Transcript Bell Ringer

Bell Ringer
When 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
6.58 g SO3 x
+
H2O
H2SO4
1 mol SO3 x 1 mol H2SO4 x 98.09 g H2SO4
=
80.07 g SO3
1 mol SO3
1 mol H2SO4
8.06 g H2SO4
1.64 g H2O x
1 mol H2O x 1 mol H2SO4 x 98.09 g H2SO4
=
18.02 g H2O
1 mol H2O
1 mol H2SO4
8.93 g H2SO4
Bell Ringer
When 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 =
7.99 g H
2SO4
Actual
Yield
Theoretical
Yield
8.06 g H2SO
4
x
100 % = 99.1 %
Limiting Factors & Percent Yield
Quiz
Homework Answers
1. D
11. Mass-mass
25. 4770 g H2O
2. A
12. Mass-volume
27. 98 g AgCl; 120 g AgNO3
3. C
13. Mole-mole
29. 700 g CO2 ; 500 g O2
4. B
14. Limiting reactant
31. 89.6 L H2
5. A
15. Volume-volume
33. 234 g ZnSO4
6. C
35. 7.75 L O2
7. C
17. 0.52 mol PBr3
8. A
19. 0.13 mol Na
9. B
21. 50 g NaClO3
10. D
23. 5000 g HCl
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 +
O2
2 H2O
How many moles of water can be formed from 0.5 mol H2?
0.5 mol H2 x 2 mol H2O = 0.5 mol H2O
2 mol H2
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 +
O2
2 H2 O
How many moles of water can be formed from 48.0 g O2?
48.0 g O2 x 1 mol O2 x 2 mol H2O = 3.00 mol H2O
32.00 g O2
1 mol O2
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 +
O2
2 H2 O
How many grams of water can be formed from 48.0 g O2?
48.0 g O2 x 1 mol O2 x 2 mol H2O x 18.02 g H2O = 54.1 g H2O
32.00 g O2
1 mol O2
1 mol 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 +
2 H2 O
O2
How many liters of oxygen are necessary to create 48.0 g H2O?
48.0 g H2O
x
1 mol H2O
1 mol O2
x
x
18.02 g H2O 2 mol H2O
22.4 L O2
1 mol O2
= 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 +
2 H2 O
O2
How many grams of water are formed by reacting 36.0 L O2?
36.0 L O2
x
1 mol O2
22.4L O2
x
2 mol H2O
1 mol O2
x
18.02 g H2O = 58.7 g
1 mol H2O
H 2O
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 +
2 H2 O
O2
How many liters of H2 are required to react with 5.0 L O2?
5.0 L O2
x
1 mol O2
22.4 L O2
x
2 mol H2
1 mol O2
x
22.4 L H2
1 mol H2
=
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
2 H2 +
O2
2 H2O
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?
1 mol O2
10.0 g O2 x
x 2 mol H2O x 18.02 g H2O =
32.00 g O2
1 mol O2
1 mol H2O
LIMITING
11.3 g H2O
THEORETICAL
REACTANT
YIELD
5.00 g H2 x 1 mol H2
EXCESS
2.02 g H2
REACTANT
x 2 mol H2O
2 mol H2
x 18.02 g H2O =
1 mol H2O
44.06 g H2O
Limiting Reactant Problems
2 H2 +
O2
2 H2O
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?
Info we know so far:
Limiting Reactant = O2
Excess Reactant = H2
10.0 g O2 x
1 mol O2
32.00 g O2
x
2 mol H2
1 mol O2
5.00 g H2 – 1.26 g H2 = 3.74 g H2
2.02 g H2
=
1 mol H2
1.26 g H2
USED
LEFT OVER
x
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
May or may not be given
You will be given one of these
Determine the actual yield of a reaction between 6.25 g H2 and
excess O2 that has a 85% percent yield.
2 H2 +
6.25 g H2 x
O2
1 mol H2
2.02 g H2
2 H2O
x 2 mol H2O x 18.02 g H2O =
2 mol H2
1 mol H2O
THEORETICAL 55.6 g H2O
?
YIELD
85 % =
x 100 %
55.6 g H2O
ACTUAL YIELD = 47.3 g H2O
Homework Answers
1. D
11. Mass-mass
25. 4770 g H2O
2. A
12. Mass-volume
27. 98 g AgCl; 120 g AgNO3
3. C
13. Mole-mole
29. 700 g CO2 ; 500 g O2
4. B
14. Limiting reactant
31. 89.6 L H2
5. A
15. Volume-volume
33. 234 g ZnSO4
6. C
35. 7.75 L O2
7. C
17. 0.52 mol PBr3
8. A
19. 0.13 mol Na
9. B
21. 50 g NaClO3
10. D
23. 5000 g HCl
39. 8.06 g H2SO4 ; 99.1 %