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 7.1 The Mole 7.2 Percent Composition of Compounds 7.3 More Complex Stoichiometric Calculations 7.4 Limiting Reagents

 7.5 Chemical Yields 7.6 Practice Questions 7.7 Explanations
Limiting Reagents
Have you ever noticed that hot dogs are sold in packages of 10, while the buns come in packages of eight? In this scenario, the buns are the limiting reactant in the sense that they limit the hot dog preparation to eight. The limiting reactant or reagent is the one that is consumed first in the chemical reaction, and its consumption halts the progress of the forward reaction.
When answering questions about limiting reagents on the exam, your first step should always be to convert all the masses you were given into moles. You should set up your table as you did before, only now you’ll have two amounts and thus two numbers of moles to get you started.
Let’s look at a specific question, involving the Haber process. Basically, this is the process of making ammonia from the reaction of nitrogen and hydrogen gases. The reaction is shown below:
Molar mass Balanced equation No. of moles (28.02) (2.02) (17.04) N2 + 3H2 2NH3
Suppose you have a total of 25.0 kg of nitrogen to react with a total of 5.00 kg of hydrogen. What mass of ammonia can be produced? Which reactant is the limiting reactant? What is the mass of the reactant that’s in excess? Insert the masses in the correct rows and find the number of moles of both.
Molar mass Balanced equation No. of moles (28.02) (2.02) (17.04) N2 + 3H2 2NH3 892 mol 2475 mol 25,000 g 5000 g
Start with nitrogen. You have 892 moles of it available, and in order for the nitrogen to react completely with hydrogen, you’d need 3(892 mol) = 2676 moles of hydrogen, which you don’t have. Therefore, hydrogen is the limiting reagent. Now let’s answer the other parts of the question. The mass of ammonia that can be produced is limited by the amount of hydrogen, so do your calculations based on the number of moles of hydrogen available. Your chart should look like the one below:
Molar mass Balanced equation No. of moles (14.02) (2.02) (17.04) N2 + 3H2 2NH3 825 mol used 2475 mol used therefore 1650 mol produced 892 mol 23,117 g used 25,000g 5000 g 1650 mol (17.04) = 28,116 g produced
 Jump to a New ChapterIntroduction to the SAT IIIntroduction to the SAT II Chemistry TestStrategies for Taking the SAT II Chemistry TestThe Structure of MatterThe States of MatterReaction TypesStoichiometryEquilibrium and Reaction RatesThermodynamicsDescriptive ChemistryLaboratoryBasic Measurement and Calculation ReviewChemical Formulas Review: Nomenclature and Formula WritingPractice Tests Are Your Best Friends
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