Chapter 11 – Stoichiometry

Chapter 11 – Stoichiometry
11.1 What is Stoichiometry? 11.2 Stoichiometric Calculations 11.3 Limiting Reactants 11.4 Percent Yield

Section 11.1 Defining Stoichiometry
The amount of each reactant present at the start of a chemical reaction determines how much product can form. Describe the types of relationships indicated by a balanced chemical equation. Explain the relationship between stoichiometry and the principle of the conservation of mass. State the mole ratios from a balanced chemical equation.

Section 11.1 Defining Stoichiometry
Key Concepts Balanced chemical equations can be interpreted in terms of moles, mass, and representative particles (atoms, molecules, formula units). The law of conservation of mass applies to all chemical reactions. Mole ratios are derived from the coefficients of a balanced chemical equation. Each mole ratio relates the number of moles of one reactant or product to the number of moles of another reactant or product in the chemical reaction.

Stoichiometry Study of quantitative relationships between amounts of reactants used and amounts of products formed by a chemical reaction Combo of 2 Greek words – “to measure the elements” Really a tool (system) for accounting for the conservation of matter What goes in must come out

Rests on principle of conservation of matter (mass)
Stoichiometry Rests on principle of conservation of matter (mass) 2 Al(s) + 3 Br2(l)  Al2Br6(s)

Stoichiometry - Questions
If I have this much reactant, how much product can I form? If I have this much of one reactant, how much of a second reactant do I need so that after the reaction is completed I won’t have any of either reactant left over?

Conversion Map for Stoichiometry Calculations
Requires balanced equation! Mass reactant Mass product Stoichiometric factor Moles reactant Moles product

Stoichiometry & Balanced Equation
4Fe(s) + 3O2(g)  2Fe2O3(s) Can interpret as 4 atoms Fe + 3 molecules O2  2 formula units Fe2O3 4 mol Fe + 3 mol O2  2 mol Fe2O3 No direct information on masses Can use molar mass to get mass info

4Fe(s) + 3O2(g)  2Fe2O3(s) 4 mol Fe  g Fe/mol Fe = g Fe 3 mol O2  g O2/mol O2 = g O2 2 mol Fe2O3  g Fe2O3/mol Fe2O3 = g Fe2O3 Total mass reactants = g Total mass products = g

Example problem 11.1, p. 370 C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Interpret as molecules, moles & mass 1 mol C3H8  g C3H8 /mol C3H8 = g C3H8 5 mol O2  g O2/mol O2 = g O2 3 mol CO2  g CO2/mol CO2 = g CO2 4 mol H2O  g H2O /mol H2O = g H2O Mass reactants = = g Mass products = = g 1 molecule C3H8 + 5 molecules O2  3 molecules CO2 + 4 molecules H2O 1 mol C3H8 + 5 mol O2  3 mol CO2 + 4 mol H2O

Practice Problems 1 (a-c), 2(a-b) page 371 Problems 44 - 45 page 392

Al:Br2 (2:3) Al: AlBr3 (2:2) Br2:AlBr3 (3:2)
Mole Ratio Ratio between the numbers of moles of any two substances in a balanced chemical equation 2Al(s) + 3Br2(l)  2AlBr3(s) Possible to write 3 unique mole ratios: Al:Br2 (2:3) Al: AlBr3 (2:2) Br2:AlBr3 (3:2) Get 3 more ratios from inverses of above for a total of 6 for this reaction

3Fe(s) + 4H2O(l)  Fe3O4(s) + 4H2(g)
Mole Ratio In general, for n species (reactant, product) in chemical equation, have n(n-1) mole ratios 3Fe(s) + 4H2O(l)  Fe3O4(s) + 4H2(g) 4 species, 4  3 = 12 possible mole ratios (6 + inverse of each) Fe:H2O Fe:Fe3O Fe:H2 H2O:Fe3O4 H2O:H2 Fe3O4: H2

Mole Ratio How many mole ratios can be written for the following reaction? 4H2(g) + O2(g) → 2H2O(l) A B C D. 2 ?

Practice Problems 3 (a-c), 4 (a-b) page 372 Problems 47 - 53, page 392

Section 11.2 Stoichiometric Calculations
The solution to every stoichiometric problem requires a balanced chemical equation. List the sequence of steps used in solving stoichiometric problems. Solve stoichiometric problems (mole-to-mole, mole-to-mass, mass-to-mass and variations including number of particles) using the factor-label method.

Section 11.2 Stoichiometric Calculations
Key Concepts Chemists use stoichiometric calculations to predict the amounts of reactants used and products formed in specific reactions. The first step in solving stoichiometric problems is writing the balanced chemical equation. Mole ratios derived from the balanced chemical equation are used in stoichiometric calculations. Stoichiometric problems make use of mole ratios to convert between mass and moles.

Stoichiometric Calculations
Start with balanced equation Determine required mole ratios Use mole-to-mass and mass-to-mole relations for 3 basic conversions: Mole to mole Mole to mass Mass to mass Once have moles, can get # particles

nA + mB  xC + yD Mass reactant Mass product Moles reactant Moles
4 1 Moles reactant Moles product 2 5 # part. reactant # part. product 3 6

Mole to Mole Conversions
Given number of moles of a reactant (product), how many moles of a product (reactant) must be formed? Use initial moles and mole ratio mol known x [mol unknown/mol known] = mol unknown mole ratio from equation

2K(s) + 2H2O(l)  2KOH(aq) + H2(g) Mol H2 if start with 4.00x10-2 mol K? 4.00x10-2 mol K  [1 mol H2/2 mol K] mole ratio from equation = 2.00x10-2 mol H2

How many moles of CO2 will be produced in following reaction if initial amount of reactants = 0.50 moles? 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) A B C D. 1.0 ?

Practice Problems 11,12 page 375 Problems 61- 62, page 393

Mole to Mass Conversions
Step 1 – Do mole to mole conversion using mole ratio Step 2 – Do mole to mass conversion using molar mass

Mole to Mass Conversions
2Na(s) + Cl2(g)  2NaCl(s) Mass NaCl produced from 1.25 mol Cl2? 1.25 mol Cl2  [2 mol NaCl/1 mol Cl2] mole ratio from equation = 2.50 mol NaCl 2.50 mol NaCl  [58.44 g NaCl/mol NaCl] molar mass of NaCl = 146 g NaCl

Practice Problems 13,14 page 376 Problems 63 - 66, page 393

Mass to Mass Conversions
Step 1 – Do mass to mole conversion using molar mass Step 2 – Do mole to mole conversion using mole ratio Step 3 – Do mole to mass conversion using molar mass

NH4NO3(s)  N2O(g) + 2H2O(g) Mass of H2O from 25.0 g NH4NO3(s)? AN= NH4NO3(s) 25.0 g AN  [1 mol AN/80.04 g AN] 1/molar mass of AN = mol AN 0.312 mol AN  [2 mol H2O/1 mol AN] mole ratio from equation = mol H2O

NH4NO3(s)  N2O(g) + 2H2O(g) Mass of H2O from 25.0 g NH4NO3(s)? 0.624 mol H2O  [18.02 g H2O/mol H2O] molar mass of H2O = 11.2 g H2O

Practice Problems 15,16 page 377 Problems 60, 67- 72, pages 393-4

Section 11.3 Limiting Reactants
A chemical reaction stops when one of the reactants is used up. Identify the limiting reactant in a chemical equation. Identify the excess reactant, and calculate the amount remaining after the reaction is complete. Calculate the mass of a product when the amounts of more than one reactant are given.

Section 11.3 Limiting Reactants
Key Concepts The limiting reactant is the reactant that is completely consumed during a chemical reaction. Reactants that remain after the reaction stops are called excess reactants. To determine the limiting reactant, the actual mole ratio of the available reactants must be compared with the ratio of the reactants obtained from the coefficients in the balanced chemical equation. Stoichiometric calculations must be based on the limiting reactant.

Limiting Reactants Typical case: reactants not present in exact stoichiometric ratios Mole ratio given by coefficients of reactants in balanced chemical equation N2 + 3H2  2NH3 Stoichiometric ratio of reactants for ammonia synthesis is 1 N2: 3 H2 If ratio  1:3 then one reactant is limiting

Limiting and Excess Reactants Ammonia Synthesis
N2(g) + 3H2(g)  2NH3(g) Case of 1:1 N2:H2 mole ratio Before Reaction After Reaction 3N2 3H2 2NH3 2N2 N2 is excess reactant H2 is limiting reactant

Limiting and Excess Reactants
Limiting reactant Limits the extent of reaction Determines the amount of product Excess reactant Always have “left over” Often a reactant is deliberately used in excess to: Drive reaction to completion (equilibrium) Speed up reaction (kinetics)

Limiting and Excess Reactants
Limiting and excess air (oxygen) in combustion reaction Air limiting – soot (carbon) Air excess – CO2 & H2O

Limiting Reactant - Identifying
Step 1 – If masses given, convert to moles Step 2 – Compare actual mole ratio of reactants to mole ratio determined from balanced equation Ratios same - no limiting reactant Ratios different - compare to identify limiting reactant

Limiting Reactant S8(l) + 4Cl2(g)  4S2Cl2(l)
200.0 g sulfur reacts with g Cl2 Part a: g of product produced? Step 1 – Convert masses to moles 100.0 g Cl2  [1 mol Cl2 /70.91 g Cl2] 1/molar mass of S8 = mol Cl2 1/molar mass of Cl2 200.0 g S8  [1 mol S8 /256.5 g S8] = mol S8

200.0 g sulfur reacts with g Cl2 Part a: g product ? Step 2 – Compare actual mole ratio to mole ratio from equation Equation: Cl2:S8 = 4:1 Actual: mol Cl2: mol S8 = 1.808:1 Compare ratios  Cl2 is limiting

200.0 g sulfur reacts with g Cl2 Part a: g product ? Mole ratio Cl2 limiting reactant, have mol Calculate mol product using mole ratio 1.410 mol Cl2  [4 mol S2Cl2 /4 mol Cl2] molar mass of S2Cl2 = mol S2Cl2 1.410 mol S2Cl2  [135.0 g S2Cl2/mol S2Cl2] = g S2Cl2

200.0 g sulfur reacts with g Cl ? Part b: g of sulfur actually reacted? Mole ratio Cl2 limiting reactant, have mol Calculate mol sulfur using mole ratio 1.410 mol Cl2  [1 mol S8 /4 mol Cl2] molar mass of S8 = mol S8 mol S8  [256.5 g S8 /mol S8] = g S8 reacts g excess

200.0 g sulfur reacts with g Cl ? Part b: g of sulfur actually reacted? Know that all Cl2(g) reacted (100.0 g) Previously calculated g product S2Cl2(l) produced Cons. mass: g = starting mass = final mass = m[S8(l)] + m[4S2Cl2(l)] m[S8(l)]= g -m[4S2Cl2(l)] =109.6 g

Limiting Reactant Excess reactant in following reaction if start with 50.0g of each reactant? P4(s) + 5 O2(g) → P4O10(s) A. O B. P4 C. None – quantities are stoichiometric ? 50.0 g P4  [1 mol P4 / g P4]= mol P4 50.0 g O2  [1 mol O2 /32.00 g O2] = 1.56 mol O2 Need 5  mol = 2.02 mol O2  O2 limiting

Practice Problems 23, 24 page 383 Problems 77 - 82, page 394

Section Percent Yield Percent yield is a measure of the efficiency of a chemical reaction. Calculate the theoretical yield of a chemical reaction from data. Explain the difference between a theoretical yield of a chemical reaction and an actual yield. Provide reasons why the theoretical yield and actual yield can be different from each other. Determine the percent yield for a chemical reaction. Explain why the percent yield can exeed 100%

Section 11.4 Percent Yield Key Concepts
The theoretical yield of a chemical reaction is the maximum amount of product that can be produced from a given amount of reactant. Theoretical yield is calculated from the balanced chemical equation. The actual yield is the amount of product produced. Actual yield must be obtained through experimentation. Percent yield is the ratio of actual yield to theoretical yield expressed as a percent. High percent yield is important in reducing the cost of every product produced through chemical processes.

Percent Yield Actual (real world) reactions don’t produce amount of product expected from masses of reactant and the balanced chemical equation (stoichiometry)

Types of Yields Theoretical Yield Actual Yield
Maximum amount of product that can be produced from given amounts of reactants Calculated using stoichiometry Actual Yield Amount of product actually obtained when the reaction is carried out

Types of Yields Theoretical Yield and Actual Yield can differ because
Reaction stops before limiting reactant all used up Product is not all collected – sticks to reaction vessel, filter paper, etc Reactions other than the intended reaction occur, producing other products Impurities in reactants

Alternate Reaction Paths
Combustion of natural gas (methane) CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Know this reaction takes place if have enough O2 – get nice blue flame Know if limit O2, get yellow flame and soot (carbon) production Same reactants yield different products

Percent Yield Percent yield is a way of expressing the efficiency of a reaction Percent yield = 100  actual yield theoretical yield Theoretical yield computed using same procedures used in previous sections (stoichiometry)

% Percent Yield Theoretical yield of chemical reaction = 25 g. % yield if actual yield = 22 g ? A. 88% B. 44% C. 50% D. 97% ? % yield = 100  22 g / 25 g = 88%

Percent Yield In commercial production of chemicals, attempt to produce as close to 100% yield as possible Optimum use of materials Minimizes or eliminates need to separate product from unused reactants and from unwanted products

Percent Yield for Multi-Stage
Attempt to produce as close to 100% yield as possible Critical for multi-stage reaction processes where product of one reaction used as one of the reactants of a subsequent reaction A three-stage reaction with 90% yield from each stage gives only 73% overall yield (0.9 x 0.9 x 0.9 = 0.73)

>100% Percent Yield Actual Yield: Amount of product actually obtained when from reaction Obtained from a laboratory measurement – most common is measuring mass on a balance Assumption made is that the only “stuff” you’re measuring is the product of interest

>100% Percent Yield Possible to get more “stuff” you think is product than predicted if: Haven’t completely separated intended product from other products in reaction or from leftover reactants and/or solvent (insufficient drying) Product reacts with air or water vapor before it is weighed

Practice Problems 28 - 30 page 387 Problem 35, page 388

Chapter 11 – Stoichiometry

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