Limiting Reactants Limiting Reactants & Percent Yield& Percent YieldHonors ChemistryHonors Chemistry

Section 11.3Section 11.3

Limiting ReactantsLimiting ReactantsRarely 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.

If we have…If we have… how many sets can we how many sets can we make?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 overWith this left over

Excess

Limiting Factor Limiting Factor NoteSheetNoteSheet

Consider the following recipe:Consider the following recipe:

3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder 3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder 12 muffins 12 muffins

A quick check of the pantry shows that the following quantities are A quick check of the pantry shows that the following quantities are available. How many muffins could be made with respect to each available. How many muffins could be made with respect to each ingredient?ingredient?

9 cups of flour could make 9 cups of flour could make ___36______36___ muffins muffins

4 eggs could make 4 eggs could make __24____24__ muffins muffins

8 cups of sugar could make 8 cups of sugar could make __96_____96___ muffins muffins

10 tsp baking powder could make 10 tsp baking powder could make __60_____60___ muffins muffins

Limiting Factor Limiting Factor NoteSheetNoteSheet

3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder 3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder 12 muffins 12 muffins

What is the maximum number of muffins that could be made considering What is the maximum number of muffins that could be made considering ALL ingredients? ALL ingredients? __24_____24___

Which ingredient would you run out of first? Which ingredient would you run out of first? _eggs_____eggs____

This ingredient would be referred to as the This ingredient would be referred to as the __limiting__limiting_ _ __reactant____.__reactant____.

The other ingredients are said to be The other ingredients are said to be __in __in __ __ excess excess ..

When you run out of an ingredient, you stop producing the When you run out of an ingredient, you stop producing the _product__._product__.

Limiting Factor Limiting Factor NotesheetNotesheet

Consider the following “recipe” (equation):Consider the following “recipe” (equation):

2 mol of Al + 3 mol of Cl2 mol of Al + 3 mol of Cl22 2 mol of AlCl 2 mol of AlCl33

A quick check of the chemistry lab shows that the A quick check of the chemistry lab shows that the following quantities are available. How much following quantities are available. How much product could be made from each? (How many product could be made from each? (How many times could you make the recipe?)times could you make the recipe?)

10 moles Al ______10 moles Al ______

12 moles Cl12 moles Cl22 _____ _____

What is the maximum amount of AlClWhat is the maximum amount of AlCl33 that could be made that could be made considering all the ingredients? ___8 moles_____considering all the ingredients? ___8 moles_____

We would say the aluminum is ___in We would say the aluminum is ___in excess______________ and the chlorine is excess______________ and the chlorine is _______limiting__________._______limiting__________.

What if you were presented with grams of each reactant? What if you were presented with grams of each reactant? Let’s say there are 100 g of Al and 125 g of ClLet’s say there are 100 g of Al and 125 g of Cl22 available. available. How much product could be formed? (Remember, the How much product could be formed? (Remember, the coefficients in the equation are mole numbers)coefficients in the equation are mole numbers)

100 g of Al would yield ________ g of product.100 g of Al would yield ________ g of product.125 g of Cl125 g of Cl22 would yield ________ g of product. would yield ________ g of product. How much product could be made considering both How much product could be made considering both

reactants? ________reactants? ________

The limiting factor would be ________.The limiting factor would be ________.

Steps to solve limiting Steps to solve limiting reactants problemsreactants problems

1.1. Split into 2 problemsSplit into 2 problems

2.2. Solve each problemSolve each problem

3.3. Pick smallest answerPick smallest answer- Smallest answer comes from limiting reactantSmallest answer comes from limiting reactant- Largest answer comes from excess reactantLargest answer comes from excess reactant

Calculating the Product Calculating the Product when the Reactant is when the Reactant is LimitedLimitedIn the following equation:In the following equation:

If 200 g of sulfur reacts with 100 g of chlorine what mass of sulfur If 200 g of sulfur reacts with 100 g of chlorine what mass of sulfur chloride will be produced?chloride will be produced?

SS88 + 4 Cl + 4 Cl22 4 S 4 S22ClCl22

SulfurSulfur

200 g S x 200 g S x 1 mole S 1 mole S x x 4 mol S4 mol S22ClCl22 x x 134 g S134 g S22ClCl22 = 3350 g = 3350 g

32 g S 1 mole S 1 mole S 32 g S 1 mole S 1 mole S22ClCl22

ChlorineChlorine

100 g Cl100 g Cl22 x x 1 mole Cl1 mole Cl22 x x 4 mol S4 mol S22ClCl22 x x 134 g S134 g S22ClCl22 = 191.42 g = 191.42 g lowest # lowest #

70 g Cl70 g Cl22 4 mol Cl 4 mol Cl22 1 mole S 1 mole S22ClCl22

Steps for Calculating Limiting Steps for Calculating Limiting ReactantsReactants

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

Calculating product when reactant is Calculating product when reactant is limitinglimiting

If we had 12 moles of nitrogen and 18 moles of hydrogen, what is the maximum number of moles of NH3 that could be produced?

NN22 + H + H22 NH NH33

Calculating product when reactant is Calculating product when reactant is limitinglimiting

If we had 112 grams of nitrogen and 18 grams of hydrogen, what is the maximum number of grams of NH3 that could be produced?

NN22 + 3H + 3H22 2 NH 2 NH33

2(14) 2(1) 14 + 2(14) 2(1) 14 + 33

2828 2 17 2 17

Nitrogen

33

3

2

3

2

22136

1

17

1

2

28

1

1

112gNH

moleNH

gNHx

moleN

moleNHx

gN

moleNx

gN

Hydrogen

33

3

2

3

2

22102

1

17

3

2

2

1

1

18gNH

moleNH

gNHx

moleH

moleNHx

gH

moleHx

gH

Pick lowest number

Therefore hydrogen is your limiting reactant

Calculating product when reactant is Calculating product when reactant is limitinglimiting

With 48 grams of magnesium and 48 grams of oxygen available, how much product can be formed?

2Mg + O2Mg + O22 2 MgO 2 MgO 24 2(16) 24 + 24 2(16) 24 +

1616

32 32 4040 Magnesium

gMgOmoleMgO

gMgOx

moleMg

moleMgOx

gMg

moleMgx

gMg80

1

40

2

2

24

1

1

48

Oxygen

gMgOmoleMgO

gMgOx

moleO

moleMgOx

gO

moleOx

gO120

1

40

1

2

32

1

1

48

22

22

Pick lowest number

Therefore magnesium is your limiting reactant and oxygen is the excess reactant

Limiting Reactants Practice 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 + Cl2 Na + Cl22 2 NaCl 2 NaCl 23 2(35) 23 + 23 2(35) 23 +

3535

70 70 5858

Sodium

gNaClmoleNaCl

gNaClx

moleNa

moleNaClx

gNa

moleNax

gNa26.378

1

58

2

2

23

1

1

150

Chlorine

gNaClmoleNaCl

gNaClx

moleCl

moleNaClx

gCl

moleClx

gCl42.331

1

58

1

2

70

1

1

200

22

22

Pick lowest number

Limiting Reactants PracticeLimiting 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 + Cl2 Na + Cl22 2 NaCl 2 NaCl 23 2(35) 23 2(35)

70 70

Sodium

neededgClmoleCl

gClx

moleNa

moleClx

gNa

moleNax

gNa2

2

2226.228

1

70

2

1

23

1

1

150

Chlorine needed = 228.26 g

Chlorine given (from problem) = 200 g

Additional Chlorine needed = 28.26 g

Percent Yield Percent Yield Actual yield = the actual amount of product Actual yield = the actual amount of product

formed during an formed during an experimentexperiment Theoretical yield = amount of product that Theoretical yield = amount of product that

could be produced according to could be produced according to calculationscalculations Percent yield = ratio of actual and theoretical Percent yield = ratio of actual and theoretical

yieldyield

100yieldltheoretica

yieldactualyieldPercent

Percent Yield Practice Percent Yield Practice 1.1. A student uses stoichiometry to A student uses stoichiometry to calculatecalculate the the

predicted yield of COpredicted yield of CO22 in a chemical reaction as in a chemical reaction as 323323

g. When the student g. When the student mixes the chemicalsmixes the chemicals to make to make the COthe CO22, he finds that the reaction only produces , he finds that the reaction only produces

308 g308 g of CO of CO22. What is the percent yield?. What is the percent yield?

100yieldltheoretica

yieldactualyieldPercent

100323

308yieldPercent

%4.95yieldPercent

2.2. In the reaction below, how many grams of In the reaction below, how many grams of potassium oxide can be produced with 78 grams potassium oxide can be produced with 78 grams of potassium and 140 grams of boron oxide?of potassium and 140 grams of boron oxide?

6 K + B6 K + B22OO33 3 K 3 K22O + 2 BO + 2 B 39 2(11)+3(16) 2(39)+16 11 70 94

Potassium

OgKOmoleK

OgKx

moleK

OmoleKx

gK

moleKx

gK2

2

2294

1

94

6

3

39

1

1

78

OgKOmoleK

OgKx

OmoleB

OmoleKx

OgB

OmoleBx

OgB2

2

2

32

2

32

3232564

1

94

1

3

70

1

1

140

Pick lower number L.R. = theoretical yield

Boron Oxide

Problem 2 continuedProblem 2 continued If the actual yield from the above reaction was 81 If the actual yield from the above reaction was 81

grams, what is the percent yield?grams, what is the percent yield? Actual yield = 81 gramsActual yield = 81 grams Theoretical yield = 94 g (from last slide)Theoretical yield = 94 g (from last slide)

100yieldltheoretica

yieldactualyieldPercent

10094

81

g

gyieldPercent

%17.86yieldPercent

Round your final answer to the nearest hundredths!

3. The actual yield in the above reaction was 39 grams of 3. The actual yield in the above reaction was 39 grams of ammonium. Using 84 grams of nitrogen and 8 grams of ammonium. Using 84 grams of nitrogen and 8 grams of hydrogen, what is the percent yield?hydrogen, what is the percent yield?

-first we must calculate the theoretical yield of -first we must calculate the theoretical yield of product product which means we need to find our limiting which means we need to find our limiting reactantreactant

NN22 + 3 H + 3 H22 2 NH 2 NH332(14) 2(1) 14+3

28 2 17Nitrogen

Hydrogen

33

3

2

3

2

22102

1

17

1

2

28

1

1

84gNH

moleNH

gNHx

moleN

moleNHx

gN

moleNx

gN

33

3

2

3

2

2233.45

1

17

3

2

2

1

1

8gNH

moleNH

gNHx

moleH

moleNHx

gH

moleHx

gH

Pick lower number

L.R. = theoretical yield

Problem 3 continuedProblem 3 continued Actual yield = 39 gramsActual yield = 39 grams Theoretical yield = 45.33 g (from last slide)Theoretical yield = 45.33 g (from last slide)

100yieldltheoretica

yieldactualyieldPercent

10033.45

39yieldPercent

%04.86yieldPercent