xin lỗi dài quá

\(a,V_{C_2H_5OH}=\dfrac{10.96}{100}=9,6\left(ml\right)\\ m_{C_2H_5OH}=9,6.0,8=7,68\left(g\right)\\ n_{C_2H_5OH}=\dfrac{7,68}{46}=\dfrac{96}{575}\left(mol\right)\)

PTHH: 2C₂H₅OH + 2Na ---> 2C₂H₅ONa + H₂

              \(\dfrac{96}{575}\)------------------------------------->\(\dfrac{48}{575}\)

\(V_{H_2}=\dfrac{48}{575}.22,4=1,87\left(l\right)\)

\(b,V_{dd}=12+10,6=20,6\left(ml\right)\\ Đ_r=\dfrac{9,6}{20,6}.100=46,6^o\)

\(a,V_{C_2H_5OH}=\dfrac{96.30}{100}=28,8\left(ml\right)\\ \rightarrow m_{C_2H_5OH}=28,8.0,8=23,04\left(ml\right)\\ \rightarrow n_{C_2H_5OH}=\dfrac{23,04}{46}=0,5\left(mol\right)\)

PTHH: 2C₂H₅OH + 2Na ---> 2C₂H₅ONa + H₂

               0,5----------------------------------->0,25

\(\rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)

b, \(n_{CH_3COOH}=\dfrac{36}{60}=0,6\left(mol\right)\)

PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)

LTL: 0,5 < 0,6 => CH₃COOH dư

Theo pthh: n_CH3COOH = n_C2H5OH = 0,5 (mol)

=> m_este = 0,5.88.70% = 30,8 (g)

\(nC_2H_5OH=\dfrac{2,9}{46}=0,06\left(mol\right)\)

\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

 0,06           0,06        0,06           0,03  (mol)

VH2 = 0,03.22,4= 0,672 (l)

V = m /D

=> V rượu etylic = 2,9 / 0,8 = 3,625 (ml)

nC₂H₅OH=\(\dfrac{23}{46}\)=0,5mol

2C₂H₂OH + 2Na → 2C₂H₅ONa + H2↑

  0,5 mol                     →               0,25 mol

=>VH₂=0,25.22,4=5,6lít

Đặt:

nC2H5OH= x mol

nCH3COOH= y mol

mhh= 46x + 60y= 7.6g (1)

nH2= 1.68/22.4=0.075 mol

C2H5OH + Na --> C2H5ONa + 1/2H2

x___________________________0.5x

CH3COOH + Na --> CH3COONa + 1/2H2

y______________________________0.5y

nH2= 0.5x + 0.5y= 0.075 mol (2)

Giải (1) và (2):

x= 0.1

y= 0.05

mC2H5OH= 4.6g

mCH3COOH= 3g

2/ VC2H5OH= 10*96/100=9.6ml

VH2O= Vhh- Vr= 10-9.6=0.4 g

mH2O= 0.4g

nH2O= 1/45 mol

mC2H5OH= 9.6*0.8=7.68g

nC2H5OH= 0.16 mol

C2H5OH + Na --> C2H5ONa + 1/2H2

0.167_________________________0.0835

Na + H2O --> NaOH + 1/2H2

______1/45___________1/90

nH2= 0.0835+1/90=0.095 mol

VH2= 22.128l

a) \(V_r=\frac{Đ_r.V_{hh}}{100}=\frac{92.25}{100}=13,8\left(ml\right)\)

\(m_r=d.V=0,8.13,8=11,04\left(g\right)\)

b) PTHH :\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\uparrow\)

Theo PT: \(n_{H_2}=\frac{1}{2}n_{C_2H_5OH}=\frac{1}{2}.\frac{11,04}{46}=0,12\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,12.22,4=2,688\left(l\right)\)

c) \(V_{hhsaukhipha}=25+20=45\left(ml\right)\)

\(\RightarrowĐ_r=\frac{13,8}{45}=\frac{92}{3}\approx30,67^0\)