Taking into account the reaction stoichiometry, 15.33 grams ethanol could be obtained from 30 grams of glucose.
<h3>Reaction stoichiometry</h3>
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
<h3>Mass of C₂H₅OH formed</h3>
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}](https://tex.z-dn.net/?f=mass%20of%20ethanol%3D%5Cfrac%7B30%20grams%20of%20glucosex%2092%20grams%20of%20ethanol%7D%7B180%20grams%20of%20glucose%7D)
<u><em>mass of ethanol= 15.33 grams</em></u>
Then, 15.33 grams ethanol could be obtained from 30 grams of glucose.
Learn more about the reaction stoichiometry:
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