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Chapter 4:

Jan 14, 2016

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Chapter 4:. Balancing Equations and Stoichiometry. Key Terms & Concepts. Stoichiometry Chemical Equations reactants and products balancing chemical equations Chemical Calculations Limiting Reactant Theoretical and Percent Yield. Stoichiometry. - PowerPoint PPT Presentation

  • Chapter 4:Balancing Equations and Stoichiometry

  • Key Terms & ConceptsStoichiometryChemical Equationsreactants and productsbalancing chemical equationsChemical CalculationsLimiting ReactantTheoretical and Percent Yield

  • StoichiometryStoichiometry is the study of the quantitative nature of chemical formulas and chemical reactions.Stoichiometry is one the the most essential tools in chemistryIt allows to quantify everything from global warming to drug manufacturing

  • Chemical EquationsChemical reactions are represented in a concise manner by chemical equationFor example, when H2 burns in O2, H2O is formed.The chemical equation for this reaction is:2 H2 + O2 2H2O

  • Chemical Equations2 H2 + O2 2 H2OThe compounds on the left of the arrow are called reactantsThe compounds on the right of the arrow are called productsH2 and O2 are reactants, H2O is the product

  • Chemical Equations2 H2 + O2 2 H2ONotice that the number of atoms or each element is equal on both sides of the equation4 H, 2 0All chemical equations must meet this requirementChemical equations must be balanced!!We balance equations by changing coefficients, not chemical formulas

  • Chemical Equations

  • Chemical EquationsConsider the following chemical equationCH4 + O2 CO2 + H2OunbalancedStart with elements that only appear in one compound on either side of the equationC and H are only in one compound on either sideC is balanced

  • Chemical EquationsCH4 + O2 CO2 + H2Ounbalanced4 Hs in reactants, 2 Hs in productsPut coefficient of 2 in front of H2O4 Hs in reactants, 4 Hs in productsCH4 + O2 CO2 + 2 H2Ounbalanced

  • Chemical EquationsCH4 + O2 CO2 + 2 H2Ounbalanced2 Os in reactants, 4 Os in productsPut coefficient of 2 in front of O24 Os in reactants, 4 Os in productsCH4 + 2 O2 CO2 + 2 H2Obalanced

  • Chemical Equations

  • Chemical EquationsConsider this equation:C3H8 + O2 CO2 + H2O3C + 8H + 2O 1C + 2H + 3O

  • Chemical EquationsBalance C and HC3H8 + O2 3CO2 + 4H2O3C + 8H + 2O 3C + 8H + 10O

  • Chemical EquationsBalance OC3H8 + 5O2 3CO2 + 4H2O3C + 8H + 10O 3C + 8H + 10O

  • Chemical EquationsExample 4.1Balance the following chemical equations(1) Mg + HCl MgCl2 + H2(2) K + H2O KOH + H2(3) CaCl2 + Na3PO4 Ca3(PO4)2 + NaCl(4) NaN3 Na + N2(5) C8H18 + O2 CO2 + H2O

  • Chemical EquationsExample 4.1Balance the following chemical equations(1) Mg + 2 HCl MgCl2 + H2(2) 2 K + 2 H2O 2 KOH + H2(3) 3 CaCl2 + 2 Na3PO4 Ca3(PO4)2 + 6 NaCl(4) 2 NaN3 2 Na + 3 N2(5) 2 C8H18 + 25 O2 16 CO2 + 18 H2O

  • Chemical EquationsExample 4.2Write a balanced chemical equation for the following reactionsammonium nitrate decomposes to nitrogen gas, oxygen gas, and wateriron reacts with oxygen gas and water to form iron(II) hydroxideammonia reacts with oxygen gas to produce nitrogen monoxide and water

  • Chemical EquationsExample 4.2Write a balanced chemical equation for the following reactions 2 NH4NO3 2 N2 + O2 + 4 H2O 2 Fe + O2 + 2 H2O 2 Fe(OH)2 4NH3 + 5 O2 4 NO + 6 H2O

  • Chemical Calculations2 H2+ O2 2 H2O2 molecules 1 molecule2 molecules2(6.022x1023) molecules 6.022x1023 molecules 2(6.022x1023) molecules2 mol 1 mol2 mol

    Stoichiometric coefficients can be interpreted as either number of molecules or number of moles.

  • Chemical CalculationsExample 4.3How many moles of water can be produced from 5.25 mol O2?

  • Chemical CalculationsExample 4.4How many moles of oxygen are required to completely react with 8.50 moles of butane, C4H10?

  • Chemical CalculationsWe cant directly measure moles. We can measure mass.We can use the stoichiometric coefficients of a reaction to determine the mass relationships.However, we must always convert mass to moles.We cannot directly compare the masses of reactants and products.We can only compare the moles of reactants and products.

  • Chemical CalculationsThe general scheme is:

  • Chemical CalculationsExample 4.5Geranyl formate is used as a synthetic rose essence in cosmetics. The compound is prepared from formic acid and geraniol:HCOOH + C10H18O C11H18O2 + H2OA chemist needs to make some geranyl formate for a batch of perfume. How many grams of geranyl formate can a chemist make from 375g of geraniol?

  • Chemical Calculations

  • Chemical CalculationsExample 4.6Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by each 1.00 g of lithium hydroxide?2 LiOH (s) + CO2 (g) Li2CO3 (g) + H2O (l)

  • Chemical CalculationsExample 4.62 LiOH (s) + CO2 (g) Li2CO3 (g) + H2O (l)

  • Theoretical and Percent YieldThe amount of product that can be produced from a given amount of reactants is the theoretical yield.However, no reaction goes to actual completion. The amount of products that is actually produced from a given amount of reactants is the actual yield.Some reactants may not reactReactants may react in an undesired way (side reactions) May be difficult to remove products from pot

  • Theoretical and Percent YieldThe extent of the desired reaction is typically reported as the percent yield.

  • Theoretical and Percent YieldExample 4.7Look back at Example 4.5. If the chemist starts with 375g of geraniol and collects 417g of purified product, what is the percent yield of the synthesis?

  • Theoretical and Percent YieldExample 4.825.0 g of sodium metal is burned in an excess of chlorine gas. What is the theoretical yield of sodium chloride? If 54.8 g of sodium chloride is actually produced, what is the percent yield of the reaction?

    2 Na + Cl2 2 NaCl

  • Theoretical and Percent Yield

  • Theoretical and Percent Yield

  • Theoretical and Percent YieldExample 4.9Titanium is a strong, lightweight, corrosion-resistant metal that is used in aeronautics and bicycle frames. It is prepared by the reaction of titanium (IV) chloride with molten magnesium between 950C and 1150C.TiCl4(g) + 2 Mg(l) Ti(s) + 2 MgCl2(l)In a certain industrial process 3.54x107 g of TiCl4 are reacted with 1.13x107 g of Mg. (a)Calculate the theoretical yield of Ti in grams. (b)Calculate the percent yield if 7.91x106 g of Ti are actually produced.

  • Theoretical and Percent Yield

  • Theoretical and Percent Yield

  • Limiting ReactantsMost reactions do not occur with stoichiometric equivalent amounts of each reactant.One reactant is used up firstThis reactant is the limiting reactant because it limits the amount of products that can be formed

  • Limiting ReactantsConsider the ham sandwich exampleone sandwich is made from one slice of ham, one slice of cheese and two slices of breadHow many ham sandwiches can be made from six slices of ham, seven slices of cheese and 14 slices of bread?What is the limiting reactant?

  • Limiting Reactants

  • Limiting ReactantsIf a problem gives specific amounts of two or more reactants it is a limiting reactant problem.Determine the amount of product that can be formed from each reactantThe reactant which produces the smallest amount of product is the limiting reactantThe remaining reactants are said to be in excess

  • Limiting ReactantsExample 4.10How many moles of water can be formed when 10.0 moles of H2 reacts with 4.50 moles of O2? What is the limiting reactant?

  • Limiting ReactantsExample 4.102 H2 + O2 2 H2O

  • Limiting ReactantsExample 4.11Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and aqueous acetic acid. If 15.0 g of sulfuric acid and 15.0 g of lead (II) acetate are mixed, calculate the number of grams of lead (II) sulfate that can be produced. Also calculate the number of grams of the excess reagent remaining after the reaction is completed.

  • Limiting ReactantsH2SO4 + Pb(CH3COO)2 PbSO4 + 2 CH3COOH

  • Limiting Reactants

  • End-of-Chapter ExercisesSuggested End-of-Chapter Exercises5, 9, 10, 14, 17, 20, 22, 28