dfrac{0,26}{0,25}\)=1,04M

\(n_{FeCl_3}=0.2\cdot0.4=0.08\left(mol\right)\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(0.08...........0.24..............0.08\)

\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)

\(0.08...........0.04\)

\(m_{Fe_2O_3}=0.04\cdot160=6.4\left(g\right)\)

\(V_{dd_{NaOH}}=\dfrac{0.24}{0.5}=0.48\left(l\right)\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\) (1)

\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\) (2)

\(n_{FeCl_3}=0,2.0,4=0,08\left(mol\right)\)

Bảo toàn nguyên tố Fe : \(n_{FeCl_3}=2n_{Fe_2O_3}=0,08\left(mol\right)\)

=> \(n_{Fe_2O_3}=0,04\left(mol\right)\)

=> \(m_{Fe_2O_3}=0,04.160=6,4\left(g\right)\)

Theo PT (1) : \(n_{NaOH}=3n_{FeCl_3}=0,08.3=0,24\left(mol\right)\)

=> \(V_{NaOH}=\dfrac{0,24}{0,5}=0,48\left(l\right)\)

\(n_{FeCl_3}=C_M.V=0,4.0,2=0,08\left(mol\right)\)

PT: FeCl₃ + 3NaOH \(\rightarrow\) 3NaCl + Fe(OH)₃ \(\downarrow\)

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

      2Fe(OH)₃ \(\underrightarrow{t^o}\) Fe₂O₃ + 3H₂O

        0,08       ->    0,04                                  (mol)

m_cr=m_Fe2O3=n.M=0,04.160=6,4(g)

\(V_{NaOH}=\dfrac{n}{C_M}=\dfrac{0,24}{0,5}=0,48\left(l\right)\)