When copper (II) reacts with silver nitrate, the number of grams of copper required to produce 432...

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

B 127 g

C 216 g

D 252 g

With your lab partner, work on the following question in pairs:

Bell Ringer

2002 VA Chemistry SOL

Cu AgNO3+ Ag22 + Cu(NO3)2?

432 g Ag x 1 mol Ag

107.87 g Agx

2 mol Ag

1 mol Cu x 63.55 g Cu

1 mol Cu

= 127.25 g Cu

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 CarbonatesCommon mistake:

BaCO3 BaO + C

How to balance? What are we missing?

To remember:

think of a soda

why do we call them “carbonated” ?

H2CO3 CO2 + H2O

CO2

Homework Answers

13. 150 L O2

14. 0.73 g Zn

15. 8.36 g NaClO3

16. 16.70 g KCl

17. 93 L H2

18. 14.2 L CO2

19. 30.0 L CO2

45.0 L H2O

The Wisdom of GallagherWhy 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 NewsTakeru Kobayashi of Japan downed 44½ hot dogs in 12 minutes.

Source: CNN.com

WHAT IF…

One hot dog = one hot dog + one bun.

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?

Hot Dogs in the News

Source: CNN.com

WHAT IF…

One hot dog = one hot dog + one bun.

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

1 hot dog pack

50 hot dogs=

5 bun packs 8 buns

1 bun pack

40 buns=x

40 possible hot dogs

• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters• 12 oz. Chocolate chips

For 1 batch:In my pantry, I have:

• 5 cups of flour

• 16 Tbsp of butter

• lots of everything else

How many batches of cookies can I make?

Let’s Revisit the Cookies (again)…

• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters• 12 oz. Chocolate chips

For 1 batch: How many batches of cookies can I make?

Let’s Revisit the Cookies (again)…

5.5 c flour x2.25 c flour

1 batch cookies =

2.4 batches

16 Tbsp butter x 1 batch cookies

8 Tbsp butter=

2.0 batches

EXCESS

LIMITING

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 x2.25 c flour

1 batch cookies =

16 Tbsp butter x 1 batch

8 Tbsp butter=

2.4 batches of cookies

2.0 batches of cookies

GONE!

SOME FLOUR LEFT OVER…

2.0 batches x 2.25 cups flour

1 batch cookies= 4.5 cups flour used

5.5 cups – 4.5 cups = 1.0 cups left

Limiting Reactants in Chemistry5.0 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) + Na NaCl2 2

2 givens = 2 equations!5.0 mol Cl2 x1 mol Cl2

2 mol NaCl = 10. mol NaCl

5.0 mol Na x 2 mol NaCl2 mol Na

= 5.0 mol NaCl

EXCESS

LIMITING

5.0 mol Na x2 mol Na

1 mol Cl2=

2.5 mol Cl2

5.0 mol Cl2 given

2.5 mol Cl2 used

2.5 mol Cl2 left

Practice Problems

1. 3 CuSO4 + 2 Al Al2(SO4)3 3 Cu+

20.0 g CuSO4

20.0 g Al

x

x26.98 g Al

1 mol Al

1 mol CuSO4

159.61 g CuSO4

3 mol Cu

3 mol Cu

3 mol CuSO4

2 mol Al

x

x =

= 0125 mol Cu

1.11 mol Cu

20.0 g CuSO4 x159.61 g CuSO4

1 mol CuSO4 x 26.98 g Al

1 mol Al=

2.25 g AlUSED20.0 g Al – 2.25 g Al = 17.8 g Al EXCESS

3 mol CuSO4

2 mol Al x

Practice Problems

2. 2 H2 (g) + O2 (g) 2 H2O

5.0 g H2 x1 mol H2

2.02 g H2

2 mol H2O

2 mol H2

x = 2.5 mol H2O

USED

5.0 g H2 – 0.63 g H2 = 4.37 g H2 EXCESS

5.0 g O2 x32.00 g O2

1 mol O2 2 mol H2Ox = 0.31 mol H2O 1 mol O2

5.0 g O2 x32.00 g O2

1 mol O2 x 2.02 g H2

1 mol H2

= 0.63 g H2

1 mol O2

2 mol H2 x

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

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!

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

183 gramsx 100 = 88.5 %

Practice #20 on Homework

100 L N2 x 1 mol N2

22.4 L N2

2 mol NH3 22.4 L NH3xx1 mol N2 1 mol NH3

=

200 L NH3

THEORETICAL YIELD

200 L NH3 x 90 % = 180 L NH3 ACTUAL YIELD

What volume of ammonia can be obtained by reacting 100 L of nitrogen gas with an excess of hydrogen, if the yield is 90%?

H2 NH3N2 + 23

Next Class:• Quiz on Volume Conversions

– Mass-Volume– Volume-Mass– Volume-Volume

• 2 questions, 10 points each

• Watch significant figures & labels!