anonymous
  • anonymous
fermentation is a complex chemical process of wine making in which glucose is converted into Ethanol and carbon dioxide. C6H12O6 -> C2H5OH + 2CO2 . starting with 925.1 g of glucose , what is the maximum amount of ethanol in GRAMS and LITRES that can be obtained by this process (density of ethanol=0.789g/mol)
Chemistry
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Zale101
  • Zale101
First Step: you balance the chemical equation. Second step: convert grams of glucose to moles of glucose Third step: use mole ratio starting from oles of glucose to get moles of ethanol Fifth step: convert moles of ethanol to grams of ethanol by using the molar mass of ethanol Last step: use the given density to convert grams of ethanol to liters of ethanol (note: I think there's a typo in the density of ethanol=0.789g/mol) \(\Large \bf Density=\frac{g}{ml}\) \(\Large \bf Density\neq \frac{g}{mols}\) Therefore, density of ethanol=0.789g/ml not ethanol=0.789g/mol
anonymous
  • anonymous
FIRST STEP ; C6H12O6 -> 2C2H5OH + 2CO2 2ND STEP : 925.1 g / 180.1559 g = 5.1349970 molesof glucose 3RD STEP : 1 mole of glucose -> 2 moles of ethanol 5TH STEP ; g= 2 moles x 46. 068 44 g = 92.13688 g last step ; ml = 92. 13688 g / 0.789 = 116.7767807 ml thus , 92 g of ethanol with 116.8 ml of ethanol .. thank you for showing me the step! helps me a lot! yah, typo's there! thank you for telling! i hope i get the correct step !
anonymous
  • anonymous
@Zale101 :)

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anonymous
  • anonymous
can i just ask other question here? @Zale101 ?
Zale101
  • Zale101
i meant to write fourth step instead of fifth step xD
Zale101
  • Zale101
what you did looks good :)

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