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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 :

Jbferradoidn

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.

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