Stoichiometry Chapter 11

Chemical Reactions/Equations REVIEW Chemical Reactions/Equations Reactant + Reactant → Product + Product Balanced chemical equations are written to show chemical changes taking place Balanced chemical equations reflect the Law of Conservation of Mass Coefficients balance chemical equations and indicate the relative amounts of substances

Conversion Factors REVIEW Avogadro’s number and moles Molar mass 1 mol = 6.02 x 1023 representative particles Representative particles: Atoms – element from periodic table Ions – charged atom (due to loss/gain of e–) Formula units – ionic compound Molecules – covalent compound Molar mass 1 mol =  g of substance where  = (atomic mass × subscript) Mole ratio of element to compound  mol element = 1 mol compound where  = subscript of element in compound

Unit Objectives PREVIEW Interpret chemical equations in terms of particles, moles, and mass Write mole ratios from balanced equations Calculate number of moles and mass of a reactant or product when given number of moles or mass of another reactant or product Identify limiting reactants in chemical reactions Determine percent yield of chemical reactions

Stoichiometry U8-5 Defined: Based on the Law of Conservation of Mass: The study of quantitative relationships between amounts of reactants used and products formed by a chemical reaction Involves calculating quantities of reactants or products in a reaction using relationships found in balanced chemical equation Based on the Law of Conservation of Mass: Mass is neither created nor destroyed in any process; it is conserved

Interpreting Equations Write the balanced chemical equation for the synthesis of ammonia from nitrogen and hydrogen gases. Interpret the equation in terms of  particles,  moles, and  mass. Then,  show the Law of Conservation of Mass is obeyed. N2 + H2 3 → NH3 2

Interpreting Equations  Interpret the equation in terms of particles. Number of particles for each substance is indicated by the coefficients Four particle types: atoms, ions, formula units, or molecules N2 + 3 H2 → 2 NH3 1 molecule N2 + 3 molecules H2 → 2 molecules NH3

Interpreting Equations  Interpret the equation in terms of moles. Number of moles for each substance is indicated by the coefficients N2 + 3 H2 → 2 NH3 1 mole N2 + 3 moles H2 → 2 moles NH3

Interpreting Equations  Interpret the equation in terms of mass. Calculate the molar mass for each reactant and product N2: H2: NH3: N2 + 3 H2 → 2 NH3 2 mol x 14.0 g/mol 28.0 g/mol N2 2 mol x 1.0 g/mol 6.0 g/mol H2 1 mol x 14.0 g/mol = 14.0 g 3 mol x 1.0 g/mol = 3.0 g 17.0 g/mol NH3

Interpreting Equations  Interpret the equation in terms of mass. Multiply number of moles of each reactant and product by molar mass N2 + 3 H2 → 2 NH3 28.0 g/mol 2.0 g/mol 17.0 g/mol 1 mol N2 28.0 g N2 g N2 = = 28.0 g N2 1 mol N2

Interpreting Equations  Interpret the equation in terms of mass. Multiply number of moles of each reactant and product by molar mass N2 + 3 H2 → 2 NH3 28.0 g/mol 2.0 g/mol 17.0 g/mol 3 mol H2 2.0 g H2 g H2 = = 6.0 g H2 1 mol H2

Interpreting Equations  Interpret the equation in terms of mass. Multiply number of moles of each reactant and product by molar mass N2 + 3 H2 → 2 NH3 28.0 g/mol 2.0 g/mol 17.0 g/mol 2 mol NH3 17.0 g NH3 g NH3 = = 34.0 g NH3 1 mol NH3

Interpreting Equations  Interpret the equation in terms of mass. N2 + 3 H2 → 2 NH3 28.0 g 6.0 g 34.0 g 28.0 g N2 + 6.0 g H2 → 34.0 g NH3

Interpreting Equations  Show that the Law of Conservation of Mass is observed. Add the masses of the reactants. 28.0 g N2 + 6.0 g H2 = 34.0 g reactants Add the masses of the products. 34.0 g NH3 = 34.0 g products When the mass of the reactants equals the mass of the products, the Law of Conservation of Mass is observed. 34.0 g reactants = 34.0 g products N2 + 3 H2 → 2 NH3 28.0 g 6.0 g 34.0 g

HW due 01/28 Textbook Practice, p. 371 #1, #2

U8-5 Mole Ratio Ratio between the numbers of moles of any two substances in a balanced chemical equation

U8-5 Mole Ratios Determine all possible mole ratios for the balanced equation showing the synthesis of ammonia. How many mole ratios are possible? This reaction has three participating species. Multiply the number of species present by the next lower number. For the synthesis of ammonia, six mole ratios are possible. N2 + 3 H2 → 2 NH3 3 x 2 = 6 mole ratios

U8-5 Mole Ratios Determine all possible mole ratios for the balanced equation showing the synthesis of ammonia. N2 + 3 H2 → 2 NH3 1 mol N2 3 mol H2 3 mol H2 2 mol NH3 2 mol NH3 1 mol N2 1 mol N2 2 mol NH3 3 mol H2 1 mol N2 2 mol NH3 3 mol H2

Mole Ratios U8-5 GIVEN Which mole ratio should be used? The needed mole ratio is the one involving the UNKNOWN and the GIVEN. UNKNOWN GIVEN

HW due 01/29 Textbook Practice, p. 372 #3

Stoichiometric Calculations Write the balanced chemical equation for the reaction. Identify the UNKNOWN and the GIVEN; draw the bridge/grid. The GIVEN must be in moles or converted to moles. Identify the conversion factor that will cancel the unit of the GIVEN. Set up conversion factors and cancel units until the only unit left standing matches the UNKNOWN. Do the math* and express the answer to the correct number of significant figures.

Mole-to-Mole Conversions Both the UNKNOWN and the GIVEN are in moles Example: How many moles of ammonia are produced when 10.0 moles of hydrogen react with excess nitrogen? 10.0 mol ? mol excess N2 + 3 H2 → 2 NH3

Mole-to-Mole N2 + 3 H2 → 2 NH3 10.0 mol ? mol UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 10.0 mol H2 mol NH3 2 mol NH3 = mol H2 3 = 6.67 mol NH3

Mole-to-Mole Conversions Practice 1 How many moles of zinc chloride will be formed when 17.0 moles of hydrochloric acid react with excess zinc metal? How many mole ratios are possible for this equation? 17.0 mol ? mol excess 2 HCl + Zn → ZnCl2 + H2 12

Mole-to-Mole Conversions Practice 1 2 HCl + Zn → ZnCl2 + H2 Write all possible mole ratios for the balanced equation. 2 mol HCl 1 mol Zn 1 mol ZnCl2 1 mol H2

Mole-to-Mole 2 HCl + Zn → ZnCl2 + H2 17.0 mol ? mol UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 17.0 mol HCl 1 mol ZnCl2 mol ZnCl2 = 2 mol HCl = 8.50 mol ZnCl2

Mole-to-Mole Conversions Practice 2 Potassium chlorate decomposes into potassium chloride and oxygen. How many moles of oxygen are formed when 3.20 moles KClO3 decompose? Write the mole ratios involving the UNKNOWN and the GIVEN in the problem. 3.20 mol ? mol 2 KClO3 → KCl 2 + O2 3 3 mol O2 2 mol KClO3 2 mol KClO3 3 mol O2

Mole-to-Mole 2 KClO3 → 2 KCl + 3 O2 3.20 mol ? mol UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 3.20 mol KClO3 3 mol O2 mol O2 = 2 mol KClO3 = 4.80 mol O2

HW due 02/04 Textbook Practice, p. 375 #11

Mole-to-Mass Conversions p. 4 of Stoichiometry for Students Notes Packet

Mole-to-Mass Conversions GIVEN in moles and UNKNOWN in grams Example: Balance the following equation for the combustion of propane. Calculate the molar mass for each substance. 3 4 5 ____ C3H8 + ____ O2 → ____ CO2 + ____ H2O 44.0 g/mol 32.0 g/mol 44.0 g/mol 18.0 g/mol

Mole-to-Mass Conversions Example: If 10.0 moles of propane are used, how many grams of water are formed? Write the mole ratios involving the UNKNOWN and the GIVEN. 10.0 mol ? g C3H8 + 5 O2 → 3 CO2 + 4 H2O 1 mol C3H8 4 mol H2O 4 mol H2O 1 mol C3H8

Mole-to-Mass C3H8 + 5 O2 → 3 CO2 + 4 H2O 10.0 mol ? g UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 10.0 mol C3H8 4 mol H2O 18.0 g H2O g H2O = 1 mol C3H8 1 mol H2O = 720. g H2O

Mole-to-Mass Conversions Practice 1: Sulfuric acid is produced when sulfur dioxide reacts with oxygen and water. Balance the equation for the synthesis reaction. Calculate the molar mass for each substance. ____ SO2 + ____ O2 + ____ H2O → ____ H2SO4 2 64.1 g/mol 32.0 g/mol 18.0 g/mol 98.1 g/mol

Mole-to-Mass Conversions Practice 1: How many grams of sulfuric acid are produced when 1.50 moles of sulfur dioxide completely reacts with oxygen and water ? Write the mole ratios involving the UNKNOWN and the GIVEN. 1.50 mol ? g 2 SO2 + O2 + 2 H2O → 2 H2SO4 2 mol H2SO4 2 mol SO2 2 mol SO2 2 mol H2SO4

Mole-to-Mass 2 SO2 + O2 + 2 H2O → 2 H2SO4 ? g 1.50 mol UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 1.50 mol SO2 2 mol H2SO4 98.1 g H2SO4 g H2SO4 = 2 mol SO2 1 mol H2SO4 = 147 g H2SO4

Mass-to-Mole Conversions GIVEN in grams and UNKNOWN in moles Example: Methane and sulfur produce carbon disulfide and hydrogen sulfide gas, as indicated by the following chemical equation. Calculate the molar mass for each substance. ____ CH4 + ____ S8 → ____ CS2 + ____ H2S 2 4 16.0 g/mol 256.8 g/mol 76.2 g/mol 34.1 g/mol

Mass-to-Mole Conversions Example: Suppose that 19.75 g sulfur react with an excess of methane. How many moles of carbon disulfide will be produced? 19.75 g ? mol 2 CH4 + S8 → 2 CS2 + 4 H2S

Mass-to-Mole 2 CH4 + S8 → 2 CS2 + 4 H2S ? mol 19.75 g UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 19.75 g S8 1 mol S8 2 mol CS2 mol CS2 = 256.8 g S8 1 mol S8 = 0.1538 mol CS2

Mass-to-Mole Conversions Practice 1: Sodium fluoride is formed when sodium metal reacts with fluorine gas. Calculate the molar mass for each substance. Balance the equation. ____ Na + ____ F2 → ____ NaF 2 23.0 g/mol 38.0 g/mol 42.0 g/mol

Mass-to-Mole Conversions Practice 1: How many moles of sodium fluoride can be formed when 4.57 grams of fluorine gas reacts completely with excess sodium? 4.57 g ? mol 2 Na + F2 → 2 NaF

Mass-to-Mole 2 Na + F2 → 2 NaF ? mol 4.57 g UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 4.57 g F2 1 mol F2 2 mol NaF mol NaF = 38.0 g F2 1 mol F2 = 0.241 mol NaF

HW due 02/05 Textbook Practice, p. 376 #14

Mass-to-Mass Conversions GIVEN in grams and UNKNOWN in grams Example: The following equation shows the combustion of butane. Calculate the molar mass for each substance. Balance the equation. ____ C4H10 + ____ O2 → ____ CO2 + ____ H2O 8 10 2 13 58.0 g/mol 32.0 g/mol 44.0 g/mol 18.0 g/mol

Mass-to-Mass Conversions Example: If 75.5 grams of carbon dioxide are produced, how many grams of butane were used? ? g 75.5 g 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Mass-to-Mass 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O ? g 75.5 g UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 75.5 g CO2 1 mol CO2 2 mol C4H10 58.0 g C4H10 g C4H10 = 44.0 g CO2 8 mol CO2 1 mol C4H10 = 24.9 g C4H10

Mass-to-Mass Conversions Practice 1: Use the balanced equation for the combustion of butane. How many grams of oxygen are necessary to react completely with 217 grams of butane? 217 g ? g 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Mass-to-Mass 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 217 g ? g UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 217 g C4H10 1 mol C4H10 13 mol O2 32.0 g O2 g O2 = 58.0 g C4H10 2 mol C4H10 1 mol O2 = 778 g O2

Mass-to-Mass Conversions Practice 2: Balance the following single replacement reaction. How many grams of iron must react in order to produce 75.9 grams of iron (III) oxide? Because grams are involved for both UNKNOWN and GIVEN, the molar mass for both must be calculated. ? g 75.9 g ____ Fe + ____ H2O → ____ Fe2O3 + ____ H2 3 2 Molar mass: UNKNOWN Molar mass: GIVEN 55.8 g/mol 159.6 g/mol

Mass-to-Mass 2 Fe + 3 H2O → Fe2O3 + 3 H2 ? g 75.9 g UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U) 75.9 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.8 g Fe g Fe = 159.6 g Fe2O3 1 mol Fe2O3 1 mol Fe = 53.1 g Fe

HW due 02/06 Textbook Practice, p. 378 #21