Question

Question #c23aa

Answer 1

Every time you have to calculate how much of a reactant is needed in order for a reaction to take place, it is assumed that you will have no limiting reagent.

For instance, you know the balanced chemical equation

`C_12H_22O_11 + 12O_2 -> 11H_2O + 12CO_2`

You started with `"2.48 g"` of sucrose, which translated into `"0.00725 moles"`. The amount of oxygen you calculated was the minimum quantity that would have allowed the reaction to have no limiting reagent.

If you'd use more oxygen than you've calculated would mean that sucrose will become a limiting reagent `->` excess oxygen; likewise, less oxygen would make oxygen the limiting reagent, since now you'd have excess sucrose.

So, to answer your question, in that particular problem neither the gummy bear nor the oxygen were limiting reagents because the amount of oxygen that you calculated matched the amount of sucrose you had perfectly.

You could say that the quantity of oxygen used was tailor-made for `"2.48 g"` of sucrose - at STP, don't forget about that...

The original problem:

A gummy bear weighs 2.48 grams. What volume of oxygen is needed to completely combust the gummy bear in the equation C12H22O11 + 12O2 = 11H2O + 12CO2 ?

http://socratic.org/questions/a-gummy-bear-weighs-2-48-grams-what-volume-of-oxygen-is-neededto-completely-comb#118286