Thursday January 31, 2013 (Mass to Mole Stoichiometry)

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Presentation transcript:

Thursday January 31, 2013 (Mass to Mole Stoichiometry)

mol HgO 5.0 mol HgO HgO æ Hg + O 2 Bell Ringer Thursday, If 5.0 moles of mercury (II) oxide decompose, how many grams of mercury will be produced? mol? grams X 1 = mol Hg g Hg 1005 X g Hg mol Hg

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Assignment Currently Open Summative or Formative? Date Issued Date Due Date Into GradeSpeed Final Day QUIZ 16S11/18 1/242/7 WS - Introduction to Stoichiometry and Mole Ratios F71/232/1 TOMORROW WS – Review of Molar Mass F81/242/1 TOMORROW Project Share Quiz – Reactions and Balancing F91/25 TOMORROW WS – Mole to Mole Stoichiometry F101/292/1 TOMORROW WS – Mole to Mass Stoichiometry F111/302/12/8

Types of Stoichiometry Problems Mass – Mole Conversions : Given is a mass in grams and the unknown is an amount in moles. When you are given the mass of one substance and asked to calculate the amount in moles of another substance in the chemical reaction, the general plan is:

Converting Mass to Moles The following is a solution plan for problems in which the given quantity is expressed in grams and the unknown quantity is expressed in moles.

Converting Mass to Moles Sample Problem The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia: NH 3 + O 2 → NO + H 2 O (unbalanced) The reaction is run using 824 g of NH 3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H 2 O are formed?

Converting Mass to Moles Sample Problem The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia: NH 3 + O 2 → NO + H 2 O (unbalanced) The reaction is run using 824 g of NH 3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H 2 O are formed? Given: mass of NH 3 = 824 g Unknown: a. amt. of NO produced (in mol) b. amt. of H 2 O produced (in mol)

Converting Mass to Moles Sample Problem The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia: NH 3 + O 2 → NO + H 2 O (unbalanced) The reaction is run using 824 g of NH 3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H 2 O are formed? Start with a balanced equation: 4NH 3 + 5O 2 → 4NO + 6H 2 O

Two conversion factors are needed to solve part (a) - the molar mass of NH3 and the mole ratio of NO to NH3. Part (b) starts with the same conversion factor as part (a), but then the mole ratio of H2O to NH3 is used to convert to the amount in moles of H2O. The first conversion factor in each part is the inverted molar mass of NH3. Converting Mass to Moles

Use the periodic table to compute the molar mass of NH 3. Converting Mass to Moles