Limiting Reagents Stoichiometry Luckett. What is a limiting reagent? The reagent (reactant) that determines the amount of product that can be formed by.

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Limiting Reactants and Percent Yield

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Presentation transcript:

Limiting Reagents Stoichiometry Luckett

What is a limiting reagent? The reagent (reactant) that determines the amount of product that can be formed by a reaction.  The reaction will continue ONLY until you run out of this reactant. The reactant that is not used up all the way is called the excess reagent.

How do you find the limiting reagent? First you need your units for each reactant to be in moles! Convert if in grams or liters. Do a mole to mole conversion from one reactant to another (you choose!)  This allows you to see how you need based on how much you have (given in the question) Determine the limiting reagent from your answer. Determine the excess reagent and the amount of excess.

Lets try it! Which is the limiting reagent when 80.0 g of Copper reacts with 25.0 g of S in the following equation: Copper reacts with sulfur to form copper (I) sulfide.

Lets Try it! First convert grams to moles for each reactant:  80.0 g Cu | 1 mole Cu = 1.26 mol Cu | g Cu  25.0 g S | 1 mole S = mol S | g S Second compare one reactant to another:  1.26 mol Cu | 1 mole S =.630 mol S | 2 mol Cu  That means we need.630 mol of Sulfur to run this reaction…. How many do we have?  Since we have more than we need Sulfur is the excess reagent, therefore Copper is the limiting reagent.

Now you try! Identify the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg in the following equation: __Mg (s) + __HCl (aq)  __MgCl 2 (aq) + __H 2 (g) If 2.70 mol C 2 H 4 is reacted with 6.30 mol O 2, identify the limiting reagent using the following equation: __C 2 H 4 (g) + __O 2 (g)  __CO 2 + __H 2 O

How do you find out how much product can be produced? Determine the limiting reagent. Convert to find the number of grams of the selected product.

How to find the amount of excess Convert from grams of the limiting reagent to grams of the excess reagent. Subtract how much you need from how much you have. This is your excess in grams.