Raider Rev Up Calculate moles in 168.0 g of HgS (mercury II sulfide) MAR 7 2016.

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Presentation transcript:

Raider Rev Up Calculate moles in g of HgS (mercury II sulfide) MAR

Learning Focus (GPS- SC3)  Students will learn to make stoichiometric calculations.  Mole – Mole  Mole – Mass & Mass-Mole  Mass-Mass

What is Stoichiometry?  In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements.

What is Stoichiometry? Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Using the Balanced Equation….  2H 2 + 1O 2  2H 2 O.

What is Stoichiometry?  2H 2 + 1O 2  2H 2 O Stoichiometry lets the chemist use information from the balanced equation to make predictions about the amount of reactants needed or products that form when a reaction is carried out.

Types of Stoichiometric Calculations I. Mole – Mole ( 1 step) II. Mole – Mass or Mass – Mole (2 steps) III. Mass-Mass (3 steps)

REMEMBER………  YOU NEED A BALANCED EQUATION to do STOICHIOMETRY!!!!

Type 1 – Mole – Mole Stoichiometry 1 How many moles of FeS will react with 6 moles of HCl? FeS + 2HCl => FeCl 2 +H 2 S THIS IS A MOLE MOLE stoichiometry problem because : a. given information is 6 moles HCl b. Unknown is asking for moles of FeS

Type 1 – Mole – Mole Stoichiometry 1 How many moles of FeS will react with 6 moles of HCl? FeS + 2HCl => FeCl 2 +H 2 S Step 1: Find the mole ratio between FeS:HCl 1mol FeS:2 mole HCl Step 2: The mole ratios become Conversion Factors 1 mol FeS OR 2 mol HCl 2 mol HCl 1 mol FeS

Type 1 – Mole – Mole Stoichiometry 1 How many moles of FeS will react with 6 moles of HCl? FeS + 2HCl => FeCl 2 +H 2 S Step 3: Solve (given information) x (Mole Ratio) 6 mol HCl x 1mol FeS = 6 mol FeS = 3 mol FeS will 2 mol HCl 2 react with 6 mol HCl

Type 1- Mole-Mole Stoichiometry 2 For the balanced equation shown below, how many moles of O 2 will react with moles of C 3 H 6 O? 2C 3 H 6 O + 5O 2 => 6CO + 6H 2 O THIS IS A MOLE MOLE stoichiometry problems because : a. given information is moles C 3 H 6 O b. Unknown is asking for moles of O 2

Type 1- Mole-Mole Stoichiometry 2 For the balanced equation shown below, how many moles of O 2 will react with moles of C 3 H 6 O? 2C 3 H 6 O + 5O 2 => 6CO + 6H 2 O Step 1: Find the mole ratio between O 2 :C 3 H 6 O 5 mol O 2 : 2 mole C 3 H 6 O Step 2: The mole ratios become Conversion Factors 5 mol O 2 OR 2 mol C 3 H 6 O 2 mol C 3 H 6 O 5 mol O 2

Type 1- Mole-Mole Stoichiometry 2 For the balanced equation shown below, how many moles of O 2 will react with moles of C 3 H 6 O? 2C 3 H 6 O + 5O 2 => 6CO + 6H 2 O Step 3: Solve (given information) x (Mole Ratio) mol C 3 H 6 O x 5 mol O 2 = 4.93 mol FeS = 2.47 mol O 2 will 2 mol C 3 H 6 O 2 react with mol C 3 H 6 O

Raider Rev Up How many moles of hydrogen gas are needed to react completely with two moles of nitrogen gas? 3H2 + N2 --> 2NH3 MAR

Work Session – Day 2 Complete the Mole-Mole Practice Problems on your own paper. (Qty: 5)

Day 3 – Mole – Mass & Mass - Mole Given the following equation: 2 NaClO 3 → 2 NaCl + 3 O moles of NaClO 3 will produce how many grams of O 2 ? THIS IS MOLE MASS because: a. Given is moles of NaClO 3 b. Unknown grams O 2

Day 3 – Mole – Mass & Mass - Mole Given the following equation: 2 NaClO 3 → 2 NaCl + 3 O moles of NaClO 3 will produce how many grams of O 2 ? Step 1: Mole Ratio for NaClO 3 and O 2 from balanced equation 2 mole NaClO 3 : 3 mole O 2

Day 3 – Mole – Mass & Mass - Mole Given the following equation: 2 NaClO 3 → 2 NaCl + 3 O moles of NaClO 3 will produce how many grams of O 2 ? Step 2: The mole ratio becomes the Conversion Factors 2 mol NaClO 3 OR 3 mol O 2 3 mol O 2 2 mol NaClO 3

Day 3 – Mole – Mass & Mass - Mole Given the following equation: 2 NaClO 3 → 2 NaCl + 3 O moles of NaClO 3 will produce how many grams of O 2 ? Step 3: We also need Molar Mass for O 2, the UNKNOWN substance (1 mol O 2 = 32 g); Set up as a conversion factor Step 4: Solve: (given information) x (Mole Ratio) x (Molar Mass of Unknown Substance) 12 mol NaClO 3 x 3 mol O 2 x 32 g O 2 = 1152 g O 2 = 576 g O 2 2 mol NaClO 3 1 mol O 2 2 is produced