# Advanced Stoichiometry - PowerPoint PPT Presentation

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Advanced Stoichiometry. Chapter 9, Section 2 (pages 312 – 318) Problem Set E (p. 314 # 1 – 3) Problem Set F (p. 317 # 1 – 3). Terms to Know and Understand:. Limiting Reactant – the substance that controls the quantity of product that can form in a chemical reaction.

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#### Presentation Transcript

Chapter 9, Section 2

(pages 312 – 318)

Problem Set E (p. 314 # 1 – 3)

Problem Set F (p. 317 # 1 – 3)

### Terms to Know and Understand:

• Limiting Reactant – the substance that controls the quantity of product that can form in a chemical reaction.

• (May not be the reactant having the lowest mass!)

• Identified through stoichiometry.

• Excess Reactant – the substance that is not used up completely in a reaction.

• Theoretical Yield – the maximum quantity of product that a reaction could theoretically make if everything about the reaction works perfectly. The theoretical yield is found by stoichiometry.

• Actual Yield – the quantity of product actually produced in a reaction.

• The actual yield is found experimentally.

• Usually less than the theoretical yield.

• Percentage Yield – the ratio relating the actual yield of a reaction to its theoretical yield.

### Identifying a limiting reactant:

Example: A chemical reaction occurs between copper (II) oxide and hydrogen according to the following balanced equation:

CuO (s) + H2 (g)  Cu (s) + H2O (g)

• What is the limiting reactant when 19.9-g of CuO react with 2.02-g H2?

• What is the theoretical yield of copper?

• First read through other questions in the problem to see if you will be asked to find the theoretical yield of one of the products – if this is the case, do stoichiometry using this product as your ‘find’.

• If you are not asked to find the mass of one of the products, choose one as your ‘find’.

• Do the stoichiometrytwice using each reactant as your ‘given’. To do the remaining work I will use Cu (s) as my find’.

### Identifying a limiting reactant:

19.9 g CuO

0.250 mol CuO

• Using CuO as the given:

1 mol CuO

79.545 g CuO

Step 1

0.250 mol CuO

1 mol Cu

1 mol CuO

0.250 mol Cu

Step 2

0.250 mol Cu

63.546 g Cu

1 mol Cu

15.9 g Cu

Step 3

Mass of copper that would be produced if CuO was the limiting reactant.

Identifying a limiting reactant:

• Using H2 as the given:

2.02 g H2

1.00 mol H2

1 mol H2

2.0158 g H2

Step 1

1.00 mol H2

1 mol Cu

1 mol H2

1.00 mol Cu

Step 2

1.00 mol Cu

63.546 g Cu

1 mol Cu

63.5 g Cu

Step 3

Mass of copper that would be produced if H2 was the limiting reactant.

### Identifying a limiting reactant:

• Compare the calculated quantities of product:

• When 15.9-g of Cu are produced, the 19.9-g of CuO will be completely consumed (used up).

• If 63.5-g of Cu were produced, 2.02 g of H2 would be completely consumed, but there is not enough CuO to do this.

• The limiting reactant is CuO, because it will be gone after 15.9-g of Cu is produced, and the reaction will stop.

• The excess reactant is H2, because the amount available would allow the reaction to produce 63.5-g Cu, but the reaction stops when the limiting reactant is used up.

• The theoretical yield of Cu will be 15.9-g, because the reaction will stop as soon as this amount of product is formed.

### Identifying a limiting reactant:

• Note that CuO is the limiting reactant despite the fact that there was a lesser mass of H2 available to react – it is the stoichiometric quantitythat is important!

• If you are only asked to find the theoretical yield of a product you must still identify the limiting reactant, as it will determine the amount of product formed!

• You can recognize limiting reactant problems because you are given amounts of two of the reactants in the problem.

### Determining Percentage Yield

actual yield

theoretical yield

x 100

percentage yield =

For example, say we conduct the reaction between CuO and H2 with the amounts given in the example problem on slide 3. While we expect to produce 15.9-g of Cu (the theoretical yield), we find that we are only able to collect 13.8-g of Cu following the reaction. We can calculate the percentage yield as follows:

percentage yield =

13.8-g

15.9-g

x 100

percentage yield = 86.8%

• What mass of silver sulfide, Ag2S, can be made from 123g of H2S obtained from a rotten egg?

4Ag + 2H2S +O2 2Ag2S + 2H2O

• What mass, in grams, of water is produced when 80 liters of hydrogen gas react with oxygen?

2H2 + O2 2H2O

• •       Practice Box E (p. 314 # 1 – 3)

• •       Chapter 9 Review, p. 331 # 31, 32, 33

### Determining Percentage Yield

• A typical problem will give you the actual yield, but require that you first identify the limiting reactant, and use stoichiometry to calculate the theoretical yield:

Calculate the percentage yield of H3PO4 if 126.2-g are recovered when 100.0-g of P4O10 react with 200.0-g H2O according to the following balanced equation:

P4O10 + 6 H2O  4 H3PO4

• Identify the information given to us by the problem:

• actual yield of H3PO4 is 126.2-g

• available amounts of reactants:

• 100.0-g of P4O10

• 200.0-g H2O

• Determine what we need to calculate first

• the limiting reactant

• the theoretical yield of H3PO4

### Determining Percentage Yield – Find the Limiting Reactant and Theoretical Yield

limiting

reactant

• Using P4O10 as the given:

• Using H2O as the given:

1 mol P4O10

283.882 g P4O10

100.0 g P4O10

0.3522 mol P4O10

Step 1

4 mol H3PO4

1 mol P4O10

1.409 mol H3PO4

0.3522 mol P4O10

Step 2

theoretical

yield

1.409 mol H3PO4

97.9927 g H3PO4

1 mol H3PO4

138.1 g H3PO4

Step 3

200.0 g H2O

11.10 mol H2O

1 mol H2O

18.0148 g H2O

Step 1

4 mol H3PO4

6 mol H2O

7.400 mol H3PO4

11.10 mol H2O

Step 2

7.400 mol H3PO4

97.9927 g H3PO4

1 mol H3PO4

725.1 g H3PO4

Step 3

### Determining Percentage Yield

percentage yield =

actual yield

theoretical yield

x 100

percentage yield =

126.2 g

138.1 g

x 100

percentage yield = 91.38%