 Diman Regional VocTech
 PreAP/Extra Credit Pages
 PreAP Cram Sheet #3: Moles and Stoichiometry

Textbook Reading:
Holt "Chemistry" Chapter 9 Sections 12, Chapter 13 Section 2
Silberberg "Chemistry" Chapter 3 Sections 34, Chapter 4 Section 1A Balanced Chemical Equation Is
 A chemical equation which has equal numbers of each kind of atoms on both sides.
 This is a consequence of conservation of mass.
 Unbalanced Equation Example: C_{2}H_{5}OH + I_{2} + Na_{2}CO_{3} + H_{2}O → CHI_{3} + NaHCO_{2} + CO_{2} + HI
 Reactants: 3 C atoms, 2 I atoms, 2 Na atoms, 5 O atoms, 7 H atoms
 Products: 3 C atoms, 4 I atoms, 1 Na atom, 4 O atoms, 3 H atoms
 Balanced Equation Example: 2 C_{2}H_{5}OH + 8 I_{2} + Na_{2}CO_{3} + H_{2}O → 2 CHI_{3} + 2 NaHCO_{2} + CO_{2} + 10 HI
 Both Sides (Reactants & Products): 5 C atoms, 16 I atoms, 2 Na atoms, 6 O atoms, 14 H atoms
 An equation is balanced by inserting coefficients before formulas as necessary.
 If there is no coefficient, a '1' is implied.
 Coefficients are generally integers (although you may see some exceptions when we deal with thermochemistry).
 Coefficients should not all share a common factor:
 6 H_{2} + 2 N_{2} → 4 NH_{3} : BAD because all coefficients are multiples of 2.
 3 H_{2} + N_{2} → 2 NH_{3} : GOOD because 3, 1, 2 have no common factor.
 Balancing is an art but there are helpful rules.
 "Double the odds": if an element is present in multiple formulas in even numbers everywhere but one place where it is odd, double the coefficient for the formula where it is odd.
 H_{2} + O_{2} → H_{2}O: "Double the odd" oxygen atom in H_{2}O to get H_{2} + O_{2} → 2 H_{2}O.
 Now balance H_{2} to get the balanced equation 2 H_{2} + O_{2} → 2 H_{2}O
 "Least common multiple": if an element is present on both sides in 2 different numbers, bring them both up to the LCM.
 KClO_{3} → KCl + O_{2}: "LCM" of 2 and 3 is 6 oxygens.
 2 KClO_{3} → KCl + 3 O_{2}
 Now balance Cl and K:2 KClO_{3} → 2 KCl + 3 O_{2}
 NEVER CHANGE A FORMULA OR SUBSCRIPT WHEN BALANCING.
 H_{2} + O_{2} → H_{2}O : do not modify H_{2}O to get H_{2} + O_{2} → H_{2}O_{2}
 This is no longer talking about the same substances, so it is wrong.
 "Double the odds": if an element is present in multiple formulas in even numbers everywhere but one place where it is odd, double the coefficient for the formula where it is odd.
 Practice your balancing. Play The Game! This is a regular class activity, but you can get ahead if you like.
Coefficients of a Balanced Chemical Equation are Ratios of Moles
 This sounds familiar. Didn't we say that Chemical Formulas are Ratios of Moles?
 The equation 3 H_{2} + N_{2} → 2 NH_{3} means 3 moles of hydrogen react with 1 mole of nitrogen producing 2 moles of ammonia.
 We can write the mole ratios (3 mol H_{2}/1 mol N_{2}), (3 mol H_{2}/2 mol NH_{3}), and (1 mol N_{2}/2 mol NH_{3}) for example.
 We can use these ratios to relate between any products and reactants in the same reaction.
 If you have 9 moles of H_{2}, you will require 3 moles of N_{2} to react with it.
 You will obtain 6 moles of NH_{3} if this reaction goes perfectly.
 The relationships between quantities of products and reactants are known as stoichiometry.
Theoretical and Percentage Yield
 The theoretical yield of a reaction is the amount of a product predicted from the stoichiometry of the balanced equation.
 The actual yield of a reaction is the amount of product you get in a real experiment.
 The percentage yield of a reaction is 100 x actual yield / theoretical yield.
 The equation 3 H_{2} + N_{2} → 2 NH_{3} predicts that 18 moles of H_{2} and 6 moles of N_{2} yield 12 moles of NH_{3}.
 If you actually do the reaction but only obtain 9 moles of NH_{3}, the theoretical yield is 100 x 9 / 12 = 75%.
Limiting and Excess Reactants
 The grocery store sells hotdogs in packs of 10, but buns only come in packs of 8.
 If you buy one of each, you can only have 8 sandwiches.
 We would say that buns are limiting but hotdogs are in excess, because there will be some hotdogs left over.
 The yield of a reaction is determined by the limiting reactant.
 Suppose we have 6 moles of H_{2} and 4 moles of N_{2} and we react them as in 3 H_{2} + N_{2} → 2 NH_{3}
 6 moles of H_{2} is enough to produce 4 moles of NH_{3}.
 4 moles of N_{2} is enough to produce 8 moles of NH_{3} but we won't get that: instead, some N_{2} will be left over.
 We get the smaller yield, 4 moles of NH_{3}. H_{2} is the limiting reactant and N_{2} is in excess.
Stoichiometry in Solution
 Chemical reactions are often conducted in solution.
 The molarity of a solution is defined as the number of moles of a substance per Liter of solution.
 M = n/V
 The symbol for molarity is a capital M.
 3M HCl means a solution containing 3 moles of HCl per Liter
 You can write the ratio (3 mol HCl/1 L) for use in stoichiometry calculations.
Use Mole Ratios from the Balanced Chemical Equation to Solve Stoichiometry Problems
Iron (III) hydroxide decomposes when heated to yield iron (III) oxide and water vapor by the following equation:
2 Fe(OH)_{3} → Fe_{2}O_{3} + 3 H_{2}O
How many moles of water vapor are released by the decomposition of 8 moles of Fe(OH)_{3}?
8 mol Fe(OH)3 x 3 mol H2O = 12 mol H2O 2 mol Fe(OH)3
Aluminum reacts with oxygen by the equation 4 Al + 3 O_{2} → 2 Al_{2}O_{3}. What mass of oxygen (O_{2}) is required to react with 18 g of aluminum?
18 g Al x 1 mol Al x 3 mol O2 x 32.0 g O2 = 16 g O2 27.0 g Al 4 mol Al 1 mol O2
Iron (Fe) dissolves in dilute sulfuric acid (H_{2}SO_{4}) yielding iron (II) sulfate and hydrogen gas. How much 3 M H_{2}SO_{4} (3 moles per Liter) solution is required to react with 100 g of Fe metal?
Write a balanced chemical equation: Fe + H_{2}SO_{4} → FeSO_{4} + H_{2}
100 g Fe x 1 mol Fe x 1 mol H2SO4 x 1 L H2SO4 solution = 0.597 L H2SO4 solution 55.85 g Fe 1 mol Fe 3 mol H2SO4
Find Limiting Reactant by Working Out Stoichiometry for each Reactant
How many grams of nitric acid, HNO_{3}, can be prepared from the reaction of 69.0 g of NO_{2} with 29.0 g H_{2}O according to the equation below? What is the limiting reactant? Which reactant is in excess?
3 NO_{2} + H_{2}O → 2HNO_{3} + NO
Work out the yield of HNO_{3} starting from NO_{2}
69.0 g NO2 x 1 mol NO2 x 2 mol HNO3 x 63.0 g HNO3 = 63.0 g HNO3
46.0 g NO2 3 mol NO2 1 mol HNO3Work out the yield of HNO_{3} starting from H_{2}O
29.0 g H2O x 1 mol H2O x 2 mol HNO3 x 63.0 g HNO3 = 203 g HNO3
18.0 g NO2 1 mol H2O 1 mol HNO3The theoretical yield is the smaller of the two, 63.0 g HNO_{3}, from NO_{2}. The limiting reactant is NO_{2} because it limits the amount of product. Not all the H_{2}O will react, so H_{2}O is in excess.
Use Reaction Stoichiometry to Calculate Theoretical Yield
Suppose the reaction above between NO_{2} and H_{2}O was performed, but instead of the theoretical yield of 63.0 g HNO_{3} , only 50.0 g of HNO_{3} was obtained. No real reaction is perfect. What is the percentage yield of this reaction?
100 x 50.0 g HNO3 = 79.3 % 63.0 g HNO3
Solve Stoichiometry Problems for Unknown Molar Masses
Remember that the molar mass of a substance is equal to the mass in grams of a sample of that substance, divided by the number of moles of that substance.
22.4 grams of an unknown metal carbonate, represented as MCO_{3}, is heated to decompose it to the metal oxide MO and carbon dioxide. 8.8 grams of carbon dioxide are produced by the reaction MCO_{3} → MO + CO_{2}. What is the identity of the metal M?
8.8 g CO2 x 1 mol CO2 x 1 mol MCO3 = 0.2 mol MCO3
44 g CO2 1 mol CO222.4 g MCO3 = 112 g/mol : this is the molar mass of MCO3
0.2 mol MCO3Subtracting the mass of CO3 = (12.0 + 3 * 16.0) = 60 g/mol, the molar mass of M must be 112  60 = 52 g/mol. The metal M must therefore be chromium (Cr).
 A chemical equation which has equal numbers of each kind of atoms on both sides.