March 21, 2012

AGENDA:1 – Bell Ringer & Part.

Log2 – CN: Multiple Step

Mass and Mole Stoichiometry

3 – Work Time4 – Quiz Grades

Today’s Goal:Students will be able to

do stoichiometry between mass and moles.

Homework1. Multiple Step Mass

and Mole Stoichiometry

2. Friday is the Last Day to make up last week’s quiz.

3. Friday is the Last Day to turn in Last Week’s Work.

Topic: Multiple Step Mass and Mole StoichiometryDate: 3/21/2012

Balanced Chemical Equation:a A b B

Coefficients = # moles of each

compound

Grams Moles Moles Grams

A A B Bx Molar Mass of A

÷ Molar Mass of A

x Molar Mass of B

÷ Molar Mass of B

x b a

x a b

2 Mg + 1 O2 2 MgO

Mole Mole Gram

How many grams of MgO would you form

from 5 mol of Mg?

1. Molar Mass Mg =

24 O = +

16

40 g

mol

2. Conversions 5 mol Mg x 2 mol MgO x 40 g

MgO 2 mol Mg 1 mol

MgO

= 5 x 2 x 40 g MgO = 400 = 200 g MgO

2 2

2 Mg + 1 O2 2 MgO

Gram Mole Mole

How many moles of O2 must be present

to react with 100 g of Mg?

1. Molar Mass Mg = 24 g

mol

2. Conversions 100 g Mg x 1 mol Mg x 1 mol O2

24 g Mg 2 mol Mg

= 100 x 1 x 1 mol O2 ≈ 2 mol O2 24 x 2