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
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
(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