Đáp án : B

Gọi độ rượu là x => trong 10 ml X có x ml C₂H₅OH và  (10 – x) ml H₂O

=> m_C2H5OH = 0,8x (g) và m_H2O = (10 – x) g

=> 2n_H2 = n_C2H5OH + n_H2O

=> 0,229 = 0,8x/46 + (10 – x)/18

=> x = 8,580 g/ml

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

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

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

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

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

Bài 1:

PTHH: \(C_2H_5OH+O_2\xrightarrow[]{mengiấm}CH_3COOH+H_2O\)

Ta có: \(n_{C_2H_5OH}=\dfrac{115\cdot0,8}{46}=2\left(mol\right)=n_{CH_3COOH\left(lýthuyết\right)}\)

\(\Rightarrow m_{CH_3COOH\left(thực\right)}=2\cdot60\cdot90\%=108\left(g\right)\)

Bài 2:

PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[H_2SO_4\left(đ\right)]{t^o}CH_3COOC_2H_5+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\n_{C_2H_5OH}=\dfrac{92}{46}=2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Rượu còn dư, Axit p/ứ hết

\(\Rightarrow n_{CH_3COOC_2H_5\left(lýthuyết\right)}=1\left(mol\right)\) \(\Rightarrow m_{CH_3COOC_2H_5\left(thực\right)}=1\cdot88\cdot80\%=70,4\left(g\right)\)

Vì dd rượu gồm rượu etylic và nước nên ta gọi :

\(\left\{{}\begin{matrix}n\left(nước\right)=x\\n\left(rượu-etylic\right)=x\end{matrix}\right.\left(mol\right)\)

PTHH :

2Na + 2H2O - > 2NaOH + H2\(\uparrow\) (1)

..........xmol.........................1/2xmol

2Na + 2C2H5OH - > 2C2H5ONa + H2\(\uparrow\) (2)

............ymol......................................1/2ymol

Ta có HPT : \(\left\{{}\begin{matrix}18x+46y=10,1\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,125\end{matrix}\right.\) => x = 0,05 ; y = 0,2

Ta có :

V(rượu nguyên chất) = \(\dfrac{m}{D}=\dfrac{0,2.46}{0,8}=11,5\left(ml\right)\)

V(nước) = \(\dfrac{m}{D}=\dfrac{10,1-9,2}{1}=0,9\left(ml\right)\)

=> V(dd rượu) = V(nước) + V(rượu nguyên chất) = 0,9 + 11,5

=> độ rượu = \(\dfrac{V\left(rượu-nguyên-chất\right)}{Vdd\left(rượu\right)}.100=\dfrac{11,5}{12,4}.100\approx92,74^o\)

PTHH: CH3COOH+C2H5OH→CH3COOC2H5+H2O

Ta có:

nCH3COOH = 120.15%/60=0,3mol

=> nC2H5OH = nCH3COOH=0,3mol

mC2H5OH = 0,3.46 = 13,8g

=> VC2H5OH = 13,8/0,8 = 17,25ml

=> mH2O = 40−13,8 = 26,2g

=> VH2O=26,2/1=26,2ml

=> Dr=17,25/26,2.100 = 65,84

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