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)

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.

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume