Bell Ringer
With your lab partner, work on the following
question in pairs:
When copper (II) reacts with silver nitrate, the
number of grams of copper required to produce
432 grams of silver is:
A 31.5 g Cu + 2 AgNO3 ?
2 Ag + Cu(NO3)2
B 127 g 432 g Ag x 1 mol Ag x 1 mol Cu x 63.55 g Cu
C 216 g 107.87 g Ag 2 mol Ag 1 mol Cu
D 252 g = 127.25 g Cu
2002 VA Chemistry SOL
Mass-Mass Quiz
Good job on:
- Calculating molar mass
- Balancing reactions with coefficients
We need to fix:
- remembering decomposition reactions
- making units cancel in stoichiometry work
- using the mole ratio
Decomposition of Carbonates
Common mistake:
BaCO3 BaO + C 2
CO
How to balance? What are we missing?
To remember:
think of a soda
why do we call them “carbonated” ?
H2CO3 CO2 + H2O
Homework Answers
13. 150 L O2 17. 93 L H2
14. 0.73 g Zn 18. 14.2 L CO2
15. 8.36 g NaClO3 19. 30.0 L CO2
16. 16.70 g KCl 45.0 L H2O
The Wisdom of Gallagher
Why are there Interstate
Highways in Hawaii?
Why are there floatation
devices under plane seats
instead of parachutes?
Why do we drive on parkways
and park on driveways?
Why do hot dogs come ten to a package
and hot dog buns only eight?
Limiting Factors and Percent Yield
Ms. Besal
3/1/06
Hot Dogs in the News
Takeru Kobayashi of
Japan downed 44½
hot dogs in 12 minutes.
One hot dog = one hot dog
+ one bun.
WHAT IF…
Mr. Kobayashi didn’t do his math correctly. He bought 5
packs of hot dogs (10 per package) and 5 packs of hot dog
buns (8 per package). How many hot dogs (according to the
official formula) could he have eaten? Source: CNN.com
Hot Dogs in the News
One hot dog = one hot dog + one bun.
WHAT IF…
Mr. Kobayashi didn’t do his math correctly.
He bought 5 packs of hot dogs (10 per
package) and 5 packs of hot dog buns (8 per
package). How many hot dogs (according to
the official formula) could he have eaten?
5 hot dog packs x 10 hot dogs = 50 hot dogs
1 hot dog pack
40 possible
5 bun packs x 8 buns = 40 buns hot dogs
1 bun pack
Source: CNN.com
Let’s Revisit the Cookies (again)…
For 1 batch:
In my pantry, I have:
• 2.25 cups flour
• 5 cups of flour
• 8 Tbsp butter
• 16 Tbsp of butter
• 0.5 cups shortening
• lots of everything else
• 0.75 cups sugar
• 0.75 cups brown sugar
• 1 tsp salt How many batches of cookies
can I make?
• 1 tsp baking soda
• 1 tsp vanilla
• 0.5 cups Egg Beaters
• 12 oz. Chocolate chips
Let’s Revisit the Cookies (again)…
For 1 batch: How many batches of cookies
• 2.25 cups flour can I make?
• 8 Tbsp butter
EXCESS
• 0.5 cups shortening 5.5 c flour x 1 batch cookies =
• 0.75 cups sugar 2.25 c flour
2.4 batches
• 0.75 cups brown sugar
LIMITING
• 1 tsp salt
16 Tbsp butter x 1 batch cookies =
• 1 tsp baking soda 8 Tbsp butter
• 1 tsp vanilla 2.0 batches
• 0.5 cups Egg Beaters
• 12 oz. Chocolate chips
Now I Want to Bake a Cake!
But do I have all the ingredients I need?
How much flour do I have left after baking all those cookies?
5.5 c flour x 1 batch cookies = 2.4 batches SOME
2.25 c flour of cookies FLOUR LEFT
GONE! OVER…
16 Tbsp butter x 1 batch = 2.0 batches
8 Tbsp butter of cookies
2.0 batches x 2.25 cups flour = 4.5 cups flour used
1 batch cookies
5.5 cups – 4.5 cups = 1.0 cups left
Limiting Reactants in Chemistry
5.0 moles of chlorine gas react with 5.0 moles of sodium to
produce sodium chloride. Which reagent is the limiting factor?
How much of the excess reactant is left over?
Cl2 (g) + 2 Na 2 NaCl
EXCESS
2 equations!
5.0 mol Cl2 x 2 2 givens == 10. mol NaCl
mol NaCl
1 mol Cl2
LIMITING
5.0 mol Na x 2 mol NaCl = 5.0 mol NaCl
5.0 mol Cl2 given
2 mol Na
2.5 mol Cl2 used
5.0 mol Na x 1 mol Cl2 = 2.5 mol Cl2 2.5 mol Cl2 left
2 mol Na
Practice Problems
1. 3 CuSO4 + 2 Al Al2(SO4)3 + 3 Cu
1 mol CuSO4 x 3 mol Cu
20.0 g CuSO4 x = 0125 mol Cu
159.61 g CuSO4 3 mol CuSO4
20.0 g Al x 1 mol Al x 3 mol Cu = 1.11 mol Cu
26.98 g Al 2 mol Al
20.0 g CuSO4 x 1 mol CuSO4 x 2 mol Al x 26.98 g Al =
159.61 g CuSO4 3 mol CuSO4 1 mol Al
2.25 g Al
20.0 g Al – 2.25 g Al = 17.8 g Al EXCESS USED
Practice Problems
2. 2 H2 (g) + O2 (g) 2 H2O
1 mol H2 2 mol H2O
5.0 g H2 x x = 2.5 mol H2O
2.02 g H2 2 mol H2
5.0 g O2 x 1 mol O2 x 2 mol H2O = 0.31 mol H2O
32.00 g O2 1 mol O2
5.0 g O2 x 1 mol O2 x 2 mol H2 x 2.02 g H2 = 0.63 g H2
32.00 g O2 1 mol O2 1 mol H2 USED
5.0 g H2 – 0.63 g H2 = 4.37 g H2 EXCESS
On Perfection
“Perfection never exists in reality, but only in our
dreams.”
- Dr. Rudolf Dreikurs
“Perfection is our goal, excellence will be
tolerated.”
- J. Yahl
Get Real!
Johnny took a quiz yesterday. He missed 4
questions and earned 63 points out of 70.
-Was he perfect?
-What was his possible score?
-What was his actual percent score?
Get Real!
Ms. Besal ran a reaction in her lab yesterday. She predicted
that 183 grams of product would be formed. The reaction
only yielded 162 grams of product. But she looked really
cool in her lab coat.
-Was her reaction perfect?
-What was the percent yield?
162 grams
x 100 = 88.5 %
183 grams
Practice #20 on Homework
What volume of ammonia can be obtained by reacting
100 L of nitrogen gas with an excess of hydrogen, if
the yield is 90%?
N2 + 3 H2 2 NH3
100 L N2 x 1 mol N2 x 2 mol NH3 x 22.4 L NH3 =
22.4 L N2 1 mol N2 1 mol NH3
200 L NH3
THEORETICAL YIELD
200 L NH3 x 90 % = 180 L NH3 ACTUAL YIELD
Next Class:
• Quiz on Volume Conversions
– Mass-Volume
– Volume-Mass
– Volume-Volume
• 2 questions, 10 points each
• Watch significant figures & labels!