1. \(\left(C_6H_{10}O_5\right)_n+nH_2O\underrightarrow{t^o,H^+}nC_6H_{12}O_6\)

\(C_6H_{12}O_6\underrightarrow{^{lenmenruou}}2C_2H_5OH+2CO_2\)

\(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄, t^o)

2. \(\left(C_6H_{10}O_5\right)_n+6nO_2\underrightarrow{t^o}6nCO_2+5nH_2O\)

\(C_6H_{12}O_6+6O_2\underrightarrow{t^o}6CO_2+6H_2O\)

3. \(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

4. \(CH_3COOC_2H_5+5O_2\underrightarrow{t^o}4CO_2+4H_2O\)

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

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= \(\frac{11,5\times100}{40}=27,85ml\)

\(n_{C_2H_6O}=\dfrac{0,46}{46}=0,01\left(mol\right)\)

PTHH: C₂H₆O + 3O₂ --t^o--> 2CO₂ + 3H₂O

            0,01---------------->0,02

=> \(V_{CO_2}=0,02.22,4=0,448\left(l\right)\)

C₂H₅OH+3O₂-to>2CO₂+3H₂O

0,01----------------------0,02

n _C2H5OH =\(\dfrac{0,46}{46}\)=0,01 mol

=>V_CO2=0,02.22,4=0,448l

\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2\uparrow+3H_2O\)

0,01           0,03       0,02         0,03

\(nC_2H_5OH=\) \(\dfrac{0,46}{46}=0,01mol\)

->  nCO\(_2\)=0.4->VCO\(_2\)( đktc)=0.02x22.4= 0,448 (lít)

a)

\((C_6H_{10}O_5)_n + nH_2O \xrightarrow{t^o,xt} nC_6H_{12}O_6\\ C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH\\ \)

m tinh bột = 1.75% = 0,75(kmol)

n tinh bột = 0,75/162n = 1/216n(kmol)

n glucozo = n. 1/216n  .81% = 3/800(kmol)

n C2H5OH = 2n glucozo . 81% = 243/40000(kmol)

m C2H5OH = 46.243/40000 = 0,27945(kg) = 279,45(gam)

V C2H5OH = 279,45/0,8 = 349,3125(ml)

V rượu 30^o = 349,3125.100/30 = 1164,375(ml)

b)

$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
n CH3COOH = 243/40000 .92% = 5,6.10^-3(kmol)

m CH3COOH = 60.5,6.10^-3 = 0,336(kg) = 336(gam)

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