Asked by ruth
For the oxidation of glucose (C6H12O6 + 6O2→6H2O + 6CO2), how many grams of oxygen gas will be consumed when 950 g of glucose is oxidized completely?
Responses
480 g
480 g
571 g
571 g
960 g
960 g
1,013g
All Answers 1
Answered by
GPT-5 mini
AI
Balanced equation: C6H12O6 + 6 O2 → 6 H2O + 6 CO2
Molar masses: glucose ≈ 180.16 g·mol⁻¹, O2 = 32.00 g·mol⁻¹.
Moles glucose = 950 g / 180.16 g·mol⁻¹ = 5.273 mol
Moles O2 required = 6 × 5.273 = 31.638 mol
Mass O2 = 31.638 mol × 32.00 g·mol⁻¹ = 1,012.4 g ≈ 1,013 g
Answer: 1,013 g O2.
Molar masses: glucose ≈ 180.16 g·mol⁻¹, O2 = 32.00 g·mol⁻¹.
Moles glucose = 950 g / 180.16 g·mol⁻¹ = 5.273 mol
Moles O2 required = 6 × 5.273 = 31.638 mol
Mass O2 = 31.638 mol × 32.00 g·mol⁻¹ = 1,012.4 g ≈ 1,013 g
Answer: 1,013 g O2.
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