Chemical Reactions

Chemical Equations 4 Al (s) + 3 O2 (g)  2 Al2O3 (s) A chemical reaction shows the formulas and relative amounts of reactants and products in a reaction. 4 Al (s) + 3 O2 (g)  2 Al2O3 (s) reactants product stoichiometric coefficients physical state s, l, g or aq

Chemical Equations 4 Al (s) + 3 O2 (g)  2 Al2O3 (s) Interpret this equation as 4 atoms solid Al react with 3 molecules gaseous O2 to form 2 formula units of solid Al2O3 4 moles solid Al react with 3 moles gaseous O2 to form 2 moles of solid Al2O3

Balancing Chemical Equations “Matter is conserved in chemical change” Antoine Lavoisier, 1789 An equation must be balanced: It must have the same number of atoms of each kind on both sides

The rules of the game Write the correct formulas of the reactants and products Do not change the formulas to balance the equation Put a coefficient in front of each formula so that the same number of atoms of each kind appear in both the reactants and products The coefficent multiplies through its formula 2 H2O shows 4 H atoms and 2 O atoms

Example 1 Balance this equation: N2 + H2  NH3 N2 + H2  2 NH3 There are 2 Ns on the left, we need 2 Ns on the right Put a coefficient of 2 in front of NH3 N2 + H2  2 NH3 Now there are 6 Hs on the right, we need 6 on the left Put a coefficient of 3 in front of H2 N2 + 3 H2  2 NH3 Balanced equation shows 2 Ns and 6 Hs on each side.

Example 2 Balance this equation: CH4 + O2  CO2 + H2O There is one C on each side. Cs are balanced There are 4 Hs on left, 2 on right. Put a 2 in front of H2O CH4 + O2  CO2 + 2 H2O There are 2 + 2 = 4 Os on right. Put a 2 in front of O2 CH4 + 2 O2  CO2 + 2 H2O Balanced equation has 1 C, 4 Hs, and 4 Os on each side

Mg (s) + HCl (aq)  MgCl2 (aq) + H2 (g) Balance these equations and list the coefficients from left to right (including coefficients of one) HgO (s)  Hg (l) + O2 (g) 2 HgO (s)  2 Hg (l) + O2 (g) Answer 2, 2, 1 Mg (s) + HCl (aq)  MgCl2 (aq) + H2 (g) Mg (s) + 2 HCl (aq)  MgCl2 (aq) + H2 (g) Answer 1, 2, 1, 1

C3H8 (g) + O2 (g)  CO2 (g) + H2O (l) Balance these equations and list the coefficients from left to right (including coefficients of one) C3H8 (g) + O2 (g)  CO2 (g) + H2O (l) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Answer 1, 5, 3, 4 Al (s) + Cl2 (g)  AlCl3 (s) 2 Al (s) + 3 Cl2 (g)  2 AlCl3 (s) Answer 2, 3, 2

C4H10 (g) + O2 (g)  CO2 (g) + H2O (l) Balance these equations and list the coefficients from left to right (including coefficients of one) H2O2 (aq)  H2O (l) + O2 (g) 2 H2O2 (aq)   H2O (l) + O2 (g) Answer 2, 2, 1 C4H10 (g) + O2 (g)  CO2 (g) + H2O (l) 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) Answer 2, 13, 8, 10

Combustion Reactions In combustion, a hydrocarbon or C–H–O fuel combines with O2 to form CO2 and H2O __ CH4 + __ O2  __ CO2 + __ H2O 1 CH4 + 2 O2  1 CO2 + 2 H2O Balanced equation shows 1 C, 4 H, and 4 O on each side If N or S are in the formula for the fuel, assume it is oxidized to NO2 or SO2

Reaction types Synthesis or combination A + B → C 2 or more formulas combine to make one product Decomposition C → A + B One formula breaks apart into 2 or more formulas Combustion C, H, O fuel + O2 → CO2 + H2O Single replacement A + BX → AX + B One element pushes another element out of a compound Double replacement AX + BY → AY + BX Two compounds switch partners

Reaction types Classify each reaction as combination, decomposition, combustion, single replacement, or double replacement: 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O 2 Na + Cl2 → 2 NaCl Cl2 + 2 KBr → 2 KCl + Br2 NH4NO3 → N2O + 2 H2O K2SO4 + BaCl2 → BaSO4 + 2 KCl

Stoichiometry Stoichiometry is chemical accounting Heart of stoichiometry is mole ratio given by coefficients of balanced equation

Stoichiometry Stoichiometry is chemical accounting Heart of stoichiometry is mole ratio given by coefficients of balanced equation moles A moles B mole ratio

Example How many moles of O2 are produced from the decomposition of 1.76 mol KClO3: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g) Question about O2 but information for KClO3 O2 and KClO3 related by coefficients of balanced equation. Coefficients are in moles: 2 mol KClO3 : 3 mol O2

Example How many moles of O2 are produced from the decomposition of 1.76 mol KClO3: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

Example How many moles of CO2 form when 0.35 mol C2H6 burn? 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Example How many moles of CO2 form when 0.35 mol C2H6 burn? 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Example How many grams of Al react with 0.15 moles HCl: 2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g) Coefficients are in moles Two steps: convert mol HCl to mol Al, then convert mol Al to grams

Example How many grams of Al react with 0.15 moles HCl: 2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g) Coefficients are in moles Two steps: convert mol HCl to mol Al, then convert mol Al to grams

Example How many moles of Ag are produced when 10.0 g Ag2O decompose: 2 Ag2O (s)  4 Ag (s) + O2 (g) Two steps: first convert g Ag2O to mol Ag2O, then use mole ratio to find moles Ag

Example How many moles of Ag are produced when 10.0 g Ag2O decompose: 2 Ag2O (s)  4 Ag (s) + O2 (g) Two steps: first convert g Ag2O to mol Ag2O, then use mole ratio to find moles Ag

Example How many grams of magnesium nitride are produced when 3.82 g Mg react with excess N2: 3 Mg (s) + N2 (g)  Mg3N2 (s)

Example How many grams of H2 (g) are needed to produce 100.0 g of CH3OH: CO (g) + 2 H2 (g)  CH3OH (l)

Example How many grams of H2 react with 3.20 grams of O2 in the reaction 2 H2 (g) + O2 (g)  2 H2O (l)

Example How many grams of O2 react with 268 grams of octane (C8H18) in the combustion of octane? 2 C8H18 + 25 O2  16 CO2 + 18 H2O

Example How many grams of O2 react with 268 grams of octane (C8H18) in the combustion of octane? 2 C8H18 + 25 O2  16 CO2 + 18 H2O

Stoichiometry Convert to moles (÷ molar mass) Use the mole ratio Convert new moles back to grams (x new molar mass) grams A moles A moles B grams B mole ratio

Example Balance the chemical equation Identify the type of reaction Calculate how many moles of AgNO3 are needed to produce 3.17 g Ag2CO3 AgNO3 + K2CO3  Ag2CO3 + KNO3

Example Balance the chemical equation Identify the type of reaction Calculate how many grams of Al are needed to produce 0.045 g H2 Al (s) + HCl (aq)  AlCl3 (aq) + H2 (g)

2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g) Example 4-6B The model problem describes an Al-Cu alloy composed of 93.7% Al and 6.3% Cu by mass, with a density of 2.85 g/cm3. The Al (but not the Cu) reacts with HCl: 2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g) How many grams of Cu are present in a sample of alloy that yields 1.31 g H2 when it reacts with HCl?

Example 4-7A The model problem describes an HCl solution which is 28% HCl by mass and has a density of 1.14 g/mL. It reacts with Al: 2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g) How many mg of H2 are produced when 1 drop (0.05 mL) of the HCl solution reacts with Al?

Example 4-7B A vinegar contains 4.0% HC2H3O2 by mass and has a density of 1.01 g/mL. It reacts with sodium hydrogen carbonate: HC2H3O2 (aq) + NaHCO3 (s)  NaC2H3O2 (aq) + H2O (l) + CO2 (g) How many grams of CO2 are produced by the reaction of 5.00 mL of this vinegar with NaHCO3?

Chemical Reactions in Solution Most reactions occur in aqueous solution SOLUTE is the substance to be dissolved in solution SOLVENT is the substance (often a liquid) the solute dissolves in The concentration of the solution is Molarity (M) = moles solute L solution

Example 4-8A If 22.3 g acetone, (CH3)2CO, are dissolved in enough water to make 1.25 L of solution, what is the concentration (M) of the solution?

Example 4-8B 15.0 mL of concentrated acetic acid, HC2H3O2 (d = 1.048 g/mL), are dissolved in enough water to produce 500.0 mL of solution. What is the concentration of the solution?

Example 4-9A At 25 °C, an aqueous solution saturated with NaNO3 is 10.8 M NaNO3. How many grams of NaNO3 are present in 125 mL of this solution?

Example 4-9B How many grams of Na2SO4 • 10 H2O are needed to prepare 355 mL of 0.445 M Na2SO4?

Dilution problems It is common to prepare a solution by diluting a more concentrated solution (the stock solution). The moles of solute taken from the stock solution are given by moles solute = volume x molarity All the solute taken from the stock appears in the diluted solution, so moles solute are constant: VstockMstock = VdiluteMdilute

Example 4-10A 15.00 mL of 0.450 M K2CrO4 solution are diluted to 100.00 mL. What is the concentration of the dilute solution?

Example 4-10B After being left out in an open beaker, 275 mL of 0.105 M NaCl has evaporated to only 237 mL. What is the concentration of the solution after evaporation?

Stoichiometry in Solution Stoichiometry in solution is just the same as for mass problems, except the conversion into or out of moles uses molarity instead of molar mass: grams A mL A moles A moles B grams B mL B mole ratio

Example 4-11A K2CrO4 (aq) + 2 AgNO3 (aq)  Ag2CrO4 (s) + 2 KNO3 (aq) How many mL of 0.250 M K2CrO4 must react with excess AgNO3 to produce 1.50 g Ag2CrO4?

Example 4-11B K2CrO4 (aq) + 2 AgNO3 (aq)  Ag2CrO4 (s) + 2 KNO3 (aq) How many mL of 0.150 M AgNO3 must react with excess K2CrO4 to produce exactly 1.00 g Ag2CrO4?

Limiting reactant In a given reaction, often there is not enough of one reactant to use up the other reactant completely The reactant in short supply LIMITS the quantity of product that can be formed

Goldilocks Chemistry Imagine reacting different amounts of Zn with 0.100 mol HCl: Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g) Rxn 1 Rxn 2 Rxn 3 Mass Zn 6.54 g 3.27 g 1.31 g Moles Zn 0.100 mol 0.0500 mol 0.0200 mol Moles HCl 0.100 mol 0.100 mol 0.100 mol Ratio mol HCl 1.00 2.00 5.00 mol Zn

Limiting reactant problems The easiest way to do these is to do two stoichiometry calculations Find the amount of product possible from each reactant The smaller answer is the amount of product you can actually make (you just ran out of one reactant) The reactant on which that answer was based is the limiting reactant

Example 4-12A How many grams of PCl3 form when 215 g P4 react with 725 g Cl2: P4 (s) + 6 Cl2 (g)  4 PCl3 (l)

Example 4-12B How many kg of POCl3 form if 1.00 kg of each reactant are allowed to react: 6 PCl3 (l) + 6 Cl2 (g) + P4O10 (s)  10 POCl3 (l)

Example 4-13A When 215 g P4 react with 725 g Cl2 P4 (s) + 6 Cl2 (g)  4 PCl3 (l) (example 4-12A) which reactant is in excess and what mass of that reactant remains after the reaction is finished?

Example 4-13B 12.2 g H2 and 154 g O2 are allowed to react. Identify the limiting reactant, which gas remains after the reaction, and what mass of it is left over. 2 H2 (g) + O2 (g)  2 H2O (l)

Percent Yield In real experiments we often do not get the amount of product we calculate we should, because the reactants may participate in other reactions (side reactions) that produce other products (by-products) The reaction often does not go to completion. Percent yield tells the ratio of actual to theoretical amount formed.

Percent Yield Suppose you calculate that a reaction will produce 50.0 g of product. This is the theoretical yield. The reaction actually produces only 45.0 g of product . This is the actual yield. Percent yield = 45.0 g (actual) x 100 = 90.0% 50.0 g (theoretical)

Example 4-14A If 25.7 g CH2O is produced per mole CH3OH that reacts, what are the theoretical, actual, and percent yield: CH3OH (g)  CH2O (g) + H2 (g)

Example 4-14B What is the percent yield if 25.0 g P4 reacts with 91.5 g Cl2 to produce 104 g PCl3: P4 (s) + 6 Cl2 (g)  4 PCl3 (l)

Example 4-15A If the observed percent yield for the formation of urea is 87.5%, what mass of CO2 must react with excess NH3 to produce 50.0 g CO(NH2)2: 2 NH3 (g) + CO2 (g)  CO(NH2)2 (s) + H2O (l)

Example 4-15B What mass of C6H11OH should you start with to produce 45.0 g C6H10 if the reaction has 86.2% yield and the C6H11OH is 92.3% pure: C6H11OH (l)  C6H10 + H2O (l)

Exercise 26 Balance these equations by inspection (NH4)2Cr2O7 (s)  Cr2O3 (s) + N2 (g) + H2O (g) NO2 (g) + H2O (l)  HNO3 (aq) + NO (g) H2S (g) + SO2 (g)  S (g) + H2O (g) SO2Cl2 + HI  H2S + H2O + HCl + I2

Exercise 30 Write balanced equations for these reactions: Sulfur dioxide gas with oxygen gas to produce sulfur trioxide gas Solid calcium carbonate with water and dissolved carbon dioxide to produce aqueous calcium hydrogen carbonate Ammonia gas and nitrogen monoxide gas to produce nitrogen gas and water vapor

Exercise 32 3 Fe (s) + 4 H2O (g)  Fe3O4 (s) + H2 (g) How many moles of H2 can be produced from 42.7 g Fe and excess steam? How many grams of H2O are consumed in the conversion of 63.5 g Fe to Fe3O4? If 7.36 mol H2 are produced, how many grams of Fe3O4 must also be produced?

Exercise 36 Silver oxide decomposes above 300 °C to yield metallic silver and oxygen gas. 3.13 g impure silver oxide yields 0.187 g O2. Assuming there is no other source of O2, what is the % Ag2O by mass in the original sample?

Exercise 42 How many grams of CO2 are produced in the complete combustion of 406 g of a bottled gas that consists of 72.7% C3H8 (propane) and 27.3% C4H10 (butane), by mass?

Exercise 45 What are the molarities of these solutes? 150.0 g sucrose (C12H22O11) in 250.0 mL aqueous solution 98.3 mg of 97.9% pure urea, CO(NH2)2, in 5.00 mL aqueous solution 12.5.0 mL methanol (CH3OH, density = 0.792 g/mL) in 15.0 L aqueous solution

Exercise 52 After 25.0 mL of aqueous HCl solution is diluted to 500.0 mL, the concentration of the diluted solution is found to be 0.085 M HCl. What was the concentration of the original HCl solution?

Exercise 56 Ca(OH)2 (s) + 2 HCl (aq)  CaCl2 (aq) + 2 H2O (l) How many grams of Ca(OH)2 will react completely with 415 mL of 0.477 M HCl? How many kilograms of Ca(OH)2 will react with 324 L of an HCl solution that is 24.28% HCl by mass, density = 1.12 g/mL?

Exercise 63 0.3126 g oxalic acid, H2C2O4, is exactly neutralized by 26.21 mL of a NaOH solution. What is the concentration of the NaOH solution? H2C2O4 + 2 NaOH  Na2C2O4 + 2 H2O

Exercise 70 Chlorine can be generated by heating calcium hypochlorite and hydrochloric acid to form chlorine gas, calcium chloride, and water. If 50.0 g Ca(OCl)2 and 275 mL 6.00 M HCl react, how many grams of Cl2 gas form? Which reactant is left over, and how much (in grams)?

Exercise 72 2 C6H5NO2 + 4 C6H14O4  (C6H5N)2 + 4 C6H12O4 + 4 H2O nitrobenzene triethylene azobenzene glycol If 0.10 L nitrobenzene (d = 1.20 g/mL) react with 0.30 L triethylene glycol (d = 1.12 g/mL) to form 55 g azobenzene, find Theoretical yield Actual yield Percent yield

Stoichiometry grams A grams B L of moles A moles B gas A gas B mL of sol’n A moles A moles B grams B sol’n B mole ratio L of gas A gas B

Stoichiometry grams A mL A moles A moles B grams B mL B mole ratio