Question

(ii) Calculate the maximum mass of ethanol that could be obtained g from 30.0 g of glucose.

Answer 1

Taking into account the reaction stoichiometry, 15.33 grams ethanol could be obtained from 30 grams of glucose.

Reaction stoichiometry

In first place, the balanced reaction is:

C₆H₁₂O₆  → 2 C₂H₅OH + 2 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• C₆H₁₂O₆: 1 mole  
• C₂H₅OH: 2 moles
• CO₂: 2 moles

The molar mass of the compounds is:

• C₆H₁₂O₆: 180 g/mole
• C₂H₅OH: 46 g/mole
• CO₂: 44 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• C₆H₁₂O₆: 1 mole ×180 g/mole= 180 grams
• C₂H₅OH: 2 moles ×46 g/mole= 92 grams
• CO₂: 2 moles ×44 g/mole= 88 grams

Mass of C₂H₅OH formed

The following rule of three can be applied: if by reaction stoichiometry 180 grams of glucose form 92 grams of ethanol, 30 grams of glucose form how much mass of ethanol?

$mass of ethanol=\frac{30 grams of glucosex 92 grams of ethanol}{180 grams of glucose}$

mass of ethanol= 15.33 grams

Then, 15.33 grams ethanol could be obtained from 30 grams of glucose.

Learn more about the reaction stoichiometry:

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