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II. Gas Stoichiometry. 1 mol of a gas=___ 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. 1 mol of a gas=___ 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

2 1 mol of a gas=___ 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 MOLES Molar Volume (22.4 L/mol) 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? CaCO 3  CaO + CO 2 ? g9.00 L

5 A. Limiting Reactants b Limiting Reactant _______in a reaction determines the amount of _______ b Excess Reactant added to ensure that the other ______ is completely used up cheaper & easier to recycle

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

7 A. Limiting Reactants 1. Write a _______ equation. 2. For each ______, calculate the amount of ________ formed. 3. Smaller answer indicates: _____________ amount of product

8 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

9 A. Limiting Reactants Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M

10 A. Limiting Reactants Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M

11 A. Limiting Reactants Zn: 27.1 L H 2 HCl: 25 L H 2 Limiting reactant: ____ Excess reactant: _____ Product Formed: ______ left over zinc

12 B. Percent Yield(pp ) ________________________ ________________________ _

13 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

14 B. Percent Yield K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g? g actual: 46.3 g Theoretical Yield:

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


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