a)PTHH: \(Ba\left(OH\right)+Na_2CO_3\rightarrow2NaOH+BaCO_3\downarrow\)

             \(BaCO_3\underrightarrow{t^o}BaO+CO_2\uparrow\)

Ta có: \(n_{Ba\left(OH\right)_2}=0,4\cdot0,2=0,08\left(mol\right)\)

\(\Rightarrow n_{NaOH}=0,16mol\) \(\Rightarrow C_{M_{NaOH}}=\dfrac{0,16}{0,4}=0,4\left(M\right)\) (Coi V_dd thay đổi không đáng kể)

b) Theo PTHH: \(n_{BaCO_3}=n_{Ba\left(OH\right)_2}=n_{BaO}=0,08mol\) \(\Rightarrow m_{BaO}=0,08\cdot153=12,24\left(g\right)\)

a. Ba(OH)₂ +Na₂CO₃ ➝ BaCO₃ + 2NaOH

BaCO₃ ➝ BaO + CO₂

nBa(OH)₂ = 0,08 mol

=> nNaOH = 2nBa(OH)₂ = 0,16 mol

=> C_M = 0,4 M

b) Bảo toàn Ba: nBaO = nBa(OH)₂ = 0,08 mol

=> m = 12,24 g

a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

b, \(n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c, \(n_{NaOH}=2n_{CuCl_2}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{200}.100\%=8\%\)

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

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

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

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

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) CuSO4+2NaOH--->Cu(OH)2+Na2SO4

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

m\(_{CuSO4}=\frac{200.8}{100}=16\left(g\right)\)

n\(_{CuSO4}=\frac{16}{160}=0,1\left(mol\right)\)

m\(_{NaOH}=\frac{120.10}{100}=12\left(g\right)\)

n\(_{NaOH}=\frac{12}{40}=0,3\left(mol\right)\)

=> NaOH duư

dd sau pư gồm NaOHdư và Na2SO4

m dd sau pư=200+150=350(g)

Theo pthh

n\(_{NaOH}=2n_{CuSO4}=0,2\left(mol\right)\)

n\(_{NaOH}dư=0,3-0,2=0,1\left(mol\right)\)

C% NaOH =\(\frac{0,1.40}{350}.100\%=1,14\%\)

Theo pthh

n\(_{Na2SO4}=n_{CuSO4}=0,1\left(mol\right)\)

C% CuSO4 =\(\frac{0,1.142}{350}.100\%=4,06\%\)