Unit 4: Reactions and Stoichiometry

Transcription

1 Unit 4: Reactions and Stoichiometry Reactions Chemical equation Expression representing a chemical reaction Formulas of reactants on the left side Formulas of products on the right side Arrow(s) connect(s) reactants and products Examples Symbols (s): in solid state (l): in liquid state (g): in gaseous state (aq): an aqueous solution (substance dissolved in H 2 O) or : heat is supplied to the reaction : use of a catalyst (MnO 2 in this case) : yields or produces : reversible reaction Catalyst Speeds up a reaction by lowering the activation energy Not used up in a chemical reaction Skeleton equation Chemical equation that does not indicate the relative amounts of reactants and products Does not necessarily follow the Law of Conservation of Mass Balanced equation Chemical equation in which mass is conserved Each side of the equation has the same number of atoms of each element. NOTE: DO NOT CHANGE FORMULAS OF COMPOUNDS TO MAKE AN EQUATION BALANCED. To balance an equation, write coefficients in front that apply to each atom of the element.

2 Iron (III) chloride reacts with sodium hydroxide to produce iron (III) hydroxide and sodium chloride. Skeleton equation Iron (III) chloride» Iron (III) ion: Fe 3+» Chloride ion: Cl» Formula: FeCl 3 Sodium hydroxide» Sodium ion: Na +» Hydroxide ion: OH» Formula: NaOH Iron (III) hydroxide» Iron (III) ion: Fe 3+» Hydroxide ion: OH» Formula: Fe(OH) 3 Sodium chloride» Sodium ion: Na +» Chloride ion: Cl» Formula: NaCl FeCl 3 + NaOH Fe(OH) 3 + NaCl Chemical equation Write coefficients in front that apply to each atom of the element. Balance to make equal numbers of same atoms on each side. DO NOT CHANGE FORMULAS OF COMPOUNDS TO MAKE AN EQUATION BALANCED! Skeleton: FeCl 3 + NaOH Fe(OH) 3 + NaCl Balanced: FeCl 3 + 3NaOH Fe(OH) 3 + 3NaCl Copper (II) sulfide reacts with oxygen gas to produce copper and sulfur dioxide. Skeleton equation Copper (II) sulfide» Copper (II) ion: Cu 2+» Sulfide ion: S 2» Formula: CuS Sulfur dioxide: SO 2 CuS + O 2 Cu + SO 2 Chemical equation Write coefficients in front that apply to each atom of the element. Balance to make equal numbers of same atoms on each side. DO NOT CHANGE FORMULAS OF COMPOUNDS TO MAKE AN EQUATION BALANCED! Skeleton: CuS + O 2 Cu + SO 2 Balanced: CuS + O 2 Cu + SO 2 Oxygen gas: O 2 Copper: Cu

3 Unit 4: Reactions and Stoichiometry Types of Chemical Reactions Classifying reactions Knowing what kind of reaction is occurring can help you to predict the products of the reaction Three general reaction types Acid-base reaction Oxidation-reduction reaction Precipitation reaction Acid-base reaction (general AND specific reaction type) Chemical change in which a salt (ionic compound) and water are produced Form: HA + BOH BA + H 2 O Hints Reactants Acid and base present Can involve transfers of hydrogen (hydronium) ions or electrons (to be discussed next term) Products One ionic compound and water Examples HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) HNO 3 (aq) + LiOH(aq) LiNO 3 (aq) + H 2 O(l) H 3 PO 4 (aq) + 3KOH(aq) K 3 PO 4 (aq) + 3H 2 O(l)

4 Precipitation reaction (general reaction type) Produces a precipitate (insoluble salt resulting from a reaction) Double-replacement reaction (specific reaction type) Chemical change that involves an exchange of positive ions between two compounds Form: AB + CD CB + AD Hints Reactants» Generally two aqueous, ionic compounds Products» Aqueous compound» Precipitate Insoluble salt that is formed based on rules of solubility (back of periodic table) Generally a solid Barium chloride and potassium carbonate solutions react.» BaCl 2 (aq) + K 2 CO 3 (aq)» BaCl 2 (aq) + K 2 CO 3 (aq) BaCO 3 (s) + 2KCl(aq)» BaCO 3 is the precipitate. Sodium hydroxide and lead (II) nitrate solutions react.» NaOH(aq) + Pb(NO 3 ) 2 (aq)» 2NaOH(aq) + Pb(NO 3 ) 2 (aq) 2NaNO 3 (aq) + Pb(OH) 2 (s)» Pb(OH) 2 is the precipitate. Oxidation-reduction reaction (general reaction type) Reaction in which one or more electrons are transferred Often referred to as redox reactions Involves assigning oxidation states/numbers Concept that provides a way to keep track of electrons in redox reactions according to certain rules Rules for assigning oxidation states/numbers The oxidation number of an element in an elementary substance is 0.» Examples The oxidation number of chlorine in Cl 2 or of phosphorus in P 4 is 0.

5 The oxidation number of Fe by itself is 0. The oxidation number of an element in a monatomic ion is equal to the charge of that ion.» Examples In the ionic compound NaCl, sodium has an oxidation number of +1 and chlorine has an oxidation number of 1. In the ionic compound FeBr 3, the oxidation number of the bromide ion (Br - ) is 1 while the oxidation number of the iron (III) ion (Fe 3+ ) is +3. The oxidation number of hydrogen in a compound is +1, except in metal hydrides (i.e., NaH) where it is 1. The oxidation number of oxygen in a compound is 2. except in peroxides (i.e., H 2 O 2 ) where it is 1. For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal 0. For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion. Meaning of oxidation and reduction Oxidation Classical definition: combination of an element with oxygen to produce oxides Modern definition: complete or partial loss of electrons or gain of oxygen Reduction Classical definition: loss of oxygen from a compound Modern definition: complete or partial gain of electrons or loss of oxygen LEO says GER (lose electrons = oxidation; gain electrons = reduction) Oxidizing agent: reactant that accepts electrons from another reactant Reducing agent: reactant that donates electrons to another reactant Example: 2PbS(s) + 3O 2 (g) 2PbO(s) + 2SO 2 (g) Identify oxidation states of each atom (each) 2PbS(s) + 3O 2 (g) 2PbO(s) + 2SO 2 (g) Oxidized element: sulfur (losing electrons, which is why it becomes more positive) Reduced element: oxygen (gaining electrons, which is why it becomes more negative) Oxidizing agent: O 2 (since it causes sulfur to lose electrons) Reducing agent: PbS (since it causes oxygen to gain electrons)

6 Types of oxidation-reduction reactions Synthesis/Combination reaction (specific reaction type) Chemical change in which two or more substances react to form a single new substance Form: A + B AB Hints Reactants» Group A metal reacting with a nonmetal, two elements reacting, metallic or nonmetallic oxides reacting with water, or element reacting with a diatomic solid/liquid/gas Products» Only one product formed 2K(s) + Cl 2 (g) 2KCl(s) S(s) + O 2 (g) SO 2 (g) 2S(s) + 3O 2 (g) 2SO 3 (g) Fe(s) + S(s) FeS(s) 2Fe(s) + 3S(s) 2Fe 2 S 3 (s) SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) CaO(s) + H 2 O(l) Ca(OH) 2 (aq) Decomposition reaction (specific reaction type) Chemical change in which a single compound is broken down into two or more simpler products Form: AB A + B Hints Reactants» Single reactant Catalysts» Typically require heat, light, or electricity Products» Two or more products formed» Difficult to predict products

7 Single-replacement reaction (specific reaction type) Chemical change in which one element replaces a second element in a compound Forms A + BC B + AC A + BC C + BA Hints Reactants» One ionic compound and an element or diatomic gas/liquid/solid Products» One ionic compound and an element or diatomic gas/liquid/solid Activity series of metals A table listing metals in order of decreasing reactivity Metals will displace other metals Activity Series of Metals (Decreasing reactivity) > Li K Ca Na Mg Al Zn Fe Pb H Cu Hg Ag Examples Mg + 2HCl MgCl2 + H2 Cu + HCl no reaction 2Fe + 6HCl 2FeCl3 + 3H2 Combustion reaction (specific reaction type) Chemical change in which oxygen reacts with another substance, often producing energy in the form of heat and light Commonly involve hydrocarbons (compounds with hydrogen and carbon) Hints Reactants» Oxygen is one reactant. 2Mg(s) + O 2 (g) 2MgO(s)» Also a synthesis reaction» Combustion reactions can also be other types. S(s) + O 2 (g) SO 2 (g) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) 2CH 3 OH(l) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(g)

8 Chapter 9: Stoichiometry The Arithmetic of Equations Chemical Calculations Limiting Reagent and Percent Yield The Arithmetic Of Equations -- The Arithmetic of Equations -- Using Everyday Equations Stoichiometry Part of chemistry that deals with numerical relationships in chemical reactions Calculation of quantities of substances involved in chemical equations Recipes: proportional amounts Interpreting chemical equations Information that can be seen or derived from chemical equations Particles Moles Mass Volume -- The Arithmetic of Equations -- Conservation Always conserved Mass Atoms Sometimes conserved Molecules Formula units Moles Volumes of gases

9 -- The Arithmetic of Equations -- Example For the reaction 2H 2 (g) + O 2 (g) 2H 2 O(g), interpret this reaction in terms of the following three relative quantities. Number of representative particles» 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to produce 2 molecules of water vapor. Number of moles» 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor. Masses of reactants and products -- The Arithmetic of Equations -- Masses of reactants and products» In order to relate masses of reactants and products, you must multiply the number of moles of each reactant and product by its molar mass.» H 2 : 2(1.0079) = g/mol» O 2 : 2(15.999) = g/mol» H 2 O: 2(1.0079) + 1(15.999) = g/mol» Reactants: 2H 2 + O 2 = 2(2.0158) = g» Products: 2H 2 O = 2( ) = g» g = g» The Law of Conservation of Mass is preserved. Chemical Calculations Chemical calculations For all calculations involving two chemicals, you need to know the BALANCED chemical reaction involved. Mole to mole calculations If you know the number of moles for a compound in one part of an equation, you can determine the number of moles for a different compound in another part of the equation (using molar ratio). For the reaction, 2H 2 (g) + O 2 (g) 2H 2 O(g), how many moles of oxygen gas must react with hydrogen gas in order to get 0.85 mol water?» For every two moles of water, there is one mole of oxygen gas.

10 Iron metal reacts with oxygen gas to produce iron (III) oxide. How many moles of oxygen gas are needed to react fully with 4.65 moles of iron metal?» Fe + O 2 Fe 2 O 3 (unbalanced)» 4Fe + 3O 2 2Fe 2 O 3 (balanced)» For every three moles of oxygen gas, there are four moles of iron metal. Mass to mole calculations; mole to mass calculations If you know the mass for a compound in one part of an equation, you can determine the mass for a different compound in another part of the equation by first converting to moles and then back to mass (using molar ratio again). Mass A Moles A Moles B Mass B Hydrochloric acid reacts with iron (III) oxide to produce iron (III) chloride and water. Calculate the number of grams of iron (III) oxide needed to react with excess hydrochloric acid in order to produce 4.68 g water.» HCl + Fe 2 O 3 FeCl 3 + H 2 O (unbalanced)» 6HCl + Fe 2 O 3 2FeCl 3 + 3H 2 O (balanced)» Since mass is involved with the two chemical described, find the molar mass of the involved chemicals. H 2 O: 2(1.0079) = g/mol Fe 2 O 3 : 2(55.847) + 3(15.999) = g/mol» Like the calculations you did in Chapter 7, always convert to moles first. For every three moles of water, there is one mole of iron (III) oxide. Tin reacts with hydrofluoric acid to produce tin (II) fluoride and hydrogen gas. Calculate the number of grams of tin needed to completely react with 10.2 grams of hydrofluoric acid.» Sn + HF SnF 2 + H 2 (unbalanced)» Sn + 2HF SnF 2 + H 2 (balanced)» Since mass is involved with the two chemicals described, find the molar mass of the involved chemicals. Sn: g/mol HF: 1(1.0079) + 1(18.998) = g/mol» Like the calculations you did in Chapter 7, always convert to moles first. For every one mole of tin, there are two moles of hydrofluoric acid.

11 Other calculations ALWAYS convert to moles first (if not already in moles). If mass is involved, use molar mass. If volume at STP is involved, use molar volume (22.4 L/mol). If representative particles are involved (i.e., atoms, ions, formula units, molecules), use Avogadro s number ( particles/mol). Units A Moles A Moles B Units B Tin reacts with hydrofluoric acid to produce tin (II) fluoride and hydrogen gas. How many grams of SnF 2 can be made by reacting molecules of hydrofluoric acid with excess tin?» Sn + HF SnF 2 + H 2 (unbalanced)» Sn + 2HF SnF 2 + H 2 (balanced)» Since mass of SnF 2 is involved, find the molar mass of SnF 2. Sn: 1(118.69) + 2(18.998) = g/mol» Since molecules (representative particles) are involved for HF, Avogardo s number will be used ( molecules HF/mol HF).» For every one mole of SnF 2, there are two moles of hydrofluoric acid.» Since volume is involved for O 2, molar volume will be used (22.4 L O 2 /mol O 2 ).» For every one mole of O 2, there are two moles of NO. Nitrogen monoxide gas and oxygen gas combine to form the brown gas nitrogen dioxide. How many liters of oxygen gas are required to react fully with 7.8 grams of nitrogen monoxide gas at standard temperature and pressure?» NO + O 2 NO 2 (unbalanced)» 2NO + O 2 2NO 2 (balanced)» Since mass of NO is involved, find the molar mass of NO. NO: 1(14.007) + 1(15.999) = g/mol

12 Limiting Reagent And Percent Yield Limiting reagent Any reactant that is used up first in a chemical reaction Determines the amount of product that can be formed in the reaction Excess reagent Any reactant that remains after the limiting reagent is used up in a chemical reaction Reagent present in a quantity that is more than sufficient to react with a limiting reagent Hints If you have more than you calculate you need, that is the excess reagent. If you calculate you need more than you have, that is the limiting reagent. Water can be prepared by the reaction of hydrogen gas with oxygen gas as shown in the equation: 2H 2 (g) + O 2 (g) 2H 2 O(l). Suppose that 6.70 moles of hydrogen gas react with 3.20 moles of oxygen gas. What is the limiting reagent, and how many moles of water are produced? First, calculate how many moles of each are needed compared to the other chemical. Second, once you have determined which is the limiting reagent, use the amount of limiting reagent to figure out the other amounts of chemicals as asked. For every one mole of O 2, there are two moles of H 2. Amount of hydrogen gas needed for the shown amount of oxygen gas: Since the calculation shows you need 6.40 mol H 2 but you have 6.70 mol H 2 (more than you need), H 2 must be the excess reagent, meaning O 2 is the limiting reagent. You could begin with the amount of H 2 instead and figure out the same answer. Since the calculation shows you need 3.35 mol O 2 but you have only 3.20 mol O 2 (less than you need), H 2 must be the excess reagent, meaning O 2 is the limiting reagent. (You can calculate either way.)

13 Since oxygen gas is the limiting reagent, any calculations predicting other chemical reactions will involve the 3.20 mol O 2. Also, you know that there are 2 moles of H 2 O for every 1 mole of O 2. Aluminum metal reacts with aqueous copper (II) sulfate to produce copper metal and aluminum sulfate grams of aluminum react with 13.5 grams of copper (II) sulfate. (A) What is the limiting reagent? (B) How many grams of excess reagent are there? (C) What is the maximum number of grams of aluminum sulfate that can be produced? Al(s) + CuSO 4 (aq) Al 2 (SO 4 ) 3 (aq) + Cu(s) (Unbalanced) 2Al(s) + 3CuSO 4 (aq) Al 2 (SO 4 ) 3 (aq) + 3Cu(s) (Balanced) First, calculate how many grams of each are needed compared to the other chemical. You will need the molar mass of each chemical to do this. Al: g/mol CuSO 4 : 1(63.546) + 1(32.06) + 4(15.999) = g/mol Amount of aluminum needed for the shown amount of copper (II) sulfate: To figure out the amount of excess reagent, take the amount you have of aluminum and minus how much you need g Al 1.52 g Al = 1.28 g Al Since the calculation shows you need 1.52 g Al but you have 2.80 g Al (more than you need), Al must be the excess reagent, meaning CuSO 4 is the limiting reagent. (You could begin with the amount of CuSO 4 instead and figure out the same answer.) Since copper (II) sulfate is the limiting reagent, any calculations predicting other chemical reactions will involve the 13.5 g CuSO 4. There is 1 mole of Al 2 (SO 4 ) 3 for every 3 moles of CuSO 4. Molar mass of Al 2 (SO 4 ) 3 : 2(26.982) + 3(32.06) + 12(15.999) = g/mol

14 (A) (B) (C) What is the limiting reagent? CuSO 4 How many grams of excess reagent are there? 1.28 g Al What is the maximum number of grams of aluminum sulfate that can be produced? 9.65 g Al 2 (SO 4 ) 3 For the reaction N 2 + 3H 2 2NH 3, you are provided 15.2 g of nitrogen gas and 16.8 g of hydrogen gas to react to produce ammonia (NH 3 ). (A) What is the limiting reagent? (B) How many grams of excess reagent are there? (C) What is the maximum number of grams of ammonia that can be produced? First, calculate how many grams of each are needed compared to the other chemical. You will need the molar mass of each chemical to do this. H 2 : 2(1.0079) = g/mol N 2 : 2(14.007) = g/mol Amount of hydrogen gas needed for the shown amount of nitrogen gas: Since the calculation shows you need 3.28 g H 2 but you have 16.8 g H 2 (more than you need), H 2 must be the excess reagent, meaning N 2 is the limiting reagent. (You could begin with the amount of H 2 instead and figure out the same answer.) 16.8 g H g H 2 = g H 2 = 13.5 g H 2 Since nitrogen gas is the limiting reagent, any calculations predicting other chemical reactions will involve the 15.2 g N 2. There are 2 moles of NH 3 for every 1 mole of N 2. Molar mass of NH 3 : 1(14.007) + 3(1.0079) = = g/mol To figure out the amount of excess reagent, take the amount you have of hydrogen gas and minus how much you need.

15 (A) (B) (C) What is the limiting reagent? N 2 How many grams of excess reagent are there? 13.5 g H 2 What is the maximum number of grams of ammonia that can be produced? 18.5 g NH 3 Theoretical yield Amount of product that could theoretically form during a reaction Calculated based on the limiting reagent and represents the maximum amount of product formed from the limiting reagent Actual yield Amount of product that forms when a reaction is carried out in a laboratory Amount usually given in a problem as actually being formed Percent yield Ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage Serves as a measure of the efficiency of a reaction Formula for percent yield or