Limiting reagents.

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Presentation transcript:

Limiting reagents

Happy Mole Day

A + 3 B  5 C In this reaction, for every mole of A you need 3 moles of B to completely react. If you only have 2 moles of B, then not all of the A can completely react to form the maximum amt of product C. B is a limiting reagent. It limits the amount of product that can be produced.

A + 3 B  5 C The fastest and easiest way to determine a limiting reagent is to calculate the amt. Of product that each reactant can produce if they are allowed to completely react. 1.0 mole A x 5 C/1A = 5.0 moles C 2.0 moles B x 5C/3B = 3.3 moles C B limits the amt of product that can form

2N2 +5 O2  2 N205 If 6.0 moles of Nitrogen are place in a container with 16 moles of oxygen, how much product can be formed? Calculate how much product each reactant can create if completely consumed. 6.0 moles N2 x 2N2O5/2N2 = 6.0 moles prod. 16 moles O2 x 2 N2O5/5O2 =