\(a,n_{C_6H_{12}O_6}=\dfrac{36}{180}=0,2\left(mol\right)\)

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

               0,2----------------->0,4----------->0,4

=> V_CO2 = 0,4.22,4 = 8,96 (l)

b, m_C2H5OH = 0,4.46.50% = 9,2 (g)

\(c,V_{C_2H_5OH}=\dfrac{9,2}{0,8}=11,5\left(ml\right)\\ \rightarrow V_{ddC_2H_5OH}=\dfrac{11,5.100}{60}=\dfrac{115}{6}\left(ml\right)\)

C6H12O6-lm->2C2H5OH+2CO2

0,2--------------------0,4-------0,4

n C6H12O6=0,2 mol

=>VCO2=0,4.22,4=8,96l

=>m C2H5OH=0,4.46=18,4g

a/

 \(n_{CaCO_3}=\dfrac{80}{100}=0,8mol\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

0,8        0,8                 0,8            0,8

\(C_6H_{12}O_6\xrightarrow[]{lên.men}2C_2H_5OH+2CO_2\)

0,4                        0,8                0,8

\(m_{C_6H_{12}O_6}=0,4.180=72g\)

b/\(V_{CO_2}=0,8.22,4=17,92l\)

c/\(m_{C_2H_5OH}=0,8.46=36,8g\)

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

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

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

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

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

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

Bài 5: 

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

Đáp án C

n G l u = 4   m o l

Quá trình lên men rượu:

Hấp thụ hết khí CO₂ vào NaOH dư thì muối thu được là Na₂CO₃