stoy-kee-ahm-eh-tree it's a big word that describes a simple idea

12
STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea. Stoichiometry: is the part of chemistry that studies amounts of substances that are involved in reactions

Upload: alexandra-zane

Post on 03-Jan-2016

18 views

Category:

Documents


0 download

DESCRIPTION

STOICHIOMETRY. STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea. Stoichiometry: is the part of chemistry that studies amounts of substances that are involved in reactions OR… - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

STOY-KEE-AHM-EH-TREE

It's a big word that describes a simple idea.

Stoichiometry: is the part of chemistry that studies amounts of substances that are involved in reactions OR… Simply the “math of chemistry”

Page 2: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

You might be looking at the amount of material that is produced by the reaction.

You might be looking at the amounts of substances before the reaction

Stoichiometry is all about amounts.

Page 3: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

PRACTICE BREAKDOWN: find the mass for these compounds

N2

NO2

2NH33(PO4) 3

14 + 14

28 grams

14 + 16 + 16

46 grams

14 + 1 + 1 + 1

17 grams

X’s 2 (the coefficient)

17 x 2= 34 grams

31 X 3= 93

285 grams

X’s 3 (the coefficient)

285 x 3= 855 grams

16X 12= 192

Page 4: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

GOAL – to find the mass of the reactants and the products in a chemical reaction

Example: find the mass of each compound or molecule for the following equation

N2 + 3H2 --- > 2NH3

( Reactants ) ( Products )

28g

6g

34g

34g

Both sides

must be equal

Page 5: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

13

27

20

40

Page 6: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

5

11

2) set is up to CANCEL the unit that you want to get rid of

Ex: If you have 50 grams of Boron, how many moles do you have?

1 mole=11g50g 1 mole

1g 11g50 g

11g

4.5 moles=x =

Page 7: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

Ex: If you have 80 grams of Sodium, how individual particles do you have?

1 mole=23g80g

1 mole 23g 23g=x =

481.6

23= 20.93913

Round to 20.9

Answer 1: 20.9 x 1023

Answer 2: 2.09 x 1024

11

23

Page 8: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

Ex: If you have 7.8 moles of magnesium, how grams do you have?

1 mole=24g

1

7.8 mole

1 mole

7.8 x 24

1=x = 187.2grams

12

24

Page 9: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

Producing a set quantity of a product using stoichiometry

Example: While working for Crest, you are instructed to generate 500grams of sodium flouride, how many grams of each of the reactants (Na & F) will you need?

Na + F2 -------- > NaF

Balance First! 2Na + F2 -------- > 2NaF

2Na + F2 -------- > 2NaF

46g

38g

84g

38g

84g

Step 1

Step 2 Find individual masses

Page 10: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

2Na + F2 -------- > 2NaF

46g

38g

84g

Step 4 Find # to multiply by

Problem asked you to find how much using 500g of sodium fluoride…so figure out how many times 500 is in NaF and use that number to recalculate the rest

500g ÷84g = 5.95

38 x 5.95 46x 5.95

273.7g 226.1g

273.7 + 226.1g = 499.8 which rounds to 500

Step 5Check your work!

Page 11: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

Producing a set quantity of a product using stoichiometry

Example: Based on the following problem, if an apple tree produces 800grams of sugar, how many grams of CO2 were used?

6CO 2 + 6H2O -------- > C6H12O6 + O 2

Looks alreadybalanced!

264g

108g

84g

92g 84g

Step 1

Step 2 Find individual masses

Sugar -

C6H12O6

6CO 2 + 6H2O ---- > C6H12O6 + 6O2

180g

Page 12: STOY-KEE-AHM-EH-TREE It's a big word that describes a simple idea

264g

108g

372g

192g

372g

Step 2 Find individual masses

6CO 2 + 6H2O ---- > C6H12O6 + 6O2

180g

Step 4 Find # to multiply by

Step 5Check your work!

800g ÷180g = 4.44

Answer for grams of CO2 = 479.52g