Đáp án B

\(m_{\text{bình tăng}}=m_{CO_2}=16.456\left(g\right)\)

\(n_{CO_2}=\dfrac{16.456}{44}=0.374\left(mol\right)\)

\(C_6H_{12}O_6\underrightarrow{^{\text{men rượu}}}2C_2H_5OH+2CO_2\)

\(0.187......................0.374.......0.374\)

\(m_{C_6H_{12}O_6}=0.187\cdot180=33.66\left(g\right)\)

\(m_{C_2H_5OH}=0.374\cdot46=17.204\left(g\right)\)

Bài 5: 

\(CO_2 + Ca(OH)2 \to CaCO_3 + H_2O\\ m_{CO_2}= m_{bình\ tăng} = 16,456(gam)\\ \Rightarrow n_{CO_2}= \dfrac{16,456}{44} = 0,374(mol)\\ C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH\\ n_{glucozo} = \dfrac{1}{2}n_{CO_2} = 0,187(mol) \Rightarrow m_{glucozo} = 0,187.180 = 33,66(gam)\\ n_{C_2H_5OH} = n_{CO_2} = 0,374(mol) \Rightarrow m_{C_2H_5OH} = 0,374.46 = 17,204(gam)\)

n_NaOH=0,1.4=0,4 mol

NaOH+CH₃COOH -->CH₃COONa+H₂O

0,4 --> 0,4

m_CH3COOH=0,4.60=24g

b/

CH₃COOH+C₂H₅OH --->CH₃COOCH₂CH₃+H₂O

0,4. --> x mol

Chỉ phản ứng 60% ta có : x=0,4.60/100=0,24mol

m_este=0,24.88=21,12g

Đáp án A

Đáp án A

Hướng dẫn

C₆H₁₂O₆  → 2C₂H₅OH + 2CO₂
V_{C2H5OH nguyên chất} = 750 x 10% = 75 ml

→ m_C2H5OH = 75 x 0,7907 = 59,3025 gam.

→ n_C2H5OH = 59,3025 : 46 = 1,289 mol

→ n_C6H12O6 = 0,645 mol → m_C6H12O6 = 0,645 x 180 = 116,1 gam.

Mà H = 60% → m = 116,1 : 60% = 193,5 gam

Đáp án: A

C₆H₁₂O₆ → 30 - 35 o C e n z i m 2C₂H₅OH + 2CO₂

V_{C2H5OH nguyên chất} = 750 x 10% = 75 ml → m_C2H5OH = 75 x 0,7907 = 59,3025 gam.

→ n_C2H5OH = 59,3025 : 46 = 1,289 mol

→ n_C6H12O6 = 0,645 mol → m_C6H12O6 = 0,645 x 180 = 116,1 gam.

Mà H = 60% → m = 116,1 : 60% = 193,5 gam