Help in calculating glucose in chem
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Help in calculating glucose in chem

[From: ] [author: ] [Date: 11-09-22] [Hit: ]
(3.26e-2 moles glucose) * (180 grams glucose/1 mole of glucose) = 5.......
Top ranking cyclists can metabolise 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 °C and 1 atm pressure).
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O



I am so confused

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1kg uses 8e-2L/min therefore

(8e-2L/min)(60min/hr) = 4.8L/hr

so find the actual # moles of oxygen used then mole to mole ratio against the theoretical values of glucose then convert moles of glucose to grams of glucose

PV=nRT

n = PV/RT
n = (1)(4.8)/(.082)(298)
n = 1.96e-1 moles of O2

(1.96e-1 moles of O2 actual) * ((1 mole glucose)/(6 moles O2)) = 3.26e-2 moles glucose

(3.26e-2 moles glucose) * (180 grams glucose/1 mole of glucose) = 5.88 grams of glucose
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