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

a) 2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

b) \(n_{CH_3COOH}=\dfrac{200.12\%}{60}=0,4\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

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

=> \(m_{Na_2CO_3}=0,2.106=21,2\left(g\right)\)

=> \(m_{dd.Na_2CO_3}=\dfrac{21,2.100}{50}=42,4\left(g\right)\)

c) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(200ml=0,2l\\ n_{Na_2CO_3}=0,5.0,2=0,1\left(mol\right)\\ PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ \left(mol\right)........0,1\rightarrow...0,2.......0,2..........0,1.........0,1\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\ b,m_{NaCl}=0,2.58,5=11,7\left(g\right)\\c, V_{ddNaCl}=V_{ddNa_2CO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ C_{M_{NaCl}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

Na₂CO₃  +  2HCl  →  2NaCl  +  H₂O  + CO₂

a) nNa₂CO₃ = 0,2.0,5 = 0,1 mol 

Theo tỉ lệ phản ứng => nHCl = 2nNa₂CO₃ = 0,1.2 = 0,2 mol

=> C_HCl = \(\dfrac{n}{V}\) = \(\dfrac{0,2}{0,2}\)=  1M

b) nNaCl = 2Na₂CO₃ = 0,2 mol

=> mNaCl = 0,2.58,5 = 11,7 gam

c) Thể tích dung dịch sau phản ứng = 0,2 + 0,2 = 0,4 lít

=> C_NaCl = \(\dfrac{0,2}{0,4}\)= 0,5M

\(n_{H_2SO_4}=\dfrac{150.9,8\%}{98}=0,15\left(mol\right)\\ H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+H_2O+CO_2\\ n_{Na_2CO_3}=n_{H_2SO_4}=0,15\left(mol\right)\\ \Rightarrow m_{ddNa_2CO_3}=\dfrac{0,15.106}{10,6\%}=150\left(g\right)\\ n_{CO_2}=n_{H_2SO_4}=0,15\left(mol\right)\\ m_{ddsaupu}=150+150-0,15.44=293,4\left(g\right)\\ n_{Na_2SO_4}=n_{H_2SO_4}=0,15\left(mol\right)\\ C\%_{Na_2SO_4}=\dfrac{0,15.142}{293,4}.100=7,26\%\)

a) 

\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

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

=> m_CH3COOH = 0,2.60 = 12 (g)

\(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,4}=0,5M\)

b) \(n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄,t^o--> CH₃COOC₂H₅ + H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH₃COOH

\(n_{CH_3COOH\left(pư\right)}=\dfrac{0,2.80}{100}=0,16\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄,t^o--> CH₃COOC₂H₅ + H₂O

                 0,16------------------------------------->0,16

=> \(m_{CH_3COOC_2H_5}=0,16.88=14,08\left(g\right)\)

a)

$CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O$

b)

n NaHCO3 = n CH3COOH = 100.12%/60 = 0,2(mol)

m dd NaHCO3 = 0,2.84/8% = 210(gam)

c)

n CO2 = n CH3COOH = 0,2(mol)

=> V CO2 = 0,2.22,4 = 4,48(lít)

d)

m dd = m dd CH3COOH + m dd NaHCO3 - m CO2 = 100  + 210 - 0,2.44 = 301,2(gam)

C% CH3COONa = 0,2.82/301,2   .100% = 5,44%

\(a,PTHH:Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+CO_2\uparrow\\ b,n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\\ \Rightarrow n_{Na_2SO_4}=0,1\left(mol\right)\\ \Rightarrow m_{Na_2SO_4}=0,1\cdot142=14,2\left(g\right)\\ c,n_{CO_2}=n_{Na_2CO_3}=0,1\left(mol\right)\\ \Rightarrow V_{CO_2\left(đktc\right)}=0,1\cdot22,4=2,24\left(l\right)\)

\(d,n_{Ca\left(OH\right)_2}=0,5\cdot0,3=0,15\left(mol\right)\\ PTHH:CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

Vì \(\dfrac{n_{Ca\left(OH\right)_2}}{1}>\dfrac{n_{CO_2}}{1}\) nên Ca(OH)₂ dư, tính theo CO₂

\(\Rightarrow n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\\ \Rightarrow m_{CaCO_3}=0,1\cdot100\cdot80\%=8\left(g\right)\)

m_Na2CO3 = \(\frac{md.C\%}{100\%}\)= \(\frac{100.10,6\%}{100\%}\)= 10,6 (g)

n_Na2CO3 = \(\frac{m}{M}\)= \(\frac{10,6}{106}\)= 0,1 (mol)

m_H2SO4 = \(\frac{md.C\%}{100\%}\)= \(\frac{200.9,8\%}{100\%}\)= 19,6 (g)

n_H2SO4 = \(\frac{m}{M}\)= \(\frac{19,6}{98}\)= 0,2 (mol)

Khi cho Na₂CO₃ tác dụng với H₂SO₄, ta có PTHH:

Na₂CO₃ + H₂SO₄  \(\rightarrow\)Na₂SO₄ + CO₂\(\uparrow\)+ H₂O

0,1          : 0,2

Xét tỉ lệ: \(\frac{0,1}{1}\)< \(\frac{0,2}{1}\)=> H₂SO₄ dư, dưa vào n_Na2CO3 để tính

Na₂CO₃ + H₂SO₄ \(\rightarrow\) Na₂SO₄ + CO₂\(\uparrow\)+ H₂O

0,1 \(\rightarrow\)0,1              : 0,1            : 0,1        : 0,1               (mol)

a. V_CO2 = n.22,4 = 0,1.22,4 = 2,24 (l)

b. Các chất sau phản ứng gồm Na₂SO₄ và H₂SO₄ dư

m_dsau = m_dNa2CO3 + m_dH2SO4 - m_CO2 = 100 + 200 - 0,1.44 = 295,6 (g)

C%_Na2CO3 = \(\frac{mt}{md}\). 100% = \(\frac{0,1.106}{295,6}\). 100% \(\approx\)3,6 %

C%_H2SO4 = \(\frac{mt}{md}\). 100% = \(\frac{\left(0,2-0,1\right)98}{295,6}\). 100% \(\approx\)3,31%