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How many moles of glucose, C6H12O6, can be burned when 60.0 mol of oxygen is available? C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Answer :

60mol O2 × 1 mol C6H12O6 / 6 moles 02 = 10 moles of Glucose

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