\(V_{C_2H_5OH}=\dfrac{1.1000.40}{100}=400\left(ml\right)\\ \rightarrow m_{C_2H_5OH\left(TT\right)}=400.0,8=320\left(g\right)\\ \rightarrow m_{C_2H_5OH\left(LT\right)}=\dfrac{320.100}{80}=400\left(g\right)\\ \rightarrow n_{C_2H_5OH\left(LT\right)}=\dfrac{400}{23}\left(mol\right)\)

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

             \(\dfrac{200}{23}\)<-----------------\(\dfrac{400}{23}\)

\(\rightarrow m_{C_6H_{12}O_6}=\dfrac{200}{23}.180=\dfrac{36000}{23}\left(g\right)\)

20l = 20000ml

\(V_{C_2H_5OH}=\dfrac{23.20000}{100}=4600\left(ml\right)\\ m_{C_2H_5OH}=4600.0,8=3680\left(g\right)\\ n_{C_2H_5OH}=\dfrac{3680}{46}=80\left(mol\right)\)

PTHH: C₆H₁₂O₆ --men rượu--> 2CO₂ + 2C₂H₅OH

              40<--------------------------------------80

\(m_{C_6H_{12}O_6}=\dfrac{40.180}{64\%}=11250\left(g\right)\)

n tinh bột = 1,62/162n = 0,01/n(kmol)

$(C_6H_{10}O_5)_n + nH_2O \xrightarrow{t^o,xt} n C_6H_{12}O_6$

n glucozo = n . n tinh bột . H% = n . 0,01/n . 85% = 0,0085(kmol)

$C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH$

n C2H5OH = 2 . n glucozo . H% = 2.0,0085.90% = 0,0153(kmol)

$C_2H_5OH + O_2 \xrightarrow{t^o,xt} CH_3COOH + H_2O$

n CH3COOH = n C2H5OH .H% = 0,0153.70% = 0,01071(kmol)

m CH3COOH = 0,01071.60 = 0,6426(kg)

Đổi 20l = 20000l

\(V_{C_2H_5OH}=\dfrac{20000.46}{100}=9200\left(ml\right)\\ m_{C_2H_5OH}=9200.0,8=7360\left(g\right)\\ n_{C_2H_5OH}=\dfrac{7360}{46}=160\left(mol\right)\)

PTHH:

C₆H₁₂O₆ --men rượu--> 2C₂H₅OH + 2CO₂

80<----------------------------160

C₆H₁₀O₅ + H₂O ---> C₆H₁₂O₆

80<----------------------80

\(\Rightarrow m_{tinh.bột}=\dfrac{80.162}{50\%.60\%}=43200\left(g\right)\)

\(n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)

\(2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\)

Theo PTHH : 

\(n_{KClO_3\ phản\ ứng} = \dfrac{2}{3}n_{O_2} = 0,2(mol)\\ \Rightarrow n_{KClO_3\ cần\ dùng} = \dfrac{0,2}{70\%} = \dfrac{2}{7}(mol)\\ \Rightarrow m_{KClO_3\ cần\ dùng} = \dfrac{2}{7}.122,5 = 35(gam)\)

Đáp án D

V(rượu) = 0,46 lít = 460ml. → m(rượu) = 460.0,8 = 368

Với H= 80%. m(glucozơ) = 368 . 180 2 . 46 . 0 , 8 = 900

C2H4+H2O->C2H5OH

0,15-----------------0,15

=>nC2H5OH=6,9\46=0,15 mol

=>Vc2H4=0,15.22,4=3,36l

H=5%

=>VC2H4=3,36.100\5=67,2l

CaCO3 ---------to------> CaO + CO2

100.................................56.........44 (g)

m<----------------------------280 (g)

=> m=\(\dfrac{280.100}{56}=500\left(g\right)\)

Vì H=80%

=> \(m=\dfrac{500}{80\%}=625\left(g\right)=0,625\left(kg\right)\)

\(n_{CaO}=\dfrac{280}{56}=5\left(mol\right)\)

PTHH: CaO + CO₂ → CaCO₃

Mol:       5                        5

\(\Rightarrow m=m_{CaCO_3}=5.100.80\%=400\left(g\right)=0,4\left(kg\right)\)