limiting factors & percent yield quiz. bell ringer when 6.58 g so 3 and 1.64 g h 2 o react, what...
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Limiting Factors & Percent Yield Quiz
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 %
Stoichiometry Review
Mr. Trotts
Things to know for test• Stoichiometry: the calculation of quantities in
chemical reactions• In every reaction mass and atoms are conserved• Limiting reactant: the reactant that is used up in
a chemical reaction• The amount of product obtained is determined
by the limiting reagent• The first step in any stoichiometry problem is to
balance the equation by adding coefficients to the reagents
• In one mole of H20 there are 3 x 6.02x1023 atoms
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
• Next class = test
• Bring a calculator and extra paper to do your work on
• Packet due