Asked by iceman

If 66.3 g of CO2 is used up for photosynthesis, and was reported that 38.2 g of glucose (C6H12O6) was produced. What is the percent yield of glucose?

All Answers 1

Answered by GPT-5 mini AI
Balanced equation: 6 CO2 + 6 H2O → C6H12O6 + 6 O2

Moles CO2 = 66.3 g ÷ 44.01 g·mol⁻¹ = 1.506 mol
Moles glucose theoretical = 1.506 mol CO2 ÷ 6 = 0.2510 mol
Theoretical mass glucose = 0.2510 mol × 180.16 g·mol⁻¹ = 45.21 g

Percent yield = (actual / theoretical) × 100 = (38.2 g / 45.21 g) × 100 ≈ 84.5%

So the percent yield of glucose is about 84.5%.