The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6 (s) + O2 (g) → CO2 (g) + H2O (l) What is the conversion factor that allows for the calculation of moles of carbon dioxide based on moles of glucose?

the conversion factor is f = 6 mol of glucose / mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6 (s) + O2 (g) → CO2 (g) + H2O (l) (unbalanced)

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose / mol of CO2

f = 6 mol of CO2 / mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose = 12 moles of CO2