HOC24
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Trong 1 tan nuoc mia chua 13% saccarozo co nghia la : 1000kg nuoc mia co chua 130 kg saccarozo
Do hieu suat phan ung thu duoc la 80% nen
130.\(\dfrac{80}{100}=\)104 (kg)
CxHyOz +(4x+y) O2 \(\underrightarrow{t^0}\) 4x CO2 + 2y H2O
Khi dot 1 mol gluxit se tao ra 1 mol CO2 va 0,5 mol H2O hay cu 1 mol gluxit bi dot chay se tao ra 44x gam CO2 va 9y gam H2O
Theo de bai \(\dfrac{9y}{44x}=\dfrac{33}{88}\)
\(\Rightarrow\dfrac{y}{x}=\dfrac{44.33}{9.88}=\dfrac{11}{6}=\dfrac{22}{12}\)
CnH2mOm + nO2 \(\underrightarrow{t^0}\) nCO2 +mH2O
\(\Rightarrow\dfrac{m}{n}=\dfrac{44.33}{9.88}=\dfrac{11}{6}\)
Vay CT phu hop la C12H22O11
1. C12H22O11 +H2O \(\xrightarrow[t^0]{Axit}\)C6H12O6 + C6H12O6
2. C6H12O6 \(\xrightarrow[30-35^0]{menruou}\)2C2H5OH + 2CO2
3.C2H5OH + O2 \(\underrightarrow{mengiam}\) CH3COOH +H2O
4. CH3COOH +2O2\(\underrightarrow{t^0}\)2CO2 +2 H2O
5.C6H12O6 + Ag2O \(\underrightarrow{NH_3}\)C6H12O7 + 2Ag
6. CH3COOH + C2H5 \(\underrightarrow{H_2SO_4dat,t^o}\)CH3COOC2H5 + H2O