The combustion of glucose, a six-carbon sugar, is represented by the following balanced reaction:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
If a reaction vessel initially contains 360 grams of C6H12O6 (MW = 180 g/mole), and if the reaction produces 440 grams of CO2, what is the limiting reagent?
A. C6H12O6
B. O2
C. CO2
D. H2O
Click for Explanation
This question is asking the examinee to determine the limiting reagent of a chemical reaction. To solve this problem, you must calculate the theoretical amount of CO2 that is produced given 360 grams of C6H12O6. If all 360 grams (2 moles) of glucose initially present react, then 528 g (12 moles) of CO2 should theoretically be produced.
360 g C6H12O6 x (1 mol C6H12O6 / 180 g C6H12O6) x (6 mol CO2 / 1 mol C6H12O6) x (44 g CO2 / 1 mol CO2) = 528 g CO2
Since only 440 grams of carbon dioxide are actually produced, not all of the glucose combusts. Thus, the oxygen must be the limiting reagent in this problem, making answer choice B correct. Choices C and D can be eliminated because products are never limiting reagents. Since there will be leftover C6H12O6, it is not the limiting reagent, eliminating answer choice A.
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