Please TURN IN C.7/C.10 worksheetObjectives: Determine the limiting reagent in a reaction.Calculate the percent yield of a product in a reaction

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A. The Concept of Limiting Reactants • Stoichiometric mixture

– N2(g) + 3H2(g) 2NH3(g)

• Limiting reactant mixture

A. The Concept of Limiting Reactants

– N2(g) + 3H2(g) 2NH3(g)

• For a Limiting reactant mixture the number of moles are not balanced to match the reaction equation N2(g) + 3H2(g) 2NH3(g)

A. The Concept of Limiting Reactants

– Limiting reactant is the reactant that runs out first

– When the limiting reactant is exhausted, then the reaction stops

LIMITING REACTANTa) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed?

4 Na (s) + O2(g) 2 Na2O (s)

Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant):

5.00g Na (1 mole Na) ( 2 mole Na2O)( 62 g Na

2O) = 6.74 g of Na2O

23 g Na 4 mole Na 1 mol Na2O

5.00g O2 (1 mole O2) ( 2 mole Na

2O)( 62 g Na

2O) = 19.38 g of Na2O

32 g O2

1 mole O2

1 mol Na2O

Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide.

Wrong answer

LIMITING REACTANTb) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)?

4 Na (s) + O2(g) 2 Na2O (s)

The amount of O2 used to make 6.74 g of Na2O is calculated by:

5.00g Na (1 mole Na) ( 1 mole O2)( 32 g O

2) = 1.74 g of O2 was used 23 g Na 4 mole Na 1 mol O

2

The amount of oxygen (O2) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g

The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

PRACTICE PROBLEMS2 C2H2 + 5 O2 4 CO2 + 2 H2O

1. How many moles of carbon dioxide could be produced from 220.0 g of C2H2 and 545.0 g of O2?

2. How many grams of CO2 can be produced by the reaction of 35.5 grams of C2H2 and 45.9 grams of O2? 50.5 g

13.63 mol

C. Percent Yield • Theoretical Yield

– The maximum amount of a given product that can be formed when the limiting reactant is completely consumed.

• The actual yield (amount produced) of a reaction is usually less than the maximum expected (theoretical yield).

• Percent Yield – The actual amount of a given product as the percentage of

the theoretical yield.

Calcium carbonate is synthesized by heating,as shown in the following equation: CaO + CO2 CaCO3

• What is the theoretical yield of CaCO3 if 24.8 g of CaO is heated with 43.0 g of CO2?

• What is the percent yield if 33.1 g of CaCO3 is produced?

Determine which reactant is the limiting and then decide

what the theoretical yield is.

24.8 g

CaO

1molCaO56g CaO

1mol CO2

1mol CaO44 g CO2

1molCO2

= 19.5gCO2

24.8gCaOmolCaOmol CO2gCO2

24.8 gCaO

1mol CaO56g CaO

1molCaCO31mol

CaO

100g CaCO3

1molCaCO3

= 44.3 g CaCO3

24.8gCaOmolCaOmol CaCO3gCaCO3

LR

• CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO3. We only produced 33.1 g.

• Our percent yield is:Percent yield= 33.1 g CaCO3

44.3 g CaCO3

_____________ x 100

Percent yield = 74.7%

Percent Yield Questions1. The electrolysis of water forms H2 and O2.

2H2O 2H2 + O2

What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O?

2. Iron pyrites (FeS2) reacts with oxygen according to the following equation:

4FeS2 + 11O2 2Fe2O3 + 8SO2

If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide?

1. The electrolysis of water forms H2 & O2. 2H2O 2H2 + O2

Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O?

• Actual yield is given: 12.3 g O2

• Next, calculate theoretical yield1 mol O2

2 mol H2Ox

# g O2=14.0 g H2O 12.43 g= 32 g O2

1 mol O2

x 1 mol H2O18.02 g H2Ox

Finally, calculate % yield12.3 g O2 12.43 g O2

= % yield = x 100% 98.9%= actualtheoretical x 100%

2. 4FeS2 + 11O2 2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g

Fe2O3 results. % yield Fe2O3?First, determine limiting reagent

2 mol Fe2O3

11 mol O2 x 200 g O2

181.48 g Fe2O3=

159.7 g Fe2O3

1 mol Fe2O3

x 1 mol O2 32 g O2

x

2 mol Fe2O3

4 mol FeS2 x

# g Fe2O3=300 g FeS2

199.7 g Fe2O3=

159.7 g Fe2O3

1 mol Fe2O3

x 1 mol FeS2 119.97 g FeS2

x

143 g Fe2O3 181.48 g Fe2O3

= % yield = x 100% 78.8%= actualtheoretical x 100%