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March 21, 2012

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 Multiple Step Mass and Mole Stoichiometry

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March 21, 2012

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  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 • Multiple Step Mass and Mole Stoichiometry • Friday is the Last Day to make up last week’s quiz. • Friday is the Last Day to turn in Last Week’s Work.

  2. Topic: Multiple Step Mass and Mole StoichiometryDate: 3/21/2012 Balanced Chemical Equation: a A b B Coefficients = # moles of each compound Grams MolesMoles Grams A AB B ÷ Molar Mass of A x Molar Mass of B xb a • xa • b x Molar Mass of A ÷ Molar Mass of B

  3. 2 Mg + 1 O2 2MgO 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

  4. 2 Mg + 1 O2 2MgO 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

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