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.
