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Bell Ringer When 6.58 g SO 3 and 1.64 g H 2 O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield? SO 3 + H 2 O H 2 SO 4 6.58 g SO 3 x 80.07 g SO 3 1 mol SO 3 x 1 mol SO 3 1 mol H 2 SO 4 x 1 mol H 2 SO 4 98.09 g H 2 SO 4 = 8.06 g H 2 SO 4 1.64 g H 2 O x 18.02 g H 2 O 1 mol H 2 O x 1 mol H 2 O 1 mol H 2 SO 4 x 1 mol H 2 SO 4 98.09 g H 2 SO 4 = 8.93 g H 2 SO 4

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Bell Ringer. When 6.58 g SO 3 and 1.64 g H 2 O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield?. SO 3. +. H 2 O. H 2 SO 4. 1 mol SO 3. 1 mol H 2 SO 4. 98.09 g H 2 SO 4. 6.58 g SO 3. x. x. x. =. 80.07 g SO 3. - PowerPoint PPT Presentation

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Bell RingerWhen 6.58 g SO3 and 1.64 g H2O react, what is the

expected yield of sulfuric acid? If the actual yield is

7.99 g sulfuric acid, what is the percent yield?

SO3 + H2O H2SO4

6.58 g SO3 x80.07 g SO3

1 mol SO3 x1 mol SO3

1 mol H2SO4 x1 mol H2SO4

98.09 g H2SO4 =

8.06 g H2SO4

1.64 g H2O x18.02 g H2O

1 mol H2O x1 mol H2O

1 mol H2SO4 x1 mol H2SO4

98.09 g H2SO4 =

8.93 g H2SO4

Bell RingerWhen 6.58 g SO3 and 1.64 g H2O react, what is the

expected yield of sulfuric acid? If the actual yield is

7.99 g sulfuric acid, what is the percent yield?

SO3 + H2O H2SO4

Info we’ve learned: Theoretical Yield = 8.06 g H2SO4

% Yield = Actual Yield

Theoretical Yieldx 100 % =

7.99 g H2SO4

8.06 g H2SO4

99.1 %

Limiting Factors & Percent Yield Quiz

2. A

3. C

4. B

5. A

6. C

7. C

8. A

9. B

10. D

11. Mass-mass

12. Mass-volume

13. Mole-mole

14. Limiting reactant

15. Volume-volume

17. 0.52 mol PBr3

19. 0.13 mol Na

21. 50 g NaClO3

23. 5000 g HCl

25. 4770 g H2O

27. 98 g AgCl; 120 g AgNO3

29. 700 g CO2 ; 500 g O2

31. 89.6 L H2

33. 234 g ZnSO4

35. 7.75 L O2

39. 8.06 g H2SO4 ; 99.1 %

Stoichiometry Review

Ms. Besal

3/7/2006

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Mole-Mole Problems• 1 conversion step

– Given: moles “A”

– Required: moles “B”

• Convert moles “A” to moles “B” using mole ratio.• The mole ratio is used in EVERY STOICHIOMETRY

PROBLEM. EVER. I PROMISE.

2 H2 + O22 H2O

How many moles of water can be formed from 0.5 mol H2?

0.5 mol H2 x2 mol H2

2 mol H2O = 0.5 mol H2O

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Mass-Mole Problems• 2 conversion steps

– Given: mass “A”– Required: moles “B”

• Step 1: convert grams “A” to moles “A” using Periodic Table

• Step 2: convert moles “A” to moles “B” using mole ratio

2 H2 + O22 H2O

How many moles of water can be formed from 48.0 g O2?

48.0 g O2 x1 mol O2

2 mol H2O = 3.00 mol H2O

32.00 g O2

1 mol O2 x

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Mass-Mass Problems• 3 conversion steps

– Given: mass “A”– Required: mass “B”

• Step 1: convert grams “A” to moles “A” using Periodic Table

• Step 2: convert moles “A” to moles “B” using mole ratio

• Step 3: convert moles “B” to grams “B” using Periodic Table

2 H2 + O22 H2O

How many grams of water can be formed from 48.0 g O2?

48.0 g O2 x1 mol O2

2 mol H2O =32.00 g O2

1 mol O2 x x18.02 g H2O

1 mol H2O54.1 g H2O

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Mass-Volume Problems• 3 – 4 conversion steps

– Given: mass “A”– Required: volume “B”

• Step 1: convert grams “A” to moles “A” using Periodic Table

• Step 2: convert moles “A” to moles “B” using mole ratio

• Step 3: convert moles “B” to liters “B”

2 H2 + O22 H2O

48.0 g H2O x2 mol H2O

1 mol O2 =18.02 g H2O

1 mol H2O x x22.4 L O2

1 mol O2

How many liters of oxygen are necessary to create 48.0 g H2O?

29.8 L O2

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Volume-Mass Problems• 3 – 4 conversion steps

– Given: volume “A”– Required: mass “B”

• Step 1: convert liters “A” to moles “A” • Step 2: convert moles “A” to moles “B” using

mole ratio• Step 3: convert moles “B” to grams “B” using

Periodic Table

2 H2 + O22 H2O

36.0 L O2 x1 mol O2

2 mol H2O =

22.4L O2

1 mol O2 x x18.02 g H2O

1 mol H2O

How many grams of water are formed by reacting 36.0 L O2?

58.7 g H2O

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Volume-Volume Problems• 3 – 5 conversion steps

– Given: volume “A”

– Required: volume “B”

• Step 1: convert liters “A” to moles “A” • Step 2: convert moles “A” to moles “B” using

mole ratio• Step 3: convert moles “B” to liters “B”

2 H2 + O22 H2O

5.0 L O2 x1 mol O2

2 mol H2 =22.4 L O2

1 mol O2 x x22.4 L H2

1 mol H2

How many liters of H2 are required to react with 5.0 L O2?

10. L H2

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Limiting Reactant Problems• Quantities are given for each reactant.

• 2 parallel equations

• Solve each equation for product desired and determine limiting reactant.

• Use Limiting Reactant to solve for amount or excess reactant used.

• Subtract amount excess reactant used from amount given to determine how much is left over.

Limiting Reactant Problems

If you start with 10.0 g of O2 and 5.00 g H2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be?

2 H2 + O2 2 H2O

10.0 g O2

5.00 g H2

x

x2.02 g H2

1 mol H2

1 mol O2

32.00 g O2

2 mol H2O

2 mol H2O

1 mol O2

2 mol H2

x

x =

=x1 mol H2O

18.02 g H2O

11.3 g H2O

x1 mol H2O

18.02 g H2O

44.06 g H2O

THEORETICAL YIELD

LIMITING REACTANT

EXCESS REACTANT

Limiting Reactant Problems

Info we know so far:

Limiting Reactant = O2

Excess Reactant = H2

10.0 g O2 x x

1.26 g H2

USED5.00 g H2 – 1.26 g H2 = 3.74 g H2 LEFT OVER

1 mol O2

2 mol H2 x

2 H2 + O2 2 H2OIf you start with 10.0 g of O2 and 5.00 g H2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be?

1 mol O2

32.00 g O2

2.02 g H2

1 mol H2

=

Types of Stoichiometry Problems

• Mole-Mole

• Mass-Mole

• Mass-Mass

• Mass-Volume

• Volume-Mass

• Volume-Volume

• Limiting Reactant

• Percent Yield

Percent Yield Problems• Critical Information:

– Theoretical Yield– Actual Yield– Percent Yield

You will be given one of these

May or may not be given

2 H2 + O2 2 H2O

Determine the actual yield of a reaction between 6.25 g H2 and excess O2 that has a 85% percent yield.

6.25 g H2 x2.02 g H2

1 mol H2 2 mol H2O

2 mol H2

x =x1 mol H2O

18.02 g H2O

55.6 g H2O85 % = x 100 %

55.6 g H2O

?

ACTUAL YIELD = 47.3 g H2O

THEORETICAL YIELD

2. A

3. C

4. B

5. A

6. C

7. C

8. A

9. B

10. D

11. Mass-mass

12. Mass-volume

13. Mole-mole

14. Limiting reactant

15. Volume-volume

17. 0.52 mol PBr3

19. 0.13 mol Na

21. 50 g NaClO3

23. 5000 g HCl

25. 4770 g H2O

27. 98 g AgCl; 120 g AgNO3

29. 700 g CO2 ; 500 g O2

31. 89.6 L H2

33. 234 g ZnSO4

35. 7.75 L O2

39. 8.06 g H2SO4 ; 99.1 %