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Limiting Reactants & Percent Yield Honors Chemistry Section 11.3.

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Presentation on theme: "Limiting Reactants & Percent Yield Honors Chemistry Section 11.3."— Presentation transcript:

1 Limiting Reactants & Percent Yield Honors Chemistry Section 11.3

2 Limiting Reactants Rarely are the reactants in a chemical equation present in the exact ratios specified by the balanced equation. Generally, there is too much of one and not enough of the other. The chemical reaction will stop once one of the reactants is used up. This leaves the remainder of the reactant that was in excess unreacted. The chemical that you ran out of was the limiting reactant.

3 If we have…how many sets can we make? Limiting reactant = the reactant that runs out first causing the reaction to stop Excess reactant = reactant that does not get completely used up during a reaction With this left over Excess

4 Limiting Factor NoteSheet Consider the following recipe: 3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder  12 muffins A quick check of the pantry shows that the following quantities are available. How many muffins could be made with respect to each ingredient? 9 cups of flour could make ___36___ muffins 4 eggs could make __24__ muffins 8 cups of sugar could make __96___ muffins 10 tsp baking powder could make __60___ muffins

5 Limiting Factor NoteSheet 3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder  12 muffins What is the maximum number of muffins that could be made considering ALL ingredients? __24___ Which ingredient would you run out of first? _eggs____ Which ingredient would you run out of first? _eggs____ This ingredient would be referred to as the __limiting_ __reactant____. This ingredient would be referred to as the __limiting_ __reactant____. The other ingredients are said to be __in __ excess. The other ingredients are said to be __in __ excess. When you run out of an ingredient, you stop producing the _product__. When you run out of an ingredient, you stop producing the _product__.

6 Limiting Factor Notesheet Consider the following “recipe” (equation): 2 mol of Al + 3 mol of Cl 2  2 mol of AlCl 3 A quick check of the chemistry lab shows that the following quantities are available. How much product could be made from each? (How many times could you make the recipe?) A quick check of the chemistry lab shows that the following quantities are available. How much product could be made from each? (How many times could you make the recipe?) 10 moles Al ______ 12 moles Cl 2 _____

7 What is the maximum amount of AlCl 3 that could be made considering all the ingredients? ___8 moles_____ We would say the aluminum is ___in excess______________ and the chlorine is _______limiting__________. We would say the aluminum is ___in excess______________ and the chlorine is _______limiting__________. What if you were presented with grams of each reactant? Let’s say there are 100 g of Al and 125 g of Cl 2 available. How much product could be formed? (Remember, the coefficients in the equation are mole numbers) What if you were presented with grams of each reactant? Let’s say there are 100 g of Al and 125 g of Cl 2 available. How much product could be formed? (Remember, the coefficients in the equation are mole numbers) 100 g of Al would yield ________ g of product. 125 g of Cl 2 would yield ________ g of product. How much product could be made considering both reactants? ________ How much product could be made considering both reactants? ________ The limiting factor would be ________.

8 Steps to solve limiting reactants problems 1.Split into 2 problems 2.Solve each problem 3.Pick smallest answer -Smallest answer comes from limiting reactant -Largest answer comes from excess reactant

9 Calculating the Product when the Reactant is Limited In the following equation: If 200 g of sulfur reacts with 100 g of chlorine what mass of sulfur chloride will be produced? S Cl 2  4 S 2 Cl 2 Sulfur 200 g S x 1 mole S x 4 mol S 2 Cl 2 x 134 g S 2 Cl 2 = 3350 g 32 g S 1 mole S 1 mole S 2 Cl 2 Chlorine 100 g Cl 2 x 1 mole Cl 2 x 4 mol S 2 Cl 2 x 134 g S 2 Cl 2 = g  lowest # 70 g Cl 2 4 mol Cl 2 1 mole S 2 Cl 2 70 g Cl 2 4 mol Cl 2 1 mole S 2 Cl 2

10 Steps for Calculating Limiting Reactants 1. Balance the equation 2. Label the known substances and the unknown substance for each reactant (note: you will have 2 problems) 3. Convert from mass to moles for each reactant 4. Use a mole ratio for each of the known substances for each reactant 5. Use the molar mass to convert from moles to mass for each reactant 6. Look for the lowest number – this is your limiting reactant

11 Calculating product when reactant is limiting   If we had 12 moles of nitrogen and 18 moles of hydrogen, what is the maximum number of moles of NH 3 that could be produced? N 2 + H 2  NH 3

12 Calculating product when reactant is limiting   If we had 112 grams of nitrogen and 18 grams of hydrogen, what is the maximum number of grams of NH 3 that could be produced? N 2 + 3H 2  2 NH 3 2(14) 2(1) (14) 2(1) Nitrogen Hydrogen Pick lowest number Therefore hydrogen is your limiting reactant

13 Calculating product when reactant is limiting   With 48 grams of magnesium and 48 grams of oxygen available, how much product can be formed? 2Mg + O 2  2 MgO 24 2(16) (16) Magnesium Oxygen Pick lowest number Therefore magnesium is your limiting reactant and oxygen is the excess reactant

14 Limiting Reactants Practice   How many grams of sodium chloride can be produced in the following reaction with 150 grams of sodium and 200 grams of chlorine? 2 Na + Cl 2  2 NaCl 23 2(35) (35) Sodium Chlorine Pick lowest number

15 Limiting Reactants Practice   In the above reaction, how many more grams of chlorine will be required to fully react with the remaining sodium?   We need to figure out how many grams of chlorine is needed to react with 150 grams of sodium 2 Na + Cl 2  2 NaCl 23 2(35) (35) 70 Sodium Chlorine needed = g Chlorine given (from problem) = 200 g Additional Chlorine needed = g

16 Percent Yield  Actual yield = the actual amount of product formed during an experiment  Theoretical yield = amount of product that could be produced according to calculations  Percent yield = ratio of actual and theoretical yield

17 Percent Yield Practice 1.A student uses stoichiometry to calculate the predicted yield of CO 2 in a chemical reaction as 323 g. When the student mixes the chemicals to make the CO 2, he finds that the reaction only produces 308 g of CO 2. What is the percent yield?

18 2.In the reaction below, how many grams of potassium oxide can be produced with 78 grams of potassium and 140 grams of boron oxide? 6 K + B 2 O 3  3 K 2 O + 2 B 39 2(11)+3(16) 2(39) Potassium Pick lower number L.R. = theoretical yield Boron Oxide

19  Problem 2 continued  If the actual yield from the above reaction was 81 grams, what is the percent yield?  Actual yield = 81 grams  Theoretical yield = 94 g (from last slide) Round your final answer to the nearest hundredths!

20 3. The actual yield in the above reaction was 39 grams of ammonium. Using 84 grams of nitrogen and 8 grams of hydrogen, what is the percent yield? -first we must calculate the theoretical yield of product which means we need to find our limiting reactant N H 2  2 NH 3 2(14) 2(1) Nitrogen Hydrogen Pick lower number L.R. = theoretical yield

21  Problem 3 continued  Actual yield = 39 grams  Theoretical yield = g (from last slide)


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