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CHEM 101-GENERAL CHEMISTRY CHAPTER 4

STOICHIOMETRY

INSTR : FİLİZ ALSHANABLEH

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CHAPTER 4 STOICHIOMETRY

• Fundamentals of Stoichiometry

. Obtaining Ratios from a Balanced Chemical Equations

• Limiting Reactants

• Theoretical and Percentage Yields

• Solution Stoichiometry

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Fundamentals of Stoichiometry

• Stoichiometry is a term used to describe quantitative relationships in chemistry. • “How much?” of a product is produced or reactant is

consumed.

• Balanced chemical equation needed.

• Conversion between mass or volume to number of moles frequently needed.

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Ratios from a Balanced Chemical Equation

• Mole ratios are obtained from the coefficients in the balanced chemical reaction. • 1 mol CH4 : 2 mol O2 : 1 mol CO2 : 2 mol H2O

• These ratios can be used in solving problems:

1 mol CH4

2 mol O2

or 2 mol H2O1 mol CH4

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Stoichiometry Involving Chemical Equation

• This flow diagram illustrates the various steps involved in solving a typical reaction stoichiometry problem.

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Example Problem 4.2 (from Text Book)

• How many grams of water can be produced if sufficient hydrogen reacts with 26.0 g of oxygen? ( H:1.01, O:16.0)

Write a balanced chemical equation: 2H2 (g) + O2 (g) 2H2O (g)

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Exercise 4.1

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Prob 4.19 (from text book)

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Prob 4.18 (from text book)

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Limiting Reactants

• In many chemical reactions, one reactant is often exhausted before the other reactants. This reactant is the limiting reactant. • Limiting reactant is determined using stoichiometry.

• The limiting reactant limits the quantity of product

produced.

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Limiting Reactants

• Reaction between 6 mols of H2 and 6 mols of O2 will produce 6 mols of H2O.

• 6 mols of H2 can produce 6 mols of H2O. • 6 mols of O2 can produce 12 mols of H2O.

• H2 is limiting reactant. • 3 mols of O2 left over.

2H2 (g) + O2 (g) → 2H2O(g)

Stoichiometry of Rxn: 2 mol H2 : 1 mol O2 : 2 mol H2O

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Concept Check

Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation: 2H2 + O2 → 2H2O

a) 2 moles of H2 and 2 moles of O2 b) 2 moles of H2 and 3 moles of O2 c) 2 moles of H2 and 1 mole of O2 d) 3 moles of H2 and 1 mole of O2 e) Each produce the same amount of product.

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Limiting Reactants

• In many cases, we manipulate the amounts of reactants to ensure that one specific compound is the limiting reactant. • For example, a more expensive or scarce reagent is

usually chosen to be the limiting reagent.

• Other times, it is best to have a stoichiometric mixture (equal ratio of moles) to prevent waste. • For example, rocket fuel is designed so that no mass is

left over, which would add unnecessary weight to the rocket.

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Problem 4.23 • Prob. 4.23. If 8.4 moles of disilane, Si2H6 is reacted with 15.1 moles of O2

to produce SiO2 and H2O, a. Which is limiting reactant? b. How many moles of SiO2 will be produced? c. How many moles of excess reactant will remain unused? ANSWERS: 2 Si2H6 + 7 O2 4SiO2 + 6 H2O 2 mol Si2H6 = 7 mol O2 = 4 mol SiO2 = 6 mol H2O 8.4 mol Si2H6 = 29.4 mol O2 = 16.8 mol SiO2 = 25.2 mol H2O 4.3 mol Si2H6 = 15.1 mol O2 = 8.6 mol SiO2 = 12.9 mol H2O a. O2 is limiting reactant b. 8.6 mol SiO2 will be produced c. (8.4 mol Si2H6 )given - (4.3 mol Si2H6 )used = 4.1 mol Si2H6 remained

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Problem 4.28

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Theoretical and Percentage Yields

Percentage Yield = actual yield

theoretical yield

×100%

• Reaction efficiency is measured with percentage yield. • The mass of product obtained is the actual yield.

• The ideal mass of product obtained from calculation is the theoretical yield.

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Problem 4.43

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Solution Stoichiometry

• For reactions occurring in solution, the concentration and volume of reactants and products are often used instead of mass to solve solution stoichiometry problems. • n = number of moles; M = mol/L; V = L

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Problem 4.49

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Problem 4.53