Đáp án: A

Vì dung dịch rượu gồm rượu etylic và nước nên ta gọi:

n H 2 O = x m o l  và n C 2 H 5 O H = y m o l

PTHH:

2 N a   +   2 H 2 O   →   2 N a O H   +   H 2 ↑     ( 1 )

            x mol         →         0,5.x mol

2 N a   +   2 C 2 H 5 O H   →   2 C 2 H 5 O N a   +   H 2 ↑

              y mol         →                  0,5.y mol

Ta có hệ phương trình:

18 x + 46 y = 10 , 1 0 , 5 x + 0 , 5 y = 0 , 125 ⇒ x = 0 , 05 y = 0 , 2

V C 2 H 5 O H  nguyên chất = m D = 0 , 2 . 46 0 , 8 = 11 , 5 m l

V H 2 O = m D = 10 , 1 - 9 , 2 1 = 0 , 9 m l

=> V d d   r ư ợ u = V H 2 O + V C 2 H 5 O H = 0,9 + 11,5 = 12,4 ml

=> Độ rượu  D 0 = V C 2 H 5 O H V d d r u o u . 100 = 11 , 5 12 , 4 . 100 = 92 , 74 0

Chọn D

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

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

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: \(\dfrac{23}{23+1,8}=92,74^o\)

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

Khối lượng riêng của rượu là:

\(D_r=\dfrac{m}{V}=\dfrac{0.8}{1}=0.8\left(kg\text{/}l\right)\)

Thể tích của nước là:

\(V_n=\dfrac{m_n}{D_n}=\dfrac{0.8}{1000}=0.0008\left(m^3\right)=0.8\left(l\right)\)

tóm tắc:

a) Khối lượng riêng của rượu là:

\(D=\dfrac{m}{V}=\dfrac{0,8}{0,001}=800kg/m^3\)

b) Thể tích của nước là:

\(TCT:D=\dfrac{m}{V}\Rightarrow V=\dfrac{m}{D}=\dfrac{0,8}{1000}=0,8\left(l\right)\)

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

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

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

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

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

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)