Limiting Reagents (Reactants) How do I recognize a problem as being a limiting reagent problem? You will be given a known amount of each reactant. Often, you’ll also be asked about the limiting reagent or the reagent in excess.

Let’s take a look at one. When 65 g of silver nitrate react with 50 g of calcium phosphate, what mass of silver phosphate can be formed? HINT: an alternative phrasing could be “what is the theoretical yield of silver phosphate?”

Now how did we know this was limiting? The phrasing gave no indication of the terms limiting reagent nor reagent in excess. The key was that you were given known quantities of each reactant.

Where do we begin? Since you have more than one substance, a balanced chemical equation is in order. To get there, we begin with correct formulas. Review your formula writing resources should you need help on this.

AgNO 3 + Ca 3 (PO 4 ) 3 ---> ?? What sort of reaction does this seem to be? Yes, double replacement is correct. Accordingly we have: AgNO 3 + Ca 3 (PO 4 ) 2 ---> Ca(NO 3 ) 2 + Ag 3 PO 4 Next, we need to balance this reaction in order to meet the Law of Conservation.

6AgNO 3 + Ca 3 (PO 4 ) 2 ---> 3Ca(NO 3 ) 2 + 2Ag 3 PO 4 Now that we have the equation balanced, where do we go next? For limiting reagent problems, perform 2 parallel calculations in order to compare. Convert each amount into an amount of product. Then compare how much product you can make from each beginning quantity. The lesser amount is your THEORETICAL YIELD.

65 g AgNO 3 to grams of Ag 3 PO 4 50 g Ca 3 (PO 4 ) 2 to grams of Ag 3 PO 4 Since we’re given grams, the only first step is to convert to moles of that substance. Once we have moles of the original, let’s see where to go. Shouldn’t we head toward our desired unit?

65 g AgNO 3 (1 mol AgNO 3 )( 2 mol Ag 3 PO 4 )( g Ag 3 PO 4 ) = (169.9g AgNO 3 ) ( 6 mol AgNO 3 ) ( 1 mol Ag 3 PO 4 ) = 53.4 g Ag 3 PO 4 theoretical IF every bit of the AgNO 3 reacts AND 50 g Ca 3 (PO 4 ) 2 (1 mol Ca 3 (PO 4 ) 2 )(2 mol Ag 3 PO 4 )(418.7g Ag 3 PO 4 ) = (310.3g Ca 3 (PO 4 ) 2 )(1mol Ca 3 (PO 4 ) 2 )(1mol Ag 3 PO 4 ) = g Ag 3 PO 4 theoretical IF every bit of the Ca 3 (PO 4 ) 2 reacts Since 53.4 g < g, this means that the maximum amount of Ag 3 PO 4 that can be formed is 53.4 g.

It also tells us that since AgNO 3 was the substance used to produce 53.4 g of Ag 3 PO 4, it must have run out first or was the limiting reagent. Since Ca 3 (PO 4 ) 2 had the ability to make more Ag 3 PO 4, it must have been in excess.

Great! Our theoretical yield of Ag 3 PO 4 was 53.4g. Let’s go on to part 2 of the question. It says: If 48.6 g of Ag 3 PO 4 were actually retrieved in lab, what was the percentage yield of Ag 3 PO 4 ? To find percentage yield, what must we do?

% yield = actual yield (100) theoretical yield Recall that actual yield is what you get in lab. Theoretical yield is what the calculator says you could have gotten, provided your lab was perfect.

% yield = 48.6 g ( 100) = 91% yield 53.4 g These folks did good work in lab, didn’t they? Cool! Now on to the next concept.

What mass of excess reagent remains? In order to determine this, we must know which reagent is in excess AND how much of it actually did react. Which is the reactant in excess? In your experiment, it was Ca 3 (PO 4 ) 2.

To find out how much of the excess chemical did react, start with something you KNOW reacted fully - the limiting reagent. 65 g AgNO 3 (1 mol AgNO 3 )(1 mol Ca 3 (PO 4 ) 2 )( g Ca 3 (PO 4 ) 2 )= ( g AgNO 3 ) (6 mol AgNO 3 ) ( 1 mol Ca 3 (PO 4 ) 2 ) = 19.8 g of Ca 3 (PO 4 ) 2 that were used in this experiment

Now that we know how much was needed, we can figure out how much was left over. 50 g Ca 3 (PO 4 ) 2 available g Ca 3 (PO 4 ) 2 needed for reaction 30.2 g Ca 3 (PO 4 ) 2 are left over or in excess

To ensure your success, go back through this presentation until it makes perfect sense to you. Then, try some of the problems in your packet to convince yourself that you know how to solve them. Practice makes PERFECT!