# Stoichiometric Calculations

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Stoichiometric Calculations
Stoichiometry – Ch. 11 Stoichiometric Calculations

Background on things you NEED to know how to do:
Name/write correct chemical formula Write chemical equations Balance chemical equations Predict Products Mole/mass conversions

Stoichiometry 74 tires 1 bicycle = 37 bicycles 2 tires
Stoichiometry uses ratios to determine relative amounts of reactants or products. For example If you were to make a bicycle, you would need one frame and two tires. 1 frame + 2 tires  1 bicycle If I had 74 tires, what is the most # of bicycles I could make? 74 tires 1 bicycle = 37 bicycles 2 tires

Proportional Relationships
2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. I have 5 eggs. How many cookies can I make? Ratio of eggs to cookies 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies

Proportional Relationships
Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation can be used to determine expected amounts of products given amounts of reactants. 2 Mg + O2  2 MgO

Stoichiometry Steps 1. Write a balanced equation.
2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass - moles  grams Molarity - moles  liters soln Molar volume - moles  liters gas Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

Molar Volume at STP LITERS OF GAS AT STP Molar Volume MASS IN GRAMS
(22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

Mole – Mole Conversions
The first type of problems we encounter will go between moles and moles. For this we need to use mole ratios. Ex: Write and balance the reaction between lead (II) nitrate and potasium iodide. Pb(NO3)2 + 2KI  2 KNO3 + PbI2 Mole ratio of potasium iodide to lead (II) iodide: 2 moles KI 1 mole PbI2

Mole to Mole Problems 2KClO3  2KCl + 3O2 9 mol O2 2 mol KClO3
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3  2KCl + 3O2 ? mol 9 mol 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3

Mole to Mass We can also convert from moles to mass, and mass to moles
For Example: 4 Al + 3 O2  2Al2O3 If you know how many grams of Al you start with, we can write a flow chart to show how to calculate the # of moles of oxygen need to fully react with the Al. g Al  moles Al  moles of oxygen

Mass to Moles: 4 Al + 3 O2  2Al2O3
If the reaction starts with .84 moles of aluminum, how many grams of aluminum oxide are produced? .84 mol Al 2 mol Al2O3 101.9 grams Al2O3 4 mol Al 1 mol Al2O3 = 42.8 grams Al2O3 0.92 g of Aluminum oxide are produced from the reaction. How much aluminum was used up? .92 g Al2O3 1 mol Al2O3 4 mol Al 26.9 g Al g Al2O3 2 mol Al2O3 1 mol Al = .49 grams Al

Mass to Mass Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g Cu 1 mol Cu 63.55
How many grams of silver will be formed from 12.0 g copper reacting with silver nitrate? Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g ? g 12.0 g Cu 1 mol Cu 63.55 g Cu 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag = 40.7 g Ag

Stoichiometry Problems – Mole-Mole
N2H4 + N2O4  N2 + H2O ? moles N2O4 = 2.72 moles N2H4 2 3 4 2.72 mol ? mol 2.72 mol N2 H4 1 mol N2O4 2 mol N2 H4 = 1.36 mol N2O4

Stoichiometry Problems – Mass/Mass
Sodium metal reacts with oxygen gas to produce solid sodium oxide. How many grams of sodium must react to produce 42.0 grams of sodium oxide? 4Na + O2  2Na2O ? g 42.0 g 42.0 g Na2O 1 mol Na2O 22.99 g Na 4 mol Na = 22.99 g Na 16.98 g Na2O 1 mol Na 2 mol Na2O

Stoichiometry Problems – Mole/Mass
In photosynthesis, carbon dioxide and water react to form glucose, C6H12O6 and oxygen gas. ___CO2 + ___H2O  ___C6H12O6 + ___O2 If 15.6 grams of carbon dioxide react, how many moles of glucose will be produced? How many grams of carbon dioxide must react to produce 0.25 moles of glucose? 6 6 6 15.6 g CO2 1 mol CO2 1 mol C2 H12O6 = mol C2H12O6 44.01 g CO2 6 mol CO2 0.25 mol C2 H12O6 6 mol CO2 44.01 g CO2 = 66 g CO2 1 mol C2 H12O6 1 mol CO2

Stoichiometry with Gases
If the pressure and temperature are constant, the ratio of moles in the balanced equation is the ratio of liters in an all gas reaction.

Standard Temperature & Pressure
Molar Volume at STP At STP 1 mol of a gas=22.4 L Standard Temperature & Pressure 0°C and 1 atm

*only in all gas reactions!
Molar Volume Hydrogen and chlorine gas react to produce hydrochloric acid. If 7.00 L of hydrogen gas react, how many liters of HCl gas are formed? H2 (g) + Cl2 (g)  2 HCl (g) 7.00 L H2 2.0 L HCl = 14.0 L HCl *only in all gas reactions! 1.0 L H2

Molar Volume = 15.7 L H2 17.0 g Mg 1 mol Mg 24.31 g 1 mol H2 Mg
In the following reaction, if 17 g of Mg react, how many L of H2 forms? Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g) 17.0 g Mg 1 mol Mg 24.31 g 1 mol H2 Mg 22.4 L H2 1 mol H2 = 15.7 L H2

Molar Volume Problems 2KClO3  2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol
How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3  2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3

Molar Volume Problems Cu + 2AgNO3  2Ag + Cu(NO3)2 1.5 L .10 mol AgNO3
How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3  2Ag + Cu(NO3)2 ? g 1.5L 0.10M 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu = 4.8 g Cu

Molarity Molarity is the number of moles of solute dissolved in one liter of solution. Units are moles per liter or moles of solute per liter of solution. Molarity abbreviated by a capital M Molarity = moles of solute liter of solution

Molarity Problems = .43 mol NH4Cl = .145 L = 2.97 M NH4Cl .145 L
As an example, suppose we dissolve 23 g of ammonium chloride (NH4Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution? 23 g NH4Cl 1 mole NH4Cl = .43 mol NH4Cl 53.5 g NH4Cl 145 mL 1 L = .145 L 1000 mL .43 mol NH4Cl = 2.97 M NH4Cl .145 L

Molarity Problems = 0.6 M NaCl = .96 mol HCl
Now, suppose we have a beaker with 175 mL of a 5.5 M HCl solution. How many moles of HCl is in this beaker? 175 mL 1 L 5.5 mol HCl = .96 mol HCl 1000 mL 1 L Suppose you had 70 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution? 1.2 mol NaCl 70 g NaCl 1 mol NaCl = 1.2 mol NaCl 2.0 L 58.44 g NaCl = 0.6 M NaCl

Volume of Solutions A 10.% HCl solution (soln) means: g HCl (pure) g HCl soln A solution with a density of 1.5 g/mL means: g soln mL soln

Impure Substances To say a substance is 75% NaCl by mass means: g NaCl (pure) g of NaCl solution or 100 g of NaCl solution 75 g of NaCL (pure) Or, Iron ore that is 15% iron by mass means: g Fe g ore 100 g of ore 15 g of Fe

Energy & Stoichiometry

Exothermic and Endothermic
Exothermic process – heat is released into the surroundings Exo = Exit Endothermic Process – heat is absorbed from the surroundings Endo = Into HEAT HEAT

Thermochemical Equations
In a thermochemical equation, the energy of change for the reaction can be written as either a reactant or a product Enthalpy: the heat content of a system at constant pressure (ΔH) Endothermic (positive ΔH) 2NaHCO kJ Na2CO3 + H2O + CO2 Exothermic (negative ΔH) CaO + H2O Ca(OH) kJ

Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= kJ Fe(s) + O2(g)→ Fe2O3(s) kJ How much heat is evolved when 10.00g of Iron is reacted with excess oxygen? Exo 4 3 2 10.00g Fe 1 mol Fe 1652 kJ =73.97 kJ of heat 55.85g Fe 4 mol Fe

Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = kJ: 2 NaHCO kJ → Na2CO3(s) + H2O + CO2 How much heat is required to break down 50.0g of sodium bicarbonate? Endo 1 mol NaHCO3 50.0 g NaHCO3 129 kJ 83.9 g NaHCO3 2 mol NaHCO3 =38.4 kJ of heat

Write the thermochemical equation for the synthesis of calcium oxide and water with a ΔH= kJ: CaO + H2O → Ca(OH) kJ How much energy is released when 100 g of calcium oxide reacts? Exo 100 g CaO 1 mol CaO 65.2 kJ 56.07 g CaO 1 mol CaO =116 kJ of heat

Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= kJ: 2 MgO → 2 Mg + O2 How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kJ of Energy? Endo 420 kJ 1 mol O2 31.98 g O2 =218 g of O2 61.5 kJ 1 mol O2

Stoichiometry in the Real World
Stoichiometry – Ch. 11 Stoichiometry in the Real World

Limiting Reactants Available Ingredients 4 slices of bread
1 jar of peanut butter 1/2 jar of jelly Limiting Reactant bread Excess Reactants peanut butter and jelly

Limiting Reactants Available Ingredients 24 graham cracker squares
1 bag of marshmallows 12 pieces of chocolate Limiting Reactant chocolate Excess Reactants Marshmallows and graham crackers

Limiting Reactants Limiting Reactant one that is used up in a reaction
determines the amount of product that can be produced Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

Limiting Reactant Steps
1. Write the balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant amount of product actually possible

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? g 68.1 g
79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed? Zn HCl  ZnCl H2 79.1 g 68.1 g ? g

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 68.1 g HCl
1 mol HCl g HCl 1 mol H2 2 mol HCl 2.02 g H2 1 mol H2 = 1.89 g H2

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 79.1 g Zn
1 mol Zn 65.39 g Zn 1 mol H2 Zn 2.02 g H2 1 mol = 2.44 g H2

Limiting Reactants Zn: 2.44 g H2 HCl: 1.89 g H2 Limiting reactant: HCl
Excess reactant: Zn Product Formed: 1.89 g H2 left over zinc

Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g
5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed? 2Mg O2  2MgO 5.42 g 4.00 g ? g

Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 5.42 g Mg
1 mol Mg 24.31 g Mg 2 mol MgO Mg 40.31 g MgO 1 mol = 8.99 g MgO

Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 4.00 g O2
1 mol O2 g O2 2 mol MgO 1 mol O2 40.31 g MgO 1 mol MgO = 10.1 g MgO

A. Limiting Reactants #2 Limiting reactant: Mg Excess reactant: O2
Mg: 8.99 g MgO O2: 10.1 g MgO Limiting reactant: Mg Excess reactant: O2 Product Formed: 8.99 g MgO Excess oxygen

Limiting Reactants What other information could you find in these problems? How much of each reactant is used – in grams, liters, moles How much of excess reactant is left over – in grams, liters, moles

Percent Yield measured in lab calculated on paper

Percent Yield 2 mol KCl 74.55 g KCl 1 mol 1 mol 45.8 g K2CO3 K2CO3
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO HCl  KCl H2CO3 1 mol K2CO3 138.2 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol 45.8 g K2CO3 = 49.4 grams KCl

46.3 grams KCl x 100 49.4 grams KCl % yield = 93.7 % Percent Yield
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. 46.3 grams KCl x 100 49.4 grams KCl % yield = 93.7 %