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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 %
Homework Answers1. D
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 %
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
Homework Answers1. D
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 %