Entry Task: March 13th-14th Block 1Entry task question:

Question:1.5 grams of zinc was added with 0.5 grams of iodine to produce zinc II iodide. Calculate the

limited reactant and the amount of excess.

You have ~10 minutes

Agenda:

• Sign off and Discuss Limited Reactant #1 ws• Notes on Theoretical and % yield• HW: Limited and yields practice ws

Limited and Excess #1 ws

Au2S3 + 3H2 2Au + 3H2S

25.0 grams of gold III sulfide reacts with 15.0 grams of hydrogen gas to create gold metal and dihydrogen monosulfide. Which reactant is limiting, which is excess, and how much product is produced?

25.0 g Au2S3 20.1 g Au490.14 g Au2S3

1 mole Au2S3 196.97 g Au

1 mol Au1 mole Au2S3

2 mol Au

15.0g H2 977.13 g Au

2.0 g H2

1 mole H2 196.97 g Au

1 mole Au3 mole H2

2 mole Au

Au2S3 + 3H2 2Au + 3H2S

Which reactant is Limited and which reactant is Excess?

25.0 g Au2S3 20.1 g Au490.14 g Au2S3

1 mole Au2S3 196.97 g Au

1 mol Au1 mole Au2S3

2 mol Au

15.0g H2 977.13 g Au

2.0 g H2

1 mole H2 196.97 g Au

1 mole Au3 mole H2

2 mole Au

Limited= Au2S3

Excess= H2

25.0 grams of gold III sulfide reacts with 15.0 grams of hydrogen gas to create gold metal and dihydrogen monosulfide. Which reactant is limiting, which is excess, and how much product is produced?

Start with the limited reactant!!

UsedIn reaction

15.0 grams H2 minus 0.31 g H2 =

Given UsedIn reaction

14.7 gEXCESS

25.0g Au2S3 0.31 g H2

490.14 g Au2S3

1 mole Au2S3 2.0 g H2

1 mole H21 mole Au2S3

3 mole H2

Au2S3 + 3H2 2Au + 3H2S

2Al(OH)3 + 3 H2SO4 Al2(SO4)3 + 6 H2O

45.0 g Aluminum hydroxide reacts 15.0 grams of dihydrogen monosulfur tetraoxide to create aluminum sulfate and water. Which reactant is limiting, which is excess, and how much product is produced?

45.0 g Al(OH)3

31.1 g H2O77.98 g Al(OH)3

1 mole Al(OH)3

18.0g H2O

1 mol H2O2 mole Al(OH)3

6 mol H2O

15.0 g H2SO4 5.51 g H2O

97.996 g H2SO4

1 mol H2SO4 18.0 g H2O

1 mol H2O3 mol H2SO4

6 mol H2O

2Al(OH)3 + 3 H2SO4 Al2(SO4)3 + 6 H2O

Which reactant is Limited and which reactant is Excess?

45.0 g Al(OH)3

31.1 g H2O77.98 g Al(OH)3

1 mole Al(OH)3

18.0g H2O

1 mol H2O2 mole Al(OH)3

6 mol H2O

15.0 g H2SO4 5.51 g H2O

97.996 g H2SO4

1 mol H2SO4 18.0 g H2O

1 mol H2O3 mol H2SO4

6 mol H2O

Limited= H2SO4

Excess= Al(OH)3

45.0 g Aluminum hydroxide reacts 15.0 grams of dihydrogen monosulfur tetraoxide to create aluminum sulfate and water. Which reactant is limiting, which is excess, and how much product is produced?

Start with the limited reactant!!

UsedIn reaction

45.0 grams Al(OH)3 minus 7.98 g Al(OH)3 =

Given UsedIn reaction

37.0 gEXCESS

15.0 g H2SO4 7.98 g Al(OH)3

97.996 g H2SO4

1 mol H2SO4 77.98 g Al(OH)3

1 mol Al(OH)33 mol H2SO4

2 mol Al(OH)3

2Al(OH)3 + 3 H2SO4 Al2(SO4)3 + 6 H2O

I can…

• Calculate the theoretical yield of a chemical reaction from data.

• Determine the percent yield for a chemical reaction

Theoretical yield

• Is the maximum amount of product that can be produced from the given amount of reactants.

• Calculated by stoichiometry

Actual yield

• Is the maximum amount of product ACTUALLY produced from the given amount of reactants in a lab experiment.

• FOR REAL

Percent Yield

• Percent yield of a product is the ratio of the actual amount of product to the theoretical amount of product expressed as a percent.

Percent Yield

Percent yield = Actual (for real)Theoretical (calculated)

X 100 =

NaBr + KCl NaCl + KBr

• This is a double replacement reaction

NaBr + KCl NaCl + KBr

• If we mixed 25 grams of sodium bromide with a large amount of potassium chloride, what would our theoretical yield of sodium chloride be? So a gram to gram stoich set up.

25g NaBr NaCl

102.98g NaBr

1 mole NaBr

1 mole NaBr

1 mole NaCl

1 mole NaCl

58.44 g NaCl

1461102.98

14 g

NaBr + KCl NaCl + KBr

If our actual yield from this reaction were 18 grams of sodium chloride, what would our percent yield be for this reaction?

Percent yield = Actual (for real)Theoretical (calculated)

18 g NaCl (actual)

14 g NaCl (calculated)X 100 =

128%

NaBr + KCl NaCl + KBr5. Is your answer in question 2 reasonable? If so, explain how this could happen in a real lab. If not, explain what is wrong with it and discuss possible reasons you might have gotten this answer in a real lab.

18 g NaCl (actual)

14 g NaCl (calculated)X 100 =

128%

There is NO WAY to have over 100% percent yield. It has to be experimental error- procedural or math error.

NaBr + KCl NaCl + KBr6. What are some factors that might cause percent yield to really be greater than 100%?

If there were added substances, dirty glassware or something on the scale when massing- experimental procedure mess up

NaBr + KCl NaCl + KBr6. What are some factors that might cause percent yield to really be less than 100%?

If the reactants were not finished reacting, or not enough reactants to move the reaction forward for more product.

Limited reactants % yields practice