a.

\(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)

b.

\(n_{BaCO_3}=n_{Na_2CO_3}=0,2.1=0,2\left(mol\right)\\ m_{kt}=197.0,2=39,4\left(g\right)\)

c.

\(n_{Ba\left(OH\right)_2}=n_{Na_2CO_3}=0,2\left(mol\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{0,2.171.100\%}{200}=17,1\%\)

\(n_{Na_2CO_3}=0,1.1=0,1\left(mol\right)\)

a. \(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)

       0,1            0,1                   0,1           0,2

b. \(m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\)

c. \(C\%_{Ba\left(OH\right)_2}=\dfrac{0,1.171.100}{200}=8,55\%\)

d. \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\)

      0,1           0,2

=> \(a=m_{dd.HCl}=\dfrac{0,2.36,5.100}{30}=\dfrac{73}{3}\left(g\right)\)

\(a)Ba\left(OH\right)_2+Na_2CO_3\rightarrow2NaOH+BaCO_3\\ n_{Ba\left(OH\right)_2}=0,2.2=0,4mol\\ n_{BaCO_3}=n_{Na_2CO_3}=n_{Ba\left(OH\right)_2}=0,4mol\\ m_{BaCO_3}=0,4.171=68,4g\\ b)V_{Na_2CO_3}=\dfrac{0,4}{1}=0,4l\\ c)n_{NaOH}=2n_{Ba\left(OH\right)_2}=0,8mol\\ C_{M\left(NaOH\right)}=\dfrac{0,8}{0,2+0,4}=\dfrac{4}{3}M\)

a, \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: CO₂ + Ca(OH)₂ → CaCO₃ ↓  +  H₂O

Mol:     0,25      0,25            0,25 

  \(C_{M_{ddCa\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5M\)

b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

c, 

PTHH: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Mol:       0,25           0,5

\(m_{ddHCl}=\dfrac{0,5.36,5.100}{20}=91,25\left(g\right)\)

\(a.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a.Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ 0,25.......0,25............0,25..........0,25\left(mol\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\ b.m_{\downarrow}=m_{CaCO_3}=100.0,25=25\left(g\right)\)

a) Na2CO3+Ba(OH)2--->BaCO3+2NaOH

b) n Na2CO3=0,1.1=0,1(mol)

Theo pthh

n BaCO3=n Na2CO3=0,1(mol)

m BaCO3=0,1.179=17,9(g)

c) Theo pthh

n Ba(OH)2=n Na2CO3=0,1(mol)

C%=\(\frac{0,1.171}{200}.100\%=8,55\%\)

C) BaCO3+2HCl---->BaCl2+H2O+CO2

Theo pthh

n HCl=2n BaCO3=0,2(mol)

m HCl=0,2.36,5=7,3(g)

a=m dd HCl=\(\frac{7,3.100}{30}=24,33\left(g\right)\)

Thiếu CM của Na2CO3 hoặc Ba(OH)2

PTHH: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

Ta có: \(n_{Ba\left(OH\right)_2}=0,5\cdot1=0,5\left(mol\right)=n_{BaSO_4}=n_{H_2SO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,5\cdot98}{15\%}\approx326,7\left(g\right)\\m_{BaSO_4}=0,5\cdot233=116,5\left(g\right)\end{matrix}\right.\)

\(n_{Ba\left(OH\right)_2}=0,5\cdot1=0,5mol\)

a)\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

    0,5                 0,5            0,5        

b) \(m_{H_2SO_4}=0,5\cdot98=49\left(g\right)\)

   \(\Rightarrow m_{ddH_2SO_4}=\dfrac{49}{15}\cdot100=326,67\left(g\right)\)

c) \(m_{BaSO_4}=0,5\cdot233=116,5\left(g\right)\)