What is the maximum mass of ethanol that can be made by fermenting 90 g of glucose? C6H12O6 → 2C2H5OH + 2CO2 (H 1, C 12, O 16)
- Mr of glucose = (6 × 12) + (12 × 1) + (6 × 16) = 180, so moles of glucose = 90 ÷ 180 = 0.50 mol.
- 1 mol of glucose gives 2 mol of ethanol, so moles of ethanol = 0.50 × 2 = 1.0 mol.
- Mr of C2H5OH = (2 × 12) + (6 × 1) + 16 = 46, so mass = 1.0 × 46 = 46 g.
Answer: 46 g of ethanol