a) 

\(V_{C_2H_5OH}=\dfrac{96.20}{100}=19,2\left(ml\right)\)

=> \(m_{C_2H_5OH}=19,2.0,8=15,36\left(g\right)\)

b) \(n_{C_2H_5OH}=\dfrac{15,36}{46}=\dfrac{192}{575}\left(mol\right)\)

\(V_{H_2O}=20-19,2=0,8\left(ml\right)\)

=> \(m_{H_2O}=0,8.1=0,8\left(g\right)\)

=> \(n_{H_2O}=\dfrac{0,8}{18}=\dfrac{2}{45}\left(mol\right)\)

PTHH: 2C₂H₅OH + 2Na --> 2C₂H₅ONa + H₂

               \(\dfrac{192}{575}\)------------------------->\(\dfrac{96}{575}\)

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

              \(\dfrac{2}{45}\)----------------------->\(\dfrac{1}{45}\)

=> \(V_{H_2}=22,4.\left(\dfrac{96}{575}+\dfrac{1}{45}\right)=4,238\left(l\right)\)

C2H5OH + Na -- > C2H5OHNa + 1/2 H2

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

Vr = 20x96/100 = 19,2ml = 0.0192 (l)

mC2H5OH = D.V = 19,2 x 0.8 = 15.36 (g)

nC2H5OH = m/M = 15.36 / 46 = 0.43 (mol) 

=> nH2 = 0.215 (mol)

VH2O = 1 ml => mH2O = 1 (g)

=> nH2O = m/M = 1/18 = 0.056 (mol)

=> nH2 = 0.028 (mol)

nH2 = 0.215 + 0.028 = 0.243 (mol)

=> VH2 = 22.4 x 0,243  = 5,4432 (l)

\(V_{C_2H_5OH}=\dfrac{100.40}{100}=40\left(ml\right)\\ \rightarrow m_{C_2H_5OH}=40.0,8=32\left(g\right)\\ \rightarrow n_{C_2H_5OH}=\dfrac{32}{46}=\dfrac{16}{23}\left(mol\right)\)

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

            \(\dfrac{16}{23}\)----------------------------------->\(\dfrac{8}{23}\)

\(\rightarrow V_{H_2}=\dfrac{8}{23}.22,4=\dfrac{896}{115}\left(l\right)\)

a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

b, \(n_{C_2H_6O}=\dfrac{23}{46}=0,5\left(mol\right)\)

Theo PT: \(n_{O_2}=3n_{C_2H_6O}=1,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

c, \(V_{C_2H_6O}=\dfrac{100.46}{100}=46\left(ml\right)\)

\(\Rightarrow m_{C_2H_6O}=46.0,8=36,8\left(g\right)\)

\(\Rightarrow n_{C_2H_6O}=\dfrac{36,8}{46}=0,8\left(mol\right)\)

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

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5ONa}=0,4\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)

`a)PTHH:`

`C_2 H_5 OH + Na -> C_2 H_5 ONa + 1/2H_2\uparrow`

   `0,08`                                                        `0,04`           `(mol)`

`b)m_[C_2 H_5 OH]=11,5 . 40/100 .0,8=3,68 (g)`

   `n_[C_2 H_5 OH]=[3,68]/46=0,08 (mol)`

  `=>V_[H_2]=0,04.22,4=0,896(l)`

\(a,C_2H_5OH+3O_2\xrightarrow{t^o}2CO_2+3H_2O\\ b,m_{C_2H_5OH}=49.0,8=39,2(g)\\ \Rightarrow n_{C_2H_5OH}=\dfrac{39,2}{46}=\dfrac{98}{115}(mol)\\ \Rightarrow n_{O_2}=\dfrac{98}{115}.3=\dfrac{294}{115}(mol)\\ \Rightarrow V_{kk}=\dfrac{\dfrac{294}{115}.22,4}{20\%}\approx286,33(l)\)

a) C₂H₅OH + 3O₂ --t^o-->2CO₂ + 3H₂O

b) \(m_{C_2H_5OH}=0,8.49=39,2\left(g\right)=>n_{C_2H_5OH}=\dfrac{39,2}{46}=\dfrac{98}{115}\left(mol\right)\)

PTHH: C₂H₅OH + 3O₂ --t^o-->2CO₂ + 3H₂O

_______\(\dfrac{98}{115}\)-->\(\dfrac{294}{115}\)

=> \(V_{O_2}=\dfrac{294}{115}.22,4=57,266\left(l\right)\)

=>  V_kk = 57,266 : 20% = 286,33(l)

Đáp án D

Chọn đáp án D

Lên men rượu:

⇒ có 2,5kg glucozơ chứa 20% tạp chất

⇔ n_glucozơ = 2500 × 0,8 ÷ 180 = 11,11 mol.

Từ tỉ lệ phản ứng

⇒ có 22,22 mol C₂H₅OH được sinh ra.

Mà rượu bị hao hụt mất 10%

⇒ thể tích rượu 40° (có d = 0,8 g/mL) thu được là