1 1. Interpreting Chemical Equations Stoichiometry Calculations using balanced equations are called stoichiometric calculations. The starting point for any problem involving quantities of chemicals in a reaction is the balanced equation. Consider the reaction in which the reactants are nitrogen gas and hydrogen gas. They produce the product ammonia gas. N 2 + H 2 NH 3 The balanced equation shows us that 1 mole of nitrogen gas joins with three moles of hydrogen gas to produce two moles of ammonia gas. N 2 + 3H 2 2NH 3 Reactants Product Conserved? Atoms 2 + 6 8 Yes Molecules 1 + 3 2 No Moles 1 + 3 2 No Volume (L@SATP) 24.8 + 74.4 49.6 No Mass (g) 28.0 + 6.0 34.0 Yes Thus, we have the Law of Conservation of Matter (Mass) - Matter (mass) is conserved during any chemical change. Prove that atoms and their mass are conserved during the following reaction: 2HCl + Ca(OH) 2 CaCl 2 + 2H 2 O 2. Mole Ratios Balanced equations tell us the number of moles of each chemical taking part in the reaction. The ratio that compares the number of moles of one chemical to another chemical is called a mole ratio. There are six mole ratios that can be made from this reaction: N 2 + 3H 2 2NH 3 (1:3:2 mole ratio) 1) 1 mol N 2 2) 3 mol H 2 3) 1 mol N 2 4) 2 mol NH 3 5) 3 mol H 2 6) 2 mol NH 3 3 mol H 2 1 mol N 2 2 mol NH 3 1 mol N 2 2 mol NH 3 3 mol H 2 These mole ratios are used to help solve mole-mole problems. It will be a one-step Solution: Moles of Chemical A Moles of Chemical B Mole ratio
2 Methanol fuel burns in air: 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 3.50 mol of methanol (CH 3 OH) are burned in lots of oxygen, then a) How many moles of oxygen are used? b) How many moles of water are produced? c) How many moles of carbon dioxide gas are produced? 3. Gravimetric Stoichiometry The procedure for calculating the masses, in grams, of reactants or products from a balanced chemical equation is called gravimetric stoichiometry. One example of this is the mass-mass problem. The measured mass of one chemical (A) is used to find the unknown mass of a second chemical (B). 1. Mass of A Moles of A formula mass (A) 2. Moles of A Moles of B mole ratio 3. Moles of B Grams of B formula mass (B) Note that our mass-mass problems have a three-step solution. Example: How many grams of hydrogen gas will be required to react with 50.0 grams of nitrogen gas in the following reaction N 2 (g) + H 2 (g) NH 3 Step 1: g mole (refer to given info) X mol N 2 = 50.0 g N 2 x 3 mol N 2 = 1.773 mol N 2 28.2 g N 2 Step 2: mol N 2 mol H 2 X mol H 2 = 1.773 mol N 2 x 3 mol H 2 = 5.352 mol H 2 1mol N 2 Step 3: mol H 2 g H 2 X g H 2 = 5.352 mol H 2 x 2.02 g H 2 = 10.7 g H 2 1 mol H 2 1. Suppose 2.85g of hydrogen gas are reacted with oxygen gas in the reaction 2H 2 (g) + O 2 (g) 2H 2 O(l) a) How many grams of oxygen gas will be needed? b) Haw many grams of water will be made? 2. The equation for the roasting of iron(ii) sulfide is as follows: 4FeS + 7 O 2 2Fe 2 O 3 + 4SO 2 If we started with 95g of iron(ii) sulfide, then a) What mass of oxygen gas will be needed? b) What mass of sulfur dioxide will be produced?
3 3. One type of antacid is magnesium hydroxide: Mg(OH) 2 + 2HCl MgCl 2 + 2H 2 O If an antacid tablet has a mass of 4.56g, then a) What mass of hydrochloric acid will it react with? b) What mass of water will it make? 4. Gas Stoichiometry 4a) Standard Conditions (At SATP) The three steps of the stoichiometry calculation are the same for solids and gases. When gases at SATP are involved in the calculation, we can use the molar volume as a conversion factor 1 mole of any gas = 24.8 L at SATP = molar volume Our three-step solution will be: 1. L (at SATP) of A mol of A molar volume 2. mol of A mol of B mole ratio 3. mol of B L (at SATP) of B molar volume Note: (i) The second step will always be a mole flip. (ii) A problem may involve both volume and mass, but the solution will still be in threesteps. 1. Tin (II) fluoride, formerly found in many toothpastes, is formed in the reaction Sn (s) + 2HF (g) SnF 2 (s) + H 2 (g) a) How many liters of H 2 will be formed from 6.5 L HF at SATP? b) What mass of tin will react with 3.45 L HF at SATP? 2. Match heads contain potassium chlorate to help burning: 2KclO 3 (s) 2KCl(s) + 3O 2 (g) a) What mass of KCl is formed when 45mL O 2 is also formed at SATP? b) What volume of oxygen at SATP can be made from 3.12g KCLO 3?
4 4b) Non-Standard Conditions (Not at SATP) If the conditions in an experiment are not at SATP, then the Ideal Gas Law rather than molar volume is used in calculations involving a gas. PV = nrt This equation can be rearranged into n = PV/RT or V = nrt/p P = pressure (kpa/atm) V = volume (L) N = moles R = gas constant *8.31) T = temp (kelvin) Our solutions for these problems (involving gases at non-satp) will follow these three steps: 1. volume (L) of A mol of A n = PV/RT 2. mol of A mol of B mole ratio 3. mol of B L of B V = nrt/p 1. What volume of ammonia gas (NH 3 ) at 450 kpa and 80 o C can be made from the complete reaction under the same conditions of 64 L H 2? N 2 (g) + 3H 2 (g) 2NH 3 (g) 2. Photosynthesis is 6CO 2 (g) + 6H 2 O (l) C 6 H 12 O 6 (s) + 6O 2 (g) Suppose a green plant produces 312g of glucose (C 6 H 12 O 6 ). a) What volume of CO 2 at 23 o C and 102 KPa is used? b) What volume of 6O 2 is made at 10 o C and 101 KPa? 5. Solutions Most reactions take place in solutions. The concentration (Molarity) of a solution is found with this formula: M = n / V Molarity moles volume (mol/l) of solute of solution (mol) (L) This formula will be used in the three-step mole method whenever solutions (aq) are involved in the problem. (aq = dissolved in water)
5 The formula can be rearranged into n = MV and V = n/m Titration is a common method used to determine an unknown concentration from the known concentration of another chemical that it reacts with. Chemical 1 Known V 1 Unknown M 1 add Chemical 2 Known V 2 Known M 2 stop at Endpoint (sudden change) Calculate M 1 1. Suppose 12.0g of zinc metal is to be reacted with 0.800 M HCl: Zn (s) + 2HCl (aq) ZnCl 2 (aq) + H 2 (g) What volume, in L, of HCl will be needed? 2. A can of pop contains 280 ml of 0.150 M solution of carbonic acid (H 2 CO 3 ) which decomposes when pop goes flat : H 2 CO 3 (aq) CO 2 (g) + H 2 O (l) How much CO 2 gas at 37 o C and 101kPa will you burp after drinking the can? 3. Calculate the molarity (mol/l) of the sulfuric acid that is made by reacting 10.0 L of SO 2 (g) at SATP with water to make 0.845 L of the acid: SO 3 (g) + H 2 O (l) H 2 SO 4 (aq) 6. Applications a) Limiting Reactant During chemical reactions, one reactant is used up first. This is the limiting reactant and it actually determines the amount of product formed. Let s assume A + B C If A = 1 mol but B = ½ mol, then B is the limiting reactant because it is used up first. (Note that only ½ mol of C can be made during this reaction.) Any problem that gives a mass for all of the reactants will require us to identify the limiting reactant. Our calculations will then involve only the limiting reactant. We will ignore the reactant(s) in excess. Example: 1. Solid ammonium chloride can be made from two gases: NH 3 (g) + HCl (g) NH 4 Cl (s) If 1.00g NH 3 is mixed with 1.00 g HCl, calculate the actual mass of the ammonium chloride that will be formed.
6 a. find moles of reactants: x mol NH 3 = 1.00 g NH 3 x 1 mol/17.0 g NH 3 = 0.0588 mol NH 3 x mol HCl = 1.00 g HCl x 1 mol/ 36.5 g HCl = 0.0274 mol HCl b. identify limiting reactant: (actual) mol NH 3 : mol HCl = 1 : 1 (measured) mol NH 3 : mol HCl = 0.0588 : 0.0274 = 1 : 0.47 c. solve: x mol HCl = 1.00 g Hcl x 1 mol / 36.5 g HCl = 0.0274 mol HCl x mol NH 4 Cl = 0.0274 mol HCl x 1 mol NH 4 Cl / 1 mol HCl = 0.0274 mol NH 4 Cl x g NH 4 Cl = 0.0274 mol NH 4 Cl x 53.5 g NH 4 Cl / 1 mol NH 4 Cl = 1.47 g NH 4 Cl 1. Calcium metal burns to form CaO (g): 2Ca + O 2 2CaO Calculate the mass of CaO that can be formed when 0.48 g Ca is burned in 0.32 g O 2. 2. What mass of HCl (g) is produced when 4.50 g H 2 (g) and 140.0 g Cl 2 (g) are reacted according to the equation: H 2 (g) + Cl 2 (g) 2HCl (g) 6b. Percentage yield Most chemical reactions do not produce the amount of product that is predicted by the balanced equation. The actual yield is usually less than the theoretical yield. % yiled = actual yield (unit) / theoretical yield (unit) x 100 Note that yield is usually measured as mass, in g. Example: Ethanol can be made from sugar: C 6 H 12 O 6 (s) + 2C 2 H 5 OH + 2CO 2 In one experiment 10.0 g of sugar produced 0.664 g of ethanol (C 2 H 5 OH). What was the percentage yield? 1. actual yield (given): 0.664 g C 2 H 5 OH 2. theoretical yield: x mol sugar = 10.0 g sugar x 1 mol sugar / 180.0 g sugar = 0.05556 mol sugar x mol C 2 H 5 OH = 0.05556 mol sugar x 2 mol C 2 H 5 OH / 1 mol sugar = 0.1111 mol C 2 H 5 OH x g C 2 H 5 OH = 0.1111 mol C 2 H 5 OH x 46.0 g C 2 H 5 OH / 1 mol C 2 H 5 OH = 5.11 g C 2 H 5 OH 3. solve: % yield = actual yield / theoretical yield x 100 = (0.664 g) / (5.11 g) x 100 = 13.0 %
7 1. One step in the formation of acid rain is the oxidation of sulfur dioxide (SO 2 ): 2SO 2 + O 2 2SO 3. If 3.55 kg of SO 2 is burned to make 4.20 kg of SO 3, calculate the percentage yield of the process. 2. Bromine gas can be made from acids: HbrO 3 + 5HBr 3H 2 O + 3Br 2. If 10.0 g of HbrO 3 was reacted and 26.3 g of Br 2 was produced, what was the percentage yield of the reaction?