Posted by Bradley on Wednesday, July 13, 2011 at 12:31am.
Here is a worked example of a simple stoichiometry problem. I work these limiting reagent problems (you know it is limiting reagent when BOTH reactants are given) by solving TWO of the simple ones.
http://www.jiskha.com/science/chemistry/stoichiometry.html
So convert L NO2 (moles = L/22.4) to moles and calculate moles HNO3 formed. Then convert L H2O to moles and calculate moles HNO3 formed. In limiting regent problems, the cxorrect answer (you have two answers) is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Then convert moles to L and you have it. BUT, there is an easier way.
When everything is in the gaseous form, you can work with L directly (you change L to moles in the first approach, then change moles back to liters---so we just shorten that and don't convert in the first place).
So 10.66 L NO2 x (2 moles HNO3/3 moles NO2) = 10.66 x (2/3) = 7.11 L HNO3.
Then 14.88 L H2O x (2 moles HNO3/1 mole H2O) = 14.88 x (2/1) = 29.7 L HNO3.
So the correct answer is you can form a maximum of 7.11 L HNO3.
Thank you Dr. Bob!
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