m C2H5OH = m.29,87% = 0,2987m(gam)

=> n C2H5OH = 0,2987m/46 (mol)

m H2O = m - 0,2987m = 0,7013m(gam)

=> n H2O = 0,7013m/18(mol)

$2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2$
$2Na + 2H_2O \to 2NaOH + H_2$

n H2 = 1/2 n C2H5OH + 1/2 H2O = 11,76/22,4 = 0,525(mol)

=> 1/2 . 0,2987m/46 + 1/2 . 0,7013m/18 = 0,525

=> m = 23,1(gam)

Suy ra : 

m C2H5OH = 0,2987.23,1 = 6,9(gam)

V C2H5OH = 6,9/0,8 = 8,625(ml)

m H2O = 0,7013.23,1 = 16,2(gam)

V H2O = 16,2/1 = 16,2(ml)

Vậy :

Đr = V C2H5OH / V(dd) .100 = 8,625/(8,625 + 16,2)  .100 = 34,74^o

Gọi \(\left\{{}\begin{matrix}n_{C_2H_5OH}=a\left(mol\right)\\n_{H_2O}=b\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH:

2C₂H₅OH + 2Na ---> 2C₂H₅ONa + H₂

a --------------------------------------------> 0,5a

2H₂O + 2Na ---> 2NaOH + H₂

b --------------------------------> 0,5b

Hệ pt \(\left\{{}\begin{matrix}46a+18b=20,2\\0,5a+0,5b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,4\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{C_2H_5OH}=0,4.46=18,4\left(g\right)\\m_{H_2O}=0,1.18=1,8\left(g\right)\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}V_{C_2H_5OH}=\dfrac{18,4}{0,8}=23\left(ml\right)\\V_{H_2O}=\dfrac{1,8}{1}=1,8\left(ml\right)\end{matrix}\right.\)

=> Độ rượu là: \(\dfrac{23}{23+1,8}=92,74^o\)

n_H2 = 85,12 : 22,4 = 3,8 (mol) ; n_H2O = V_H2O.D = 108 (g) => n_H2O = 108/18 = 6 (mol)

PTHH:

2Na + 2C₂H₅OH → 2C₂H₅ONa + H₂↑

                x          → 0,5x    (mol)

2Na + 2H₂O → 2NaOH + H₂↑

           6       → 3       (mol)

Ta có: n_H2 = 0,5x + 3 = 3,8

=> x = 1,6 (mol) = n_C2H5OH

m_C2H5OH = 1,6.46 = 73,6 (g)

Chọn D

Đáp án B

Đáp án A

v H 2 = |1/2.0,08+1/2.0,3| .22,4 =4,256 lít

Chọn D

Lời giải:

n_Ancol = 10.46% . 0,8 : 46

n_H2O = (10 – 10.46%) : 18

n_H2 = (n_Ancol + n_H2O ) : 2 = 0,19

⇒ V = 4,256

Đáp án C.