Chapter 9-Stoichiometry 9.1-Introduction to Stoichiometry 9.2-Ideal Stoichiometric Calculations 9.3-Limiting Reactants & Percent Yield

Introduction to Stoichiometry

Reaction Stoichiometry  Involves the mass relationships between reactants and products in a chemical reaction.  4 types of problems (given, unknown) Mole to mole Mole to mass Mass to mole Mass to mass

Conversion Factors  There are only 2 equalities that you will use to solve these problems!!  Always start with a balanced chemical equation.  From this, you can derive the mole ratio A conversion factor that relates the amounts in moles of any two substances involved in a chemical equation.

Mole Ratio 2Al 2 O 3(l)  4Al (s) + 3O 2(g)  We can write mole ratios that relate any two substances involved in the reaction.  For example:  Which substances you use and which in on top/bottom is determined by what you are converting to/from.

Molar Mass  The other conversion factor you may use is the molar mass.  Used to convert between moles and grams of the same substance.  Numbers are off of the periodic table and rounded off at the hundredths position before use.  Ex: Molar mass of Al 2 O 3 = 2(26.98 g/mol) + 3(16.00 g/mol)= g/mol

Ideal Stoichiometric Calculations

4 types of problems  Mole to mole Mole to mole  Mole to mass Mole to mass  Mass to mole Mass to mole  Mass to mass Mass to mass  General pathway: Mass given  mole given  mole unknown  mass unknown

Mole to mole Mass given  mole given  mole unknown  mass unknown  Given quantity is in moles  Unknown quantity is in moles  Need 1 conversion factor to solve Mole ratio to convert between mole given & mole unknown

Mole to mole example  CO 2(g) + 2LiOH (s)  Li 2 CO 3(s) + H 2 O (l) How many moles of lithium hydroxide are required to react with 20. mol of CO 2 ?  Given: 20. mol CO 2  Unknown: ? mol LiOH

Mole to mole example  The elements lithium and oxygen react explosively to form lithium oxide. How many moles of lithium oxide will form if 2 mole of lithium react with unlimited oxygen?  Balanced eqn: 4Li + O 2  2Li 2 O  Answer: 1 mol Li 2 O

Mole to mass Mass given  mole given  mole unknown  mass unknown  Given quantity is in moles  Unknown quantity is mass (g, kg, etc.)  Need 2 conversion factors to solve: Mole ratio to convert from mole given to mole unknown Molar mass to convert from mole unknown to mass unknown

Mole to mass example  6CO 2(g) + 6H 2 O (l)  C 6 H 12 O 6(s) + 6O 2(g) What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?  Given: 3.00 mol water  Unknown: ? grams glucose

Mole to mass example  2NaN 3(s)  2Na (s) + 3N 2(g) If mol of NaN 3 react, what mass in grams of N 2 would result?  Given: mol NaN 3  Unknown: ? grams N 2  Answer: 21.0 g N 2

Mass to mole Mass given  mole given  mole unknown  mass unknown  Given quantity is a mass (g, kg, etc.)  Unknown quantity is in moles  Need 2 conversion factors to solve: Molar mass to convert from mass given to mole given Mole ratio to convert from mole given to mole unknown

Mass to mole example  4NH 3(g) + 5O 2(g)  4NO (g) + 6H 2 O (g) This reaction is run using 824 g NH 3 & excess O 2. How many moles of NO are formed?  Given: 824 g NH 3  Unknown: ? mol NO

Mass to mole example  4NH 3(g) + 5O 2(g)  4NO (g) + 6H 2 O (g) This reaction is run using 824 g NH 3 & excess O 2. How many moles of H 2 O are formed?  Given: 824 g NH 3  Unknown: ? mol H 2 O  Answer: 72.6 mol H 2 O

Mass to mass Mass given  mole given  mole unknown  mass unknown  Given quantity is a mass (g, kg, etc.)  Unknown quantity is a mass  Need 3 conversion factors to solve: Molar mass to convert from mass given to mole given Mole ratio to convert from mole given to mole unknown Molar mass to covert from mole unknown to mass unknown

Mass to mass example  NH 4 NO 3(s)  N 2 O (g) + 2H 2 O (l) How many grams of NH 4 NO 3 are required to produce 33.0 g N 2 O?  Given: 33.0 g N 2 O  Unknown: ? g NH 4 NO 3

Mass to mass example  What mass of aluminum is produced by the decomposition of 5.0 kg of Al 2 O 3 ?  Given: 5.0 kg Al 2 O 3  Unknown: ? g Al  Balanced eqn: 2Al 2 O 3  4Al + 3O 2  Answer: 2.6 kg

Ideal conditions  These problems tell us the amount of reactants or products under ideal conditions. All reactants are completely converted into products. Give the maximum yield we could expect, but this is rarely attained in the field because we don’t have ideal conditions.

Limiting Reactants & Percent Yield

Limiting & excess reactants  Once one of the reactants is completely used up in a rxn, it doesn’t matter how much of the other reactant(s) you have. The reaction cannot continue.  Limiting reactant: The reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction.

Limiting & excess reactants  Excess reactant(s) Substance(s) that are not completely used up an a rxn and do not limit the amount of product that can be formed. These are the reactants that are “left over” at the end of a rxn.

Which is limiting? Excess?  To decide which reactant is limiting in a rxn, use one of your givens to solve for the other.  Then compare how much you have (given) to how much you need under ideal conditions (solved for). If you have more than you need, the 1 st substance is limiting. If you have less than you need, the 2 nd substance is limiting.

Limiting reactant example  CO (g) + 2H 2(g)  CH 3 OH If 500. mol of CO and 750. mol of H 2 are present, which is the limiting reactant? Solve to determine how much H 2 would be needed to completely react 500. mol CO. Do you have 1000 mol H 2 ? No, you only have 750. mol. H 2 is the limiting reactant.

Limiting reactant example 3ZnCO 3(s) + 2C 6 H 8 O 7(aq)  Zn 3 (C 6 H 5 O 7 ) 2(aq) + 3H 2 O (l) + 3CO 2(g)  If there is 1 mol of ZnCO 3 & 1 mol of C 6 H 8 O 7, which is the limiting reactant?  Answer: 1 mol ZnCO 3 could react with 0.67 mol C 6 H 8 O 7, which is less than is available. ZnCO 3 is limiting.

Limiting reactant example  Aspirin, C 9 H 8 O 4, is synthesized by the rxn of salicylic acid, C 7 H 6 O 3, with acetic anhydride, C 4 H 6 O 3. 2C 7 H 6 O 3 + C 4 H 6 O 3  2C 9 H 8 O 4 + H 2 O When 20.0g of C 7 H 6 O 3 and 20.0g of C 4 H 6 O 3 react, which is the limiting reactant? How many moles of the excess reactant are used when the rxn is complete? What mass in grams of aspirin is formed?  C 7 H 6 O 3, mol, 26.1 g

Percent Yield  Theoretical yield-maximum amount of product that can be produces from a given amount of reactant. This is what we calculate using ideal stoichiometric calculations.  Actual yield-the measured amount of a product obtained from a rxn.

Percent Yield  Ratio of the actual yield to the theoretical yield, multiplied by 100.

Percent Yield example  C 6 H 6(l) + Cl 2(g)  C 6 H 5 Cl (s) + HCl (g) When 36.8g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8g. What is the percent yield?  Given: 36.8g C 6 H 6, excess Cl 2, actual yield=38.8g C 6 H 5 Cl  Unknown: ? g C 6 H 5 Cl, percent yield

Percent Yield example  Methanol can be produced through the rxn of CO and H 2 in the presence of a catalyst.  If 75.0 g of CO reacts to produce 68.4 g CH 3 OH, what is the percent yield?  Answer: 79.7%

Percent Yield example  2ZnS (s) + 3O 2(g)  2ZnO (s) + 2SO 2(g) If the typical yield is 86.78%, how much SO 2 should be expected if 4897 g of ZnS are used?