1. March 21, 2012 AGENDA: 1 – Bell Ringer & Part. Log 2 – CN: Multiple Step Mass and Mole Stoichiometry 3 – Work Time 4 – Quiz Grades Today’s Goal: Students will be able to do stoichiometry between mass and moles. Homework 1. 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.

2. Topic: Multiple Step Mass and Mole Stoichiometry Date: 3/21/2012 Balanced Chemical Equation: a A  b B Coefficients = # moles of each compound GramsMolesMolesGrams AABBAABB x Molar Mass of A ÷ Molar Mass of A x Molar Mass of B ÷ Molar Mass of B x b a x b a x a b x a b

3. 2 Mg + 1 O 2  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

4. 2 Mg + 1 O 2  2 MgO Gram  Mole  Mole How many moles of O 2 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 O 2 24 g Mg 2 mol Mg = 100 x 1 x 1 mol O 2 ≈ 2 mol O 2 24 x 2