\(V_{C_2H_5OH}=\dfrac{8.100000}{100}=8000\left(ml\right)\\ m_{C_2H_5OH}=8000.0,8=6400\left(ml\right)\\ n_{C_2H_5OH}=\dfrac{6400}{46}=\dfrac{3200}{23}\left(mol\right)\)

PTHH: C₆H₁₂O₆ --t^o, men rượu--> 2C₂H₅OH + 2CO₂

            \(\dfrac{1600}{23}\)<----------------------------\(\dfrac{3200}{23}\)

\(n_{C_6H_{12}O_6}=\dfrac{\dfrac{3200}{23}}{95\%}=146,453\left(mol\right)\\ m_{C_6H_{12}O_6}=146,453.180=26361,54\left(g\right)\)

Giúp mình với 

nAg= 2.16/108=0.02 mol

C6H12O6 + Ag2O -NH3-> C6H12O7 + 2Ag

0.01______________________________0.02

CM C6H12O6= 0.01/0.2=0.05 M

C6H12O6-mr-> 2C2H5OH + 2CO2

0.01____________0.02

mC2H5OH= 0.02*46=0.92g

mC2H5OH thực có= 0.92*90/100=0.828g

C₆H₁₂O₆ \(\underrightarrow{lênmen}\) 2 C₂H₅OH + 2 CO₂

O,25(mol) ← 0,5(mol) ← 0,5 (mol)

❉\(m_{C_6H_{12}O_6}=0,25.180=45\left(g\right)\)

❉\(m_{C_2H_5OH}=0,5.46=23\left(g\right)\)

từ ct m = V.D => \(V=\dfrac{m}{D}=\dfrac{23}{0,8}=28,75\left(l\right)\)

rượu 23 độ => V = 28,75 . 23% = 6,6125 (l)

Đáp án D

C₆H₁₂O₆→ 2CO₂+ 2C₂H₅OH

0,1mo      l→            0,2mol     →m_{rượu ng/chất} =9,2g  →V_ng/chất=\(\frac{m}{D}=\frac{9,2}{0,8}=11,5ml\)

Độ rượu=40=\(\frac{Vruou\left(ngchat\right)}{Vdd}\)\(\times100\)→Vdd/thu đk= \(\frac{11,5\times100}{40}=27,85ml\)

C6H1206|+AGNO3+NH3\(\rightarrow\)C5H14COONH4+NH4NO3+AG

nAg=1,08:108=0,01 mol

theo pt n C6H12O6=nAg=0,01 mol

suy ra Cm =0,01:0,01=1M

PT C6H12O6\(\rightarrow\)2C2H5OH+2CO2

theo pt nC2H5OH=2nC6H12O6=0,02 MOL

SUY RA mC2H5OH=0,02*46=0,92 G

SUY RA khối lượng etylen thu đc là m C2H5OH=90*0,92:100=0,828 G

\(m_{C_2H_5OH}=88.0,8=70,4\left(kg\right)\)

\(C_2H_5OH+O_2\rightarrow\left(t^o,men.giấm\right)CH_3COOH+H_2O\)

      46g `->`   32g           `->`                60g

    70,4kg `->` ?kg            `->`              ?kg

\(m_{O_2}=\dfrac{70,4.32}{46}=48,97\left(g\right)\)

\(V_{O_2}=\dfrac{48,97}{32}.22,4.90\%=30,85\left(m^3\right)\)

\(m_{CH_3COOH}=\dfrac{70,4.60.90}{46.100}=82,64\left(kg\right)\)