Answered

Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

The formation of ethyl alcohol (C2H5OH) by the fermentation of glucose (C6H12O6) may be represent by the following: C6H12O6 --> 2 C2H5OH 2 CO2 If a particular glucose fermentation process is 70.0% efficient, how many grams of glucose would be required for the production of 51.0 g of ethyl alcohol (C2H5OH)

Sagot :

jbferrado

Answer:

142.5 g

Explanation:

According to the chemical reaction:

C₆H₁₂O₆ --> 2 C₂H₅OH + 2 CO₂

1 mol of glucose (C₆H₁₂O₆) forms 2 moles of ethyl alcohol (C₂H₅OH) and 2 moles of carbon dioxide (CO₂).

We first convert the moles to grams by using the molecular weight (Mw) of each compound:

Mw (C₆H₁₂O₆) = (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6)= 180 g/mol

1 mol C₆H₁₂O₆ = 180 g/mol x 1 mol = 180 g

Mw(C₂H₅OH) = (12 g/mol x 2) + (1 g/mol x 5) + 16 g/mol + 1 g/mol= 46 g/mol

2 mol C₂H₅OH = 2 mol x 46 g/mol = 92 g

Thus, when the process is 100% efficient, 180 grams of glucose produce 92 grams of ethyl alcohol. To form 51.0 grams of ethyl alcohol, we will need:

51.0 g C₂H₅OH x (180 g C₆H₁₂O₆/92 g C₂H₅OH) = 99.8 g C₆H₁₂O₆

As the process has a lower efficiency (70.0%), we will need more glucose to obtain the required yield. So, we divide the mass of glucose required for a process 100% efficient by the actual efficiency:

mass of glucose required = 99.8 g C₆H₁₂O₆/(70%) = 99.8 g C₆H₁₂O₆ x 100/70 = 142.5 g

Therefore, it would be required 142.5 grams of glucose to obtain 51.0 grams of ethyl alcohol.

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.