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II. Gas Stoichiometry Stoichiometry Continued…. 1 mol of a gas=22.4 L at STP A. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm.

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Presentation on theme: "II. Gas Stoichiometry Stoichiometry Continued…. 1 mol of a gas=22.4 L at STP A. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm."— Presentation transcript:

1 II. Gas Stoichiometry Stoichiometry Continued…

2 1 mol of a gas=22.4 L at STP A. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm

3 A. Molar Volume at STP Molar Mass (g/mol) 6.02  particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molar Volume (22.4 L/mol) LITERS OF GAS AT STP Molarity (mol/L)

4 B. Gas Stoichiometry Problem b How many grams of CaCO 3 are req’d to produce 9.00 L of CO 2 at STP? 9.00 L CO 2 1 mol CO L CO 2 = 40.2 g CaCO 3 CaCO 3  CaO + CO 2 1 mol CaCO 3 1 mol CO g CaCO 3 1 mol CaCO 3 ? g9.00 L

5 StoichiometryStoichiometry III.Adjusting to Reality

6 A. Limiting Reactants b Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly

7 A. Limiting Reactants b Limiting Reactant used up in a reaction determines the amount of product b Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

8 A. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant amount of product

9 A. Limiting Reactants b 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M

10 A. Limiting Reactants 79.1 g Zn 1 mol Zn g Zn = 27.1 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M

11 A. Limiting Reactants 22.4 L H 2 1 mol H L 2.5 mol HCl 1 L = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M

12 A. Limiting Reactants Zn: 27.1 L H 2 HCl: 25 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H 2 left over zinc

13 B. Percent Yield calculated on paper measured in lab

14 B. Percent Yield b When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g? g actual: 46.3 g

15 B. Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g? g actual: 46.3 g Theoretical Yield:

16 B. Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g49.4 g actual: 46.3 g


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