\(n_{NaOH}=0,2.1,5=0,3\left(mol\right)\)

PTHH: CH₃COOH + NaOH ---> CH₃COONa + H₂O

              0,1---------->0,1

=> m_NaOH = 0,1.40 = 4 (g)

=> \(C\%_{NaOH}=\dfrac{4}{80}.100\%=5\%\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO_4(đặc), t^o--> CH₃COOC₂H₅ + H₂O

              0,3------------------------------------------------>0,3

=> m_este = 0,3.88.80% = 21,12 (g)

CH3COOH+NaOH->CH3COONa+H2O

0,3--------------0,3-------0,3--------------0,3

n CH3COOH=0,3 mol

=>C% NaOH=\(\dfrac{0,3.40}{80}.100\)=15%

H=80%

=>m CH3COOC2H5=0,3.88.80%=21,12g

\(n_{H_2SO_4}=\dfrac{200.7,35\%}{98}=0,15\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

               0,3<----0,15-------->0,15

=> m_NaOH = 0,3.40 = 12 (g)

\(m_{dd.NaOH}=\dfrac{12.100}{8}=150\left(g\right)\)

m_{dd sau pư} = 200 + 150 = 350 (g)

m_Na2SO4 = 0,15.142 = 21,3 (g)

=> \(C\%_{dd.Na_2SO_4}=\dfrac{21,3}{350}.100\%=6,086\%\)

a, \(HCl+NaOH\rightarrow NaCl+H_2O\)

b, \(n_{HCl}=0,3.2=0,6\left(mol\right)\)

Theo PT: \(n_{NaOH}=n_{NaCl}=n_{HCl}=0,6\left(mol\right)\)

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

\(\Rightarrow m_{ddNaOH}=\dfrac{24}{20\%}=120\left(g\right)\)

c, \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)

\(n_{NaOH}=C_M.V=0,3\left(mol\right)\)

\(n_{Na2CO3}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

..0,15...........0,3.............0,15............

\(H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+H_2O+CO_2\)

...0,1............0,1..............0,1...............

\(TheoPTHH:n_{H2SO4}=0,1+0,15=0,25\left(mol\right)\)

\(\Rightarrow C_{MH2SO4}=\dfrac{n}{V}=1M\)

- Thấy sau các phản ứng xảy ra thu được muối Na2SO4 ( 0,25mol )

=> mM = n.M = 35,5g

H2SO4 +2 NaOH =>Na2SO4 +2H2O

0,15 mol<=0,3 mol=>0,15 mol

Na2CO3 + H2SO4 =>Na2SO4 + CO2 + H2O

0,1 mol=>0,1 mol=>0,1 mol

nNaOH bđ=0,15.2=0,3 mol

nNa2CO3 thêm=10,6/106=0,1 mol

TỪ pthh=>tổng nH2SO4=0,15+0,1=0,25 mol

CM dd H2SO4 bđ=0,25/0,25=1M

Dd A chỉ chứa Na2SO4 tạo ra từ 2 pt 

Tổng nNa2SO4=0,15+0,1=0,25 mol

=>m muối khan Na2SO4=0,25.142=35,5g

H2SO4 + 2NaOH \(\rightarrow\) Na2SO4 + H2O

a) Ta có: m H2SO4=120.10%=12 gam \(\rightarrow\) nH2SO4=\(\frac{12}{98}\) mol

Theo ptpu: nNa2SO4=nNa2SO4=12/98 mol

\(\rightarrow\)mNa2SO4=\(\frac{12}{98}\) .(23.2+96)=17,39 gam

b)Ta có: nNaOH=2nH2SO4=\(\frac{12}{49}\) mol

\(\rightarrow\) V NaOH=\(\frac{12}{49}\) :2=\(\frac{6}{49}\) lít