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

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.

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

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.

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

a) Na₂CO₃  +   Ba(OH)₂   →   BaCO₃↓  +   2NaOH

nNa₂CO₃ = 0,2.2 = 0,4 mol = nBaCO₃

=> mBaCO₃ = 0,4.197 = 78,8 gam

b) BaCO₃   +  2HCl  →  BaCl₂   +  CO₂    +   H₂O

Theo tỉ lệ phản ứng => nHCl cần dùng = 2nBaCO₃ = 0,4.2 =0,8 mol

=>mHCl = 0,8.36,5 = 29,2 gam

<=> mdung dịch HCl 30% cần dùng  = \(\dfrac{29,2}{30\%}\)= 97,3 gam

\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)

\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

a, \(n_{H_2}=n_{Zn}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

\(n_{Na_2SO_4}=\dfrac{142.10}{100.142}=0,1(mol)\\ Na_2SO_4+Ba(OH)_2\to BaSO_4\downarrow+2NaOH\\ \Rightarrow n_{BaSO_4}=n_{Ba(OH)_2}=0,1(mol);n_{NaOH}=0,2(mol)\\ a,m_{BaSO_4}=0,1.233=23,3(g)\\ b,m_{dd_{Ba(OH)_2}}=\dfrac{0,1.171}{15\%}=114(g)\\ c,C\%_{NaOH}=\dfrac{0,2.40}{142+114-23,3}.100\%=3,44\%\)

Ta có: \(n_{Na_2SO_4}=\dfrac{\dfrac{10\%.142}{100\%}}{142}=0,1\left(mol\right)\)

\(PTHH:Na_2SO_4+Ba\left(OH\right)_2--->BaSO_4\downarrow+2NaOH\)

a. Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=n_{Na_2SO_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

b. Ta có: \(m_{Ba\left(OH\right)_2}=0,1.171=17,1\left(g\right)\)

Mà: \(C_{\%_{Ba\left(OH\right)_2}}=\dfrac{17,1}{m_{dd_{Ba\left(OH\right)_2}}}.100\%=15\%\)

\(\Leftrightarrow m_{dd_{Ba\left(OH\right)_2}}=114\left(g\right)\)

c. Ta có: \(m_{dd_{NaOH}}=114+14,2-23,3=104,9\left(g\right)\)

Theo PT: \(n_{NaOH}=2.n_{Ba\left(OH\right)_2}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)

\(\Rightarrow C_{\%_{NaOH}}=\dfrac{8}{104,9}.100\%=7,63\%\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{20\%}==73\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,2}{2,5}=0,08\left(l\right)\)

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

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

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