Professional cyclists can metabolize more than 80 mL of oxygen per kg of body weight per minute (sometimes called the V-O₂ max) . Calculate the mass of glucose (in grams) that could theoretically be consumed by the following reaction by a 70 kg cyclist in an hour (assume 25^oC and 1 atm pressure) .
C₆H₁₂O₆ + 6 O₂ $\rarr$ 6 CO₂ + 6 H₂O

Question

Professional cyclists can metabolize more than 80 mL of oxygen per kg of body weight per minute (sometimes called the V-O₂ max) . Calculate the mass of glucose (in grams) that could theoretically be consumed by the following reaction by a 70 kg cyclist in an hour (assume 25^oC and 1 atm pressure) .
C₆H₁₂O₆ + 6 O₂

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6 CO₂ + 6 H₂O

Quizplus · Accepted Answer

The correct answer is C.The balanced chemical equation for the reaction is:C6H12O6 + 6 O2 → 6 CO2 + 6 H2OFrom the equation, we can see that 1 mole of glucose (180 g) reacts with 6 moles of oxygen (192 g).Therefore, the mass of glucose that could theoretically be consumed by the cyclist in an hour is:(80 mL O2/kg body weight/min) * (60 min/hr) * (70 kg body weight) * (192 g O2/6 moles O2) * (1 mole glucose/180 g glucose) = 410 g glucose

Professional Cyclists Can Metabolize More Than 80 ML of Oxygen →\rarr→

Professional Cyclists Can Metabolize More Than 80 ML of Oxygen $\rarr$