Chapter 9 Stoichiometry Chemistry

The Mathematics of Chemical Reactions: STOICHIOMETRY I. Balanced Chemical Equations A. Provide qualitative and quantitative information B. Supports the Law of Conservation of Matter 2H2 + O2  2H2O The above equation is interpreted in terms of particles as follows: 1. 2 molecules of H2 react with 1 molecule of O2 to produce 2 molecules of water. The ratio of H2 to O2 is 2:1. or 2 moles of H2 react with 1 mole of O2 to produce 2 moles of water.

2. It is more convenient to interpret the coefficients as number of moles, because we measure amounts of substances by massing. C. Stoichiometry 1. The study of the quantitative relationships that exist in a formula or a chemical reaction. 2. Importance a. Provides for the safe, economical and reproducible manufacture of chemicals. b. Provides for the safe administration of pharmaceuticals. D. Proof of the conservation of matter in a balanced equation 1. Convert all reactants and products to their mass equivalents. 2. Sum up the mass of reactants and compare the sum of the mass of products.

II. Stoichiometry Problems A. Steps to Solve Problems Write a balanced equation. Identify the given ( ) and the unknown or required substance (?). Convert mass of given into moles. Use the mole (molar) ratio to convert from given to required substance. If needed, convert moles of required into mass of required substance.

B. Examples 1. How many moles of oxygen are required to react with 16 moles of hydrogen in the production of water? H2 + O2  H2O  ? 2 2 16 moles H2 1 mole O2 2 moles H2 = 8.0 moles O2 Mole ratio links 1 substance to another in a reaction. Required in problem solving.

Antimony reacts with water to produce antimony(III) oxide and hydrogen. How many moles of hydrogen are produced from 7.5 moles of antimony? Sb + H2O  Sb2O3 + H2 2 3 3 11 mol H2

What mass of aluminum oxide can be prepared by the reaction of 67 What mass of aluminum oxide can be prepared by the reaction of 67.5 g of aluminum in a synthesis reaction? Al + O2  Al2O3 4 3 2 128 mol Al2O3

Sodium bicarbonate, a. k. a Sodium bicarbonate, a.k.a. baking soda, can be used to extinguish a fire. When heated, it decomposes to give carbon dioxide gas which smothers the fire. It also produces sodium carbonate and water. If a sample contains 4.0 g of sodium bicarbonate, what mass of carbon dioxide is produced? 2 NaHCO3  Na2CO3 + H2O + CO2

% Yield = (actual yield/expected yield) x 100 III. Percent Yield A. Expected Yield: the amount of product that should be produced (theoretical) B. Actual Yield: the amount of product that is actually produced (experimental) C. Percent Yield: percent of expected yield that was obtained % Yield = (actual yield/expected yield) x 100

D. Steps to Solving Percent Yield Problems Write a balanced equation Identify the given () which is the mass of reactant, and identify the actual yield. Solve for the expected mass of product using the given mass of reactant. Calculate the % yield. % Yield = actual yield x 100 expected yield

Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3 E. Examples 1. A reaction between 2.80 g aluminum nitrate and excess sodium hydroxide produced 0.966 g of aluminum hydroxide in this double replacement reaction. Calculate the % yield. Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3 1.03 g Al(OH)3

2. Determine the % yield for the reaction between 3.74 g of sodium and excess oxygen if 4.24 g of sodium oxide is recovered in the direct combination reaction. 4Na + O2  2Na2O

IV. Limiting Reactants Definition: the reactant that determines, or limits, the amount of product(s) formed in a chemical reaction Problem Solving Tips The limiting reactant is not necessarily the reactant present in the smallest amount When you are given the amounts of 2 or more reactants, you should suspect that you are dealing with a limiting reactant problem.

C. Steps Write a balanced equation Calculate the number of moles of each reactant Compare the mole ratios of the reactants as available ratio (from the given masses) and the required ratio (from the coefficients) Identify the limiting reactant, and use it to calculate the mass of product formed.

Oxygen is the limiting reactant D. Examples 1. What mass of CO2 could be formed by the combustion of 16.0 g CH4 with 48.0 g O2? CH4 + 2O2  CO2 + 2H2O 33.0 g CO2 Oxygen is the limiting reactant

NiCl2 + 2NaOH  Ni(OH)2 + 2NaCl Sodium Hydroxide is the LR 2. What is the maximum mass of nickel(II) hydroxide that could be prepared by mixing 25.9 g nickel(II) chloride with 10.0 g sodium hydroxide? NiCl2 + 2NaOH  Ni(OH)2 + 2NaCl 11.6 g Ni(OH)2 Sodium Hydroxide is the LR

V. Sequential Reactions Definition: A chemical process in which several reactions are required to convert starting materials into product(s). Main Concept: The amount of desired product from each reaction is taken as the starting material for the next reaction. Consideration is given to steps that produce less than 100%.

Ni + 4CO  Ni(CO)4 C. Examples 1. Given the following sequence of equations, calculate the mass of Ni(CO)4 produced from 75.0 g of carbon. Assume 100% yields. C + H2O  CO + H2 Ni + 4CO  Ni(CO)4

2. Hydrogen, obtained by the electrical decomposition of water, is combined with chlorine to produce 84.2 g of hydrogen chloride. Calculate the mass of water decomposed. Assume 100% yield. 2H2O  2H2 + O2 H2 + Cl2  2 HCl

VI. Solution Stoichiometry A. Many reactants are introduced to a reaction chamber as a solution. B. The most common solution concentration is molarity. molarity = mol/liter C. Examples Excess lead(II) carbonate reacts with 27.5 mL of 3.00M nitric acid. Calculate the mass of lead(II) nitrate formed PbCO3 + 2HNO3  Pb(NO3)2 + H2CO3

2. Calculate the volume, in mL, of a 0 2. Calculate the volume, in mL, of a 0.324 molar solution of sulfuric acid required to react completely with 2.792 g of sodium carbonate according to the equation below. H2SO4 + Na2CO3  Na2SO4 + CO2 + H2O

Problems C3H8 + 5O2 → 3CO2 + 4H2O ΔH = -2.22x103 kJ How much heat is released when 22.0g of propane is burned? How much carbon dioxide is produced, in grams, when 2,500 kJ of energy is released? -1.11x103 kJ (released) 150 g CO2

Hrxn = ∑ nHf,products - ∑ nHf,reactants ΔH from ΔHf The standard enthalpy change of a reaction is equal to the sum of the standard molar enthalpies of formation of the products multiplied by its coefficient, n, in the balanced equation, minus the corresponding sum of standard molar enthalpies of formation of reactants. Hrxn = ∑ nHf,products - ∑ nHf,reactants ***By definition, the standard enthalpy of formation of the most stable form of any element is zero because there is no formation reaction needed when the element is already in its standard state.***

SF6(g) + 3H2O(l)  6HF(g) + SO3(g) ΔH from ΔHf Problem: Using the Heats of Formation Table, calculate the H for the following reaction: SF6(g) + 3H2O(l)  6HF(g) + SO3(g) Write the thermochemical equation for this reaction: SF6 + 3H2O + 45.1kJ  6HF + SO3 45.1 kJ

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) ΔH from ΔHf Problem: Using the Heats of Formation Table, calculate the standard heat of combustion for propane. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Write the thermochemical equation for this reaction: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) + 2043.9kJ -2043.9 kJ