CHEM 101_L3_Chemical Reactions & Stoichiometry (1) PDF

Chemical Reactions & Stoichiometry Dr Disah Mpadi Office: Block 1, Room 16 Email: [email protected] 1  Outline: ▪ Chemical reactions & balancing chemical equations ▪ Chemical equations & stoichiometric calculations: ▪ Percent compo...

Chemical Reactions & Stoichiometry Dr Disah Mpadi Office: Block 1, Room 16 Email: [email protected] 1  Outline: ▪ Chemical reactions & balancing chemical equations ▪ Chemical equations & stoichiometric calculations: ▪ Percent composition of compounds & determining the formula of compound ▪ Chemical reactions in solution ▪ The concept of limiting reactant ▪ Gas stoichiometry ▪ The extent of reaction 2  Introduction ▪ Basic processes that are chemical reactions: ▪ Rusting iron & burning of natural gas (cooking gas). ▪ Reaction Stoichiometry: Establishes a quantitative NaOH(aq) + HCl(aq) → NaCl(aq) + H2O (numerical) relationships between the substances involved in a chemical reaction. We plan experiments, ▪ It provides the relationships required to relate macroscopic amounts (grams or litres; at ordinary lab scale) of substances analyse their to microscopic (atoms, ions, or molecules; can’t measure for results, & make ordinary lab experiments) view of chemical reactions. predictions using ▪ Solution stoichiometry: quantitative analysis of chemical stoichiometric reactions that occur in solution. principles. ▪ Solution molarity: A method of describing the composition of a solution. 3  Chemical Reactions & Balancing Chemical Equations Chemical equation: Normally formulas ▪ Chemical reaction: involves a process in which one (symbols) for compounds or elements are set of substances (called reactants), is converted to a used, also names can be used: new set of substances (products) or simply a process by which a chemical change or transformation The two sides (reactants & products) are occurs. joined by an arrow (reaction go only in the forward direction). ▪ Some physical evidence if a reaction has happened: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O ▪ a change in colour Reactants on the Left Products on the Right ▪ evolution of a gas ▪ formation of a precipitate (solid) within a clear solution Reversible reaction(s): If products ▪ evolution or absorption of heat react (partially or completely) to re- form the original reactants, then a double arrow is used ( or ⇌ or ⇆): ▪ A detailed chemical analysis can be conducted with some special techniques to elucidate all substances N2(g) + 3H2(g) ⇌ 2NH3(g) in a reaction mixture, in the absence of physical signs. 4  Cont’d Balancing chemical equations: ▪ Law of conservation: Atoms are neither created nor destroyed in a chemical reaction, therefore, chemical equations need to be balanced. ▪ Balanced chemical equation: the total number of atoms of each element present on both sides of the equation should be the same. ▪ Stoichiometric coefficients: the coefficients needed to balance a chemical reaction; & are essential in a variety of calculations. ▪ A chemical/reaction equation can be balanced only by adjusting the coefficients of formulas. 5  Cont’d Balancing chemical/reaction equations by inspection: ▪ Balancing by inspection: means we adjust stoichiometric coefficients by trial & error until a balanced condition is found, & the following are some useful strategies: ▪ Check or inspect if the reaction equation (whether chemical or mathematical-must have the left & right sides equal) a) NaOH(aq) + HCl(aq) ⟶ NaCl(aq) + H2O needs to be balanced. b) 2 NO(g) + O2 ⟶ 2 NO2 ▪ If an element occurs in only one cpd on each side of the equation, try balancing this element first. ▪ When one of the reactants or products exists as the free The term chemical equation automatically element, balance this element last. signifies that the balance exists ▪ In some reactions, certain groups of atoms (e.g., polyatomic ions) remain unchanged & in such cases, balance these 𝟏 [H2 (g) + 𝟐O2(g) ⟶ H2O (g)] x2; eqn 1 groups as a unit or start with the most complicated chemical formula. 2 H2 (g) + O2(g) ⟶ 2 H2O (g); eqn 2 ▪ It is allowed to use fractional or integer numbers (can be mixed) as coefficients; if desired fractions can be cleared by multiplying ALL coefficients by a common multiplier. 6  Cont’d Overall, C & H: C6H14O4 + O2 → 6 CO2 + 7 H2O Now, the right side of the expression has 19 O Example: Writing & balancing an equation atoms (12 in six CO2 molecules & 7 in seven H2O molecules); the left-hand side has only 4 O ▪ Liquid triethylene glycol, C6H14O4, used as solvent & atoms (in C6H14O4). To obtain 15 more O atoms plasticizer for vinyl & polyurethane plastics. Write a 𝟏𝟓 requires a fractional coefficient of 𝟐 for O2 balanced chemical equation for the combustion of this cpd in a plentiful supply of oxygen. (remember we balance elemental reactant or product last): Analyse: 𝟏𝟓 Balance O: C6H14O4 + 𝟐 O2 → 6 CO2 + 7 H2O When a carbon-hydrogen-oxygen cpd is burned in excess oxygen, O2, the products are CO2 & H2O. (the equation is balanced) N.B: To remove the fractional coefficient, Solve: multiply all coefficients by two: Having identified the reactants & products, write down an unbalanced chemical expression for the reaction, showing 2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O, Balanced all reactants & products, & then balance the expression with respect to each kind of atom: Assess: Check if the equation is balanced by determining the numbers of C, H, & O atoms Starting expression: C6H14O4 + O2 → CO2 + H2O that appear on each side of equation are equal. Balance C: C6H14O4 + O2 → 6 CO2 + H2O Balance H: C6H14O4 + O2 → CO2 + 7 H2O 7  Cont’d States of Matter & Reaction Conditions: Decomposition reaction: a substance is broken down into simpler substances (i.e., its ▪ State of matter: a physical form ((g)-gas, elements): (l)-liquid, (s)-solid, & (aq)-aqueous solution) of reactants & products is ∆ 2 Ag2O(s) → 4 Ag(s) + O2: essential to include in a chemical equation & are generally shown by (∆, delta, means heat/high temperature) symbols in parentheses. Synthesis reaction: a new cpd is formed from the reaction of two or more simpler substances ▪ Reaction conditions: often written above (reactants or starting materials): or below the arrow in a chemical equation. 350oC CO(g) + 2 H2(g) 340 atm CH3OH(g) ZnO, Cr2 O3 ▪ This information is necessary to carry out the chemical reaction in a lab or chemical plant. 8  Chemical Equations & Stoichiometric Calculations 2 H2(g) + O2(g) ⟶ 2 H2O(l) ▪ Stoicheion (Greek word)-means Therefore; element. 2𝒙 molecules H2 + 𝒙 molecules O2 ⟶ 2𝒙 molecules H2O; ▪ Stoichiometry: literally means to say 𝒙 = 6.02214 𝑥 1023 particles/mol(Avogadro’s measure the elements, but number); practically it includes all the quantitative relationships involving Then 𝒙 molecules represent 1 mole; thus, the chemical atomic & formula masses, equation also means that: chemical formulas, & chemical 2 mol H2 + 1 mol O2 ⟶ 2 mol H2O equations. Therefore, we can make the following statements from the coefficients of the chemical equation: ▪ Stoichiometric factor (or Mole ratio): relates the amounts, on a ✓ Every 2 moles of H2 consumed, produces 2 moles of H2O mole basis, of any two substances ✓ Every 1 mole of O2 consumed, produces 2 moles of H2O involved in a chemical reaction. ✓ Every 1 mole of O2 consumed, consume 2 moles of H2 9  Cont’d Balanced chemical equation to find appropriate mole ratio Information given Further about about known known A A&& stoichiometric coefficient of B step(s) Moles of A 𝑥 Moles of B Preliminary step(s) stoichiometric coefficient of A leading to leading to moles of A the answer Stoichiometric factor or Mole ratio i.e., using mass of known substance i.e., obtaining mass of desired substance Note: Include units & work from a balanced chemical equation when solving stoichiometric problems. Generalised stoichiometry diagram: ▪ The conversion pathway is from the known A to the unknown B. 10  Cont’d Example: Relating the numbers of moles of reactant & product Calculate the number moles of CO2 that are produced in the combustion of 2.72 mol of triethylene glycol, C6H14O4, in an excess of O2. Analyse: An excess of O2 means the complete conversion of the organic matter, C6H14O4, to CO2 & H2O. The factor for converting from moles of C6H14O4 to moles of CO2 is obtained from the balanced equation for the combustion reaction. Solve: The first step in a stoichiometric calculation is to write a balanced equation for the reaction: 2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O Thus, 12 mol CO2 are produced for every 2 mol C6H14O4 burned or consumed; therefore, this is used as the conversion factor to obtain the required moles of CO2: 12 mol CO2 ? mol CO2 = 2.72 mol C6H14O4 x = 𝟏𝟔. 𝟑 𝐦𝐨𝐥 𝐂𝐎𝟐 2 mol C6H14O4 11 N.B: Can also be solved using two equal ratios approach, with one unknown  Cont’d Example: Relating the mass of a reactant & product Compute the mass of CO2 that is formed in the reaction of 4.16 g of triethylene glycol, C6H14O4, with an excess of O2. Analyse: The solution will involve several conversion factors: (1) to moles, (2) between moles, & (3) from moles. Chemical equation: 2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O Solve: The conversions can be performed by using either a stepwise approach or the conversion pathway approach. Using a conversion pathway approach (individual steps are combined), we proceed as follows: Converts to moles of CO2 Converts to grams of CO2 1 mol C6H14O4 12 mol CO2 44.009 g CO2 ? g CO2 = 4.16 g C6H14O4 x x x 150.173 g C6H14O4 2 mol C6H14O4 1 mol CO2 = 𝟕. 𝟑𝟏 𝐠 𝐂𝐎𝟐 Converts to moles 12 of C6H14O4  Cont’d Example: Relating the masses of two reactants to each other Calculate the mass of O2 consumed in the complete combustion of 6.86 g of triethylene glycol, C6H14O4. Analyse: The solution will involve several conversion factors: (1) to moles, (2) between moles, & (3) from moles. Chemical equation: 2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O Solve: Using a conversion pathway approach (individual steps are combined), we proceed as follows: 1 mol C6H14O4 15 mol O2 31.998 g O2 ? g O2 = 6.86 g C6H14O4 x x x = 𝟏𝟏. 𝟎 𝐠 𝐎𝟐 150.173 g C6H14O4 2 mol C6H14O4 1 mol O2 13  Cont’d Example: Conversion factors in a stoichiometric calculation: Volume, Density, & % composition An alloy used in aircraft structures consists of 93.7% Al & 6.3% Cu by mass. The alloy has a density of 2.85 g/cm3. A 0.691 cm3 piece of the alloy reacts with an excess of HCl(aq). If we assume that all the Al but none of the Cu reacts with HCl(aq), then calculate the mass of H2 obtained. Analyse: A simple approach is to note all conversion factors & use the conversion pathway approach. Chemical equation: 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) Solve: Using a conversion pathway approach (which reduces the likelihood of rounding errors), we proceed as follows: 2.85 g alloy 93.7 g Al 1 mol Al 3 mol H2 2.016 g H2 ? g H2 = 0. 691 cm alloy x 3 x x x x = 𝟎. 𝟐𝟎𝟕 𝐠 𝐇𝟐 1 cm3 alloy 100 g alloy 26.982 g Al 2 mol Al 1 mol H2 14  Cont’d Example: Conversion factors in a stoichiometric calculation: Volume, Density, & % composition of a solution A hydrochloric, HCl(aq), solution consists of 28.0% HCl by mass and has a density of 1.14 g/mL. Compute the volume of this solution required to react completely with 1.87 g Al. Analyse: First determine where to begin & the appropriate starting point is with the given information- 1.87 g Al. The goal of our calculation is a solution volume- mL HCl solution. Note all conversion factors (molar mass, stoichiometric coefficients, percent composition, & density) & use the conversion pathway approach. Chemical equation: 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) Solve: Using a conversion pathway approach (which reduces the likelihood of rounding errors), we proceed as follows: 1 mol Al 6 mol HCl 36.458 g HCl 100.0 g HCl soln 1 mL HCl soln ? mL HCl soln = 1.87 g Al x x x x x 26.982 g Al 2 mol Al 1 mol HCl 28.0 g HCl 1.14 g HCl soln = 𝟐𝟑. 𝟖 𝐦𝐋 𝐇𝐂𝐥 𝐬𝐨𝐥𝐧 N.B: Solution = soln 15  Chemical Reactions in Solution ▪ Solution = Solvent + Solute ▪ Solvent determines whether the solution exists as a solid, liquid, or gas. ▪ Solute dissolves in a solvent. ▪ We’ll focus our discussion to aqueous solutions where water is the solvent (a common solvent in general chemistry lab). ▪ The relative proportions of a solute & solvent in solution is expressed by a property called molarity. 16  Cont’d 𝐦𝐨𝐥𝐚𝐫𝐢𝐭𝐲 = amount of solute (in moles) Molarity or Molar concentration: volume of solution (in litres) 𝑚𝑜𝑙 𝑛 (𝑚𝑜𝑙) 𝑐 = ▪ Is defined as the amount of solute (in 𝐿 𝑉(𝐿) 𝑚𝑜𝑙 𝑛 (𝑚𝑜𝑙) moles, mol) per liter (L) of solution: 𝑀 𝐿 = 𝑉(𝐿) ▪ Units of molarity: (mol/L) or M (stands for Note: Molarity, 𝑐 or 𝑀, is a conversion factor for the term molar). converting from litres of solution to moles of solute (from which mass of solute required to produce a given solution can be calculated). ▪ The amount of solute in moles can be determined from measured quantities such mass (grams, g) 𝑚(𝑔) as mass of solute (use molar mass as a 𝑛 𝑚𝑜𝑙 = g = 𝑔 conversion factor) or volume of a liquid Molar mass ( mol ) 𝑀𝑀( 𝑚𝑜𝑙 ) solute (use density & molar mass as a conversion factor). 𝐝𝐞𝐧𝐬𝐢𝐭𝐲 = mass volume ▪ Note: the solution volume is converted 𝑔 𝑚(𝑔) from millilitres (mL) to litres (L). 𝜌 𝑚𝐿 = 𝑉(𝑚𝐿) 17  Cont’d Example: Calculating molarity from measured quantities A solution is prepared by dissolving 25.0 mL ethanol ethanol, CH3CH2OH (𝜌=0.789 g/mL), in enough water to produce 250.0 mL solution. Calculate the molarity of ethanol in the solution. Analyse: First, calculate how many moles of ethanol are in a 25.0 mL of pure ethanol using conversion 𝒏 factors (1st density, 2nd molar mass). The molarity of the solution is then calculated by using: 𝑴 =. 𝑽 Solve: ▪ The number of moles of ethanol in a 25.0 mL Using a conversion pathway approach (in a single line): 0.789 g CH3CH2OH 1 mol CH3CH2OH ? mol CH3CH2OH = 25.0 mL CH3CH2OH x x 1 mL CH3CH2OH 46.069 g CH3CH2OH = 𝟎. 𝟒𝟐𝟖 𝐦𝐨𝐥 CH3CH2OH 0.428162104 mol CH3CH2OH 𝐌𝐨𝐥𝐚𝐫𝐢𝐭𝐲 = = 𝟏. 𝟕𝟏 𝐌 CH3CH2OH 0.2500 L Soln 18  Cont’d Example: Calculating the mass of solute in a solution of known molarity Compute the mass of K2CrO4 needed to prepare exactly 0.2500 L (250.0 mL) of a 0.250 M K2CrO4 solution in water. Analyse: Use the conversion factors (1st molarity of solution, 2nd molar mass of K2CrO4). Calculate the result using the conversion pathway approach. Solve: 0.250 mol K2CrO4 194.188 g K2CrO4 ? g K2CrO4 = 0. 2500 L Soln x x 1 L Soln 1 mol K2CrO4 = 𝟏𝟐. 𝟏 𝐠 K2CrO4 19  Cont’d Example: Calculating the molarity of ions in a solution of known molarity Calculate the concentration of each type of ion in 0.50 M Co(NO3)2 solution. Analyse: Write a balanced chemical equation for dissolving the ions. Using stoichiometric ratios and molarity, one can obtain the molarity of ions. H2 O Chemical equation: Co(NO3)2(s) Co2+(aq) + 2NO3-(aq) Solve: 1 mol Co2+ ? M Co2+ = 0. 50 M Co(NO3)2 Soln x = 𝟎. 𝟓𝟎 𝐌 𝐂𝐨𝟐+ 1 mol Co(NO3)2 Soln − 2 mol NO 3 ? M NO3 = 0. 50 M Co(NO3)2 Soln x − = 𝟏. 𝟎 𝐌 𝐍𝐎− 1 mol Co(NO3)2 Soln 𝟑 20  Cont’d Example: Calculating the molarity of ions in a solution of known molarity If 500 mL of a 0.250 M CaCl2(aq) solution is mixed with 200 mL of a 0.500 M HCl(aq)solution, calculate the molarity (M) of the Cl-(aq) ion in the resulting solution. Analyse: Write a balanced chemical equation for dissolving the ions. Using stoichiometric ratios and molarity, one can obtain the total number of moles of Cl- ions contributed by each solution and then divide this moles by the total volume of mixed solutions to obtain molarity of ions. Chemical equation 1: CaCl2(aq) → Ca2+(aq) + 2Cl-(aq) Chemical equation 2: HCl(aq) → H+(aq) + Cl-(aq) Solve: 2 mol Cl− ?𝑛 Cl− = (0. 250 M x 500 x 10−3 L) CaCl2 Soln x = 𝟎. 𝟐𝟓𝟎 𝐦𝐨𝐥 𝐂𝐥− 1 mol CaCl2 Soln 1 mol Cl− ?𝑛 Cl− = (0. 500 M x 200 x10−3 L) HCl Soln x = 𝟎. 𝟏𝟎𝟎 𝐦𝐨𝐥 𝐂𝐥− 1 mol HCl Soln 21  Cont’d Solve: Total number of moles, 𝑛, of Cl-= (0.250 + 0.100) mol Cl- = 0.350 mol Cl- Total volume of the solution = 500 mL + 200 mL = 700 mL soln − 0.350 mol Cl ? M Cl− = = 𝟎. 𝟓𝟎𝟎 𝐌 𝐂𝐥 − 700 x 10−3 L Soln 22  Cont’d For the initial (i) undiluted solution: Solution dilution: 𝑛𝑖 = 𝑐𝑖 𝑉𝑖 𝑛𝑖 = 𝑀𝑖 𝑉𝑖 ▪ Concentrated solution: Has a relatively large amount of dissolved solute. For the final (f) diluted solution: 𝑛𝑓 = 𝑐𝑓 𝑉𝑓 ▪ Dilute solution: Has a relatively small 𝑛𝑓 = 𝑀𝑓 𝑉𝑓 amount of dissolved solute. Since 𝑛𝑖 is equal to 𝑛𝑓 , then: ▪ Most labs store concentrated solutions, 𝑴𝒊 𝑽𝒊 = 𝑴𝒇 𝑽𝒇 referred to as stock solutions, from which a dilute solution can be prepared by adding a 𝑀1 𝑉1 = 𝑀2 𝑉2 solvent (i.e., water). 𝑴𝒅𝒊𝒍 𝑽𝒅𝒊𝒍 = 𝑴𝒄𝒐𝒏𝒄 𝑽𝒄𝒐𝒏𝒄 ▪ During a dilution, the amount of solute remains constant (show a demonstration using marble/sweets contained in 250 mL beaker to 100 mL beaker). 23  Cont’d Example: Preparing a solution by dilution. A particular analytical chemistry procedure requires 0.0100 M K2CrO4. Calculate the volume of 0.250 M K2CrO4 must be diluted with water to prepare 0.2500 L of 0.0100 M K2CrO4. Analyse: Use the serial dilution method: 𝑴𝒊 𝑽𝒊 = 𝑴𝒇 𝑽𝒇. Solve: 𝑴𝒇 𝑽𝒇 0.0100 M x 0.2500 L ? L Soln,𝑽𝒊 = = = 𝟎. 𝟎𝟏𝟎𝟎 𝐋 = 𝟏𝟎. 𝟎 𝐦𝐋 𝑴𝒊 0.250 M 24  Cont’d Stoichiometry of reactions in solution: ▪ In case of a solution; we use molarity as a conversion factor from solution volume to number of moles of reactant in a preliminary step : Balanced chemical equation to find appropriate mole ratio Information given about known solution Further A (Molarity & stoichiometric coefficient of B step(s) Moles of A 𝑥 Moles of B volume) & stoichiometric coefficient of A leading to Preliminary step(s) the answer leading to moles of A Stoichiometric factor or Mole ratio i.e., obtaining mass, volume, or molarity Note: Include units & work from a of desired substance balanced chemical equation when solving stoichiometric problems. 25  Cont’d Example: Relating the mass of a product to the volume & molarity of a reactant solution A 25.00 mL pipetful of 0.250 M K2CrO4 is added to an excess of AgNO3(aq). Compute the mass of Ag2CrO4 that will precipitate from the solution. K2CrO4(aq) + 2 AgNO3(aq) → Ag2CrO4(s) + 2 KNO3(aq) Analyse: AgNO3(aq) is in excess and all the K2CrO4 in the 25.00 mL sample of K2CrO4 is consumed. Use these conversion factors (mL → L, mole ratio, & molar mass of Ag2CrO4). Solve: Using a conversion pathway approach, we proceed as follows: 1L 0.250 mol K2CrO4 1 mol Ag2CrO4 331.732 g Ag2CrO4 ? g Ag2CrO4 = 25.00 mL x x x x 1000 mL 1L 1 mol K2CrO4 1 mol Ag2CrO4 = 𝟐. 𝟎𝟕 𝐠 Ag2CrO4 26  The Concept of Limiting Reactant In a chemical reaction, you can have the following outcomes: ▪ Stoichiometric proportions: where reactants are completely & simultaneously consumed (proportional to the mole ratio as per coefficients in the balanced reaction equation). ▪ At other times, one of the reactants is completely converted into products by using an excess of all other reactant(s). ▪ Thus, the amount of the product(s) formed is dictated by the reactant that is completely consumed or reacted, known as the limiting reactant. ▪ Chemicals are often referred to as reagents, therefore, the limiting reactant can be called limiting reagent. ▪ If quantities of 2 or more reactants are given, you must find out which is the limiting reactant. ▪ An excess reactant: means that more of the reactant is present than is consumed in the reaction (left over). 27  Cont’d Note: The amount of a product formed when the limiting reactant is completely consumed is called the theoretical yield of that product. 28  Cont’d Grams of P4 Grams of Cl2 Example 1: Determining the Limiting Reactant in a Reaction Use molar mass Use molar mass Phosphorus trichloride, PCl3 is a commercially important in the conversion in the conversion factor: factor: compound used in the manufacture of pesticides, gasoline 1 mol Cl2/70.91 g Cl2 1 mol P4/123.896 g P4 additives, & several other products. Liquid is made by the direct combination of phosphorus and chlorine: Moles of P4 Moles of Cl2 P4(s) + 6 Cl2(g) → 4 PCl3(l) Calculate mole ratio of Cl2 to P4 moles ofCl2 Mole ratio= Calculate the maximum mass of PCl3 that can be obtained from moles ofP4 P4 125 g and 323 g Cl2. 6 If calculated mole ratio < 1, Cl2 is limiting 6 If calculated mole ratio > , P4 is limiting Analyse: 1 The key to solving this problem is to correctly identify the limiting reactant. The approach is outlined in the figure on left panel. 29  Cont’d Solve: 1 mol Cl2 ? mol Cl2 = 323 g Cl2 x = 𝟒. 𝟓𝟔 𝐦𝐨𝐥 Cl2 70.91 g Cl2 1 mol P4 ? mol P4 = 125 g P4 x = 𝟏. 𝟎𝟏 𝐦𝐨𝐥 P4 123.896 g P4 4.56 mol Cl2 Mole ratio = = 𝟒. 𝟓𝟏 1.01 mol P4 6 mol Cl2 Since 𝐌𝐨𝐥𝐞 𝐫𝐚𝐭𝐢𝐨, 𝐭𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 = = 𝟔 Cl2/P4, for instance, there is less than 6 mol 1.mol P4 Cl2 per mole of P4, Cl2 is the limiting reactant. Having identified Cl2 as the limiting reactant, we can complete the calculation using a conversion pathway approach: 1 mol Cl2 4 mol PCl3 137.324 g PCl3 ? g PCl3 = 323 g Cl2 x x x = 𝟒𝟏𝟕 𝐠 PCl3 70.91 g Cl2 6 mol Cl2 1 mol PCl3 30  Cont’d Example 2: Determining the quantity of excess reactant(s) remaining after a reaction Calculate the mass of P4 remaining following the reaction in the previous slide, Example 1. P4(s) + 6 Cl2(g) → 4 PCl3(l) Analyse: We know that the limiting reactant is Cl2 & P4 is the excess reactant. The key issue is to calculate the mass of P4 that is consumed, we base this on either on the mass of Cl2 consumed or on the mass of PCl3 produced. The mass of P4 that remains is calculated by subtracting the mass of P4 that is consumed from the total mass of P4. Solve: Using a conversion pathway approach, we proceed as follows: 1 mol Cl2 1 mol P4 123.896 g P4 ? g P4 = 323 g Cl2 x x x = 𝟗𝟒. 𝟏 𝐠 P4 70.91 g Cl2 6 mol Cl2 1 mol P4 The mass of P4 remaining after the reaction: 125 g P4 initially – 94.1 g P4 consumed= 31 g remaining. 31  Practical Matters in Reaction Stoichiometry Theoretical yield, Actual Yield, & Percent Yield: ▪ These concept are applicable in the lab or manufacturing industry. ▪ Theoretical yield of a reaction is the calculated quantity of product expected from given quantities of reactants. ▪ The quantity of product that is actually produced is called the actual yield. ▪ Therefore, the percent yield is defined as: actual yield Percent yield = 𝑥 100% theoretical yield ▪ If the actual yield is almost exactly equals the theoretical yield, the reactions are said to be quantitative (such reactions can be used in quantitative chemical analyses). ▪ The actual yield for some reactions can be less than the theoretical yield (< 100%) due to purification steps, reverse reaction, or side reactions (unintended products called by-products). ▪ If the actual yield gives an apparent yield greater than 100%, this indicates an error in technique, when product is weighed wet with solvent, or contamination with an excess reactant or a by-product. 32  Cont’d Example: Determining Theoretical, Actual, & Percent Yields Billions of kilograms of urea, CO(NH2)2, are produced annually for use as a fertilizer. The reaction used is given below: 2 NH3(g) + CO2 → CO(NH2)2(s) + H2O(l) The typical starting reaction mixture has a 3:1 mole ratio of NH3 to CO2. If 47.7 g urea forms per mole of CO2 that reacts, calculate the (a) theoretical yield; (b) actual yield; & (c) percent yield. Analyse: The reaction mixture contains fixed amounts of NH3 & CO2, 1st find the limiting reactant. The stoichiometric proportions are 2 mol NH3: 1 mol CO2, but the reaction mixture has the mole ratio of NH3 to CO2 3:1, therefore, CO2 is the limiting reactant & NH3 is the excess reactant. Solve: (a) Use the conversion pathway approach: 1 mol CO(NH2)2 60.056 g CO(NH2)2 ? Theoretical yield, urea = 1.00 mol CO2 x x = 𝟔𝟎. 𝟏 𝐠 CO(NH2)2 1 mol CO2 1 mol CO(NH2)2 (b) Actual yield = 47.7 g CO(NH2)2 47.7 g CO(NH2)2 (c) Percent yield = 𝑥 100% = 𝟕𝟗. 𝟒% 60.1 g CO(NH2)2 33  Cont’d Example: Adjusting the quantities of reactants in accordance with the percent yield of a reaction When heated with sulfuric or phosphoric acid, cyclohexanol, C6H11OH, is converted to cyclohexene, C6H10. The balanced chemical equation for the reaction is shown below: C6H11OH(l) → C6H10 + H2O(l) If the percent yield is 83%, calculate the mass of cyclohexanol that must be used to obtain 25 g of cyclohexene. Analyse: Given the percent yield (83%) & actual yield of C6H10 (25 g), then the theoretical yield of C6H10 can be calculated. The quantity of C6H11OH required to produce the theoretical yield of C6H10 can be determined using a series of conversion factors (molar mass & mole ratio) Solve: 25 g x 100% First, calculate the theoretical yield: 𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐲𝐢𝐞𝐥𝐝 = = 𝟑. 𝟎 𝐱 𝟏𝟎𝟏 𝐠 83% (𝒏𝒐𝒕 𝟑𝟎, 𝑡𝑜 𝑒𝑚𝑝ℎ𝑎𝑠𝑖𝑧𝑒 𝑡ℎ𝑎𝑡 𝑖𝑠 2 𝑠. 𝑓. ). Using a conversion pathway approach to complete computation, we proceed as follows: 1 mol C6H10 1 mol C6H11OH 100.161 g C6H11OH ? g C6H11OH = 30. g C6H10 x x x = 𝟑𝟕. 𝐠 C6H11OH 82.146 g C6H10 1 mol C6H10 1 mol C6H11OH 34  Cont’d Consecutive, Simultaneous, & Overall Reactions: ▪ Both in laboratory work & in manufacturing, the preferred processes are those that yield a product through a single reaction. ▪ Often such processes give a higher yield becoz there is no need to remove products from one reaction mixture for further processing in subsequent reactions. ▪ In some cases, a multi-step process is inevitable. ▪ Consecutive reactions: Are reactions carried out one after another in sequence to yield a final product. ▪ In a calculation, use a stoichiometric factor for each reaction. ▪ Simultaneous reactions: two or more substances react independently of one another in separate reactions occurring at the same time. 35  Cont’d Example: Calculating the quantity of a substance produced by reactions occurring consecutively Titanium dioxide, TiO2, is the most widely used white pigment for paints, having displaced most lead-based pigments, which are environment hazards. Before it can be used, however, naturally occurring TiO2 must be freed of coloured impurities. One process for doing this converts impure TiO2(s) to TiCl4(g), which is then converted back to pure TiO2(s). The process is based on the following reactions, the first of which generates TiCl4. 2 TiO2(s) (impure) + 3 C(s) + 4 Cl2(g) → 2 TiCl4(g) + CO2(g) + 2 CO(g) (eqn 1) TiCl4(g) + O2(g) → TiO2(s) + 2 Cl2(g) (eqn 2) Compute the mass of carbon consumed to produce 1.00 kg of pure TiO2(s) in this process. Analyse: Begin with the product, TiO2, focus on the second reaction & work backward to one of the reactant, C. Use conversion factors (molar mass & mole ratio). Solve: Using a conversion pathway approach to complete computation, we proceed as follows: 1000 g TiO2 1 mol TiO2 1 mol TiCl4 3 mol C 12.011 g C ? g C = 1.00 kg TiO2 x x x x x = 𝟐𝟐𝟔 𝐠 C 1 kg TiO2 79.865 g TiO2 1 mol TiO2 2 mol TiCl4 1 mol C 36  Cont’d Example: Calculating the quantity of a substance produced by reactions occurring simultaneously Magnesium-aluminum alloys are widely used in aircraft construction. One particular alloy contains 70.0% Al & 30.0% Mg, by mass. Predict how many grams of H2(g) are produced in the reaction of a 0.710 g sample of this alloy with excess HCl(aq). Balanced chemical equations are given below for the reactions that occur. 2 Al(s) + 6 HCl(aq) + 4 Cl2(g) → 2 AlCl3(aq) + 3 H2(g) (eqn 1) Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) (eqn 2) Analyse: These two reactions are simultaneous reactions & occur independently. The total amount of H2 produced is obtained by adding together the amounts produced by each reaction. Use a stoichiometric factor (or mole ratio) to convert from mol Al to mol H2 & from mol Mg to mol H2. Convert percent masses to masses by assuming 100.0 g sample, & then convert from gram alloy to mol (Al & Mg) using molar masses. Solve: Using a conversion pathway approach to complete computation, we proceed as follows: 70.0 g Al 1 mol Al 3 mol H2 2.016 g H2 ? g H2 = 0.710 g alloy x x x x + 100.0 g alloy 26.982 g Al 2 mol Al 1 mol g H2 30.0 g Mg 1 mol Mg 1 mol H2 2.016 g H2 0.710 g alloy x x x x = 𝟎. 𝟎𝟕𝟑𝟒 𝐠 H2 100.0 g alloy 24.305 g Mg 1 mol Mg 1 mol H2 37  Gas Stoichiometry ▪ So far, we dealt with masses of reactants & products. ▪ Now, let’s discuss how volume enters the stoichiometry of gases. ▪ From ideal gas law, the volume of 1 mole of ideal gas at 0oC (273.15 K) & 1 atm is given by: L ∙ atm 𝑛𝑅𝑇 (1.000 mol) (0.08206 K ∙ mol)(273.15 K) 𝑷𝑽 = 𝒏𝑹𝑻; 𝑉 = = = 𝟐𝟐. 𝟒𝟐 𝐋 𝑃 1.000 atm ▪ The conditions 0 oC & 1 atm, called standard temperature & pressure (STP), are common reference conditions for the properties of gases. ▪ The molar volume of an ideal gas is 22.42 L at STP. 38  Cont’d Example 1: A sample of nitrogen gas has a volume of 1.75 L at STP. Calculate how many moles of N2 are present in the sample. Analyse: We can solve this problem by using the ideal gas equation, but we can take a shortcut by using the molar volume of an ideal gas at STP: Molar volume = 22.42 L/mol. Solve: Using the molar volume, we proceed as follows: 1 mol N2 ? mol N2 = 1.75 L N2 x = 𝟕. 𝟖𝟏 𝒙 𝟏𝟎−𝟐 𝐦𝐨𝐥 𝐍𝟐 22.42 L N2 39  Cont’d Example 2: Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3). Compute the volume of CO2 at STP produced from the decomposition of 152 g by the CaCO3 reaction: CaCO3(s) → CaO (s) + CO2(g) Analyse: We can solve this problem by using the ideal gas equation, but we can take a shortcut by using the molar volume of an ideal gas at STP: Molar volume = 22.42 L/mol. Also, consider the use of stoichiometry. Solve: Using the molar volume & mole ratio, we proceed as follows: 1 mol CaCO3 1 mol CO2 ? mol CO2 = 152 g CaCO3 x x = 𝟏. 𝟓𝟐 𝐦𝐨𝐥 𝐂𝐎𝟐 100.086 g CaCO3 1 mol CaCO3 22.42 L CO2 ? L CO2 @ STP = 1.51869392 mol CO2 x = 𝟑𝟒. 𝟏 𝐋 𝐂𝐎𝟐 1 mol CO2 40  The Extent of Reaction ▪ If a reaction proceeds to completion, one or more of the reactants is completely consumed or reacted. ▪ But, for some reactions, the reactants are not completely consumed & the extent to which a reaction proceeds from reactants toward products can be quantified. ▪ The progress of a chemical reaction can be monitored by specifying how much of a given reactant is consumed or how much of a given product is formed (or simply a change in the amount of any reactant or product). Note: ▪ For reactants, the changes in amount are negative (their amounts decrease), whereas for products, the changes in amount are positive (their amounts increase). ▪ The changes in amount are not independent of each other. They are each multiple of the same change 41  Cont’d Examples: Calculating the Extent of Reaction (a) State the stochiometric coefficients for the following reaction, which occurs to a limited extent at 25 oC: 4 KO2(s) + 2 CO2(g) → 2K2CO3(s) + 3O2(g) (b) In an experiment carried out at 25 oC, a reaction mixture containing 2.478 mol KO2 and 0.979 mol CO2 yields 1.128 mol O2. Calculate the extent of reaction, as well as the final amount of all reactants and products. Analyse: (a) For a given reactant or product, the stochiometric coefficient (𝑣) is set equal to the stochiometric coefficient with a + or – sign included (a + sign for a product & a – sign for a reactant). Δ𝑛𝑘 (b) The extent of reaction (𝜉) is expressed as ξ =. Construct a table with three 𝑣𝑘 rows below the chemical equation, showing the initial amounts, the changes in amount & the final amounts, respectively. 𝚫𝒏 is related to change in amount in mol. 42  Cont’d Solve: (a) The stochiometric ratios are: 𝑣KO2 = −4 𝑣CO2 = −2 𝑣K2 CO3 = +2 𝑣O2 = +3 (a) Construct a table with three rows below the chemical equation, showing the initial amounts, the changes in amount & the final amounts, respectively. 4 KO2(s) + 2 CO2(g) → 2K2CO3(s) + 3O2(g) Initial amounts: 2.478 mol 0.979 mol 0 mol 0 mol Changes: -4𝜉 -2𝜉 +2𝜉 +3𝜉 Final amounts: 2.478-4𝜉 0.979-2𝜉 2𝜉 3𝜉 43  Cont’d Solve: Note: The final amount of O2 is given as 1.128 mol, such that: 3𝜉 = 1.128 mol 1.128 𝜉= = 𝟎. 𝟑𝟕𝟔 𝐦𝐨𝐥 3 Now we can calculate the final amounts of other substances using the value of 𝜉: 𝑛KO2 ,final = 2.478 mol − 4𝜉 = 2.478 mol − 4 𝑥 0.376 = 𝟎. 𝟗𝟕𝟒 𝐦𝐨𝐥 𝑛CO2 ,final = 0.979 mol − 2𝜉 = 0.979 mol − 2 𝑥 0.376 = 𝟎. 𝟐𝟐𝟕 𝐦𝐨𝐥 𝑛K2 CO3 ,final = 2𝜉 = 2 𝑥 0.376 = 𝟎. 𝟕𝟓𝟐 𝐦𝐨𝐥 𝑛K2 CO3 ,final = 3𝜉 = 3 𝑥 0.376 = 𝟏. 𝟏𝟐𝟖 𝐦𝐨𝐥 44 The EnD of Slides 45