K2SO4(aq) + BaCl2(aq) 2 KCl + BaSO4 ionic. yes ionic. yes ionic K2SO4(aq) + BaCl2(aq) 2 KCl + BaSO4 ionic? yes ionic? yes ionic? yes ionic? yes soluble? yes soluble? yes soluble? yes soluble? no Key comment: The subscript label “aq” tells you that the particular compound is soluble. If the answer to both questions is yes, then split the compound into ions.

2 K+(aq) + SO42-(aq) + Ba2+(aq) + 2 Cl-(aq) 2 K+(aq) + 2 Cl-(aq) + BaSO4(s)

2 K+(aq) + SO42-(aq) + Ba2+(aq) + 2 Cl-(aq) 2 K+(aq) + 2 Cl-(aq) + BaSO4(s) Now cancel the common species that are exactly the same on both sides of the equation.

2 K+(aq) + SO42-(aq) + Ba2+(aq) + 2 Cl-(aq) 2 K+(aq) + 2 Cl-(aq) + BaSO4(s) Now cancel the common species that are exactly the same on both sides of the equation. SO42-(aq) + Ba2+(aq) BaSO4(s)

2 K+(aq) + SO42-(aq) + Ba2+(aq) + 2 Cl-(aq) 2 K+(aq) + 2 Cl-(aq) + BaSO4(s) Now cancel the common species that are exactly the same on both sides of the equation. SO42-(aq) + Ba2+(aq) BaSO4(s) This is the required net ionic equation.

Quantitative Aspects of Chemical Reactions

Quantitative Aspects of Chemical Reactions Objective: The study of the quantitative relations between amounts of reactants and products.

Quantitative Aspects of Chemical Reactions Objective: The study of the quantitative relations between amounts of reactants and products. Stoichiometry: The stoichiometry of a reaction is the description of the relative quantities by moles of the reactants and products as given by the coefficients of the balanced equation for the reaction.

Stoichiometry Calculations Coefficients in a balanced chemical equation give relative quantities of reactants and products by moles as well as by molecules.

Stoichiometry Calculations Coefficients in a balanced chemical equation give relative quantities of reactants and products by moles as well as by molecules. Example: 2 H2 + O2 2H2O This can be read as: 2 molecules of H2 react with 1 molecule of O2 to give 2 molecules of water.

Or multiply both sides by 6 Or multiply both sides by 6.02 x 1023 (Avogadro’s number), so we can read the equation as:

Or multiply both sides by 6 Or multiply both sides by 6.02 x 1023 (Avogadro’s number), so we can read the equation as: 2 moles of H2 react with 1 mole of O2 to give 2 moles of water.

Or multiply both sides by 6 Or multiply both sides by 6.02 x 1023 (Avogadro’s number), so we can read the equation as: 2 moles of H2 react with 1 mole of O2 to give 2 moles of water. This statement in terms of moles will be the most useful one for solving stoichiometry problems.

Or multiply both sides by 6 Or multiply both sides by 6.02 x 1023 (Avogadro’s number), so we can read the equation as: 2 moles of H2 react with 1 mole of O2 to give 2 moles of water. This statement in terms of moles will be the most useful one for solving stoichiometry problems. The essential idea is that the ratio of reactants and products is 2 : 1: 2 in both statements.

Solving Stoichiometry Problems using the mole method

Four steps: (1) Obtain the correct formulas for all reactants and products and balance the chemical equation.

Four steps: (1) Obtain the correct formulas for all reactants and products and balance the chemical equation. (2) Convert all of the known amounts of the substances into moles.

Four steps: (1) Obtain the correct formulas for all reactants and products and balance the chemical equation. (2) Convert all of the known amounts of the substances into moles. (3) Use the coefficients of the substances in the balanced equation to calculate the number of moles of the unknown quantities in the problem.

Four steps: (1) Obtain the correct formulas for all reactants and products and balance the chemical equation. (2) Convert all of the known amounts of the substances into moles. (3) Use the coefficients of the substances in the balanced equation to calculate the number of moles of the unknown quantities in the problem. (4) Convert the number of moles of the unknown quantities to grams (or other mass units).

Example: Nitrogen dioxide is a major air pollutant Example: Nitrogen dioxide is a major air pollutant. It may be formed by the direct combination of dinitrogen and dioxygen.

Example: Nitrogen dioxide is a major air pollutant Example: Nitrogen dioxide is a major air pollutant. It may be formed by the direct combination of dinitrogen and dioxygen. (a) How many moles of NO2 can be formed by the reaction of 4.75 moles of dinitrogen with sufficient dioxygen?

Example: Nitrogen dioxide is a major air pollutant Example: Nitrogen dioxide is a major air pollutant. It may be formed by the direct combination of dinitrogen and dioxygen. (a) How many moles of NO2 can be formed by the reaction of 4.75 moles of dinitrogen with sufficient dioxygen? (b) How many grams of NO2 can be formed by the reaction of 0.568 g of dioxygen with sufficient dinitrogen?

Part (a): Step (1) N2 + 2 O2 2 NO2

Part (a): Step (1) N2 + 2 O2 2 NO2 Step (2) No conversion is needed here, since the amount of starting material, N2, is given in moles.

Part (a): Step (1) N2 + 2 O2 2 NO2 Step (2) No conversion is needed here, since the amount of starting material, N2, is given in moles. Step (3) From the balanced equation (step (1)), 1 mol of N2 produces 2 mol of NO2, hence mol of NO2 produced =

Part (a): Step (1) N2 + 2 O2 2 NO2 Step (2) No conversion is needed here, since the amount of starting material, N2, is given in moles. Step (3) From the balanced equation (step (1)), 1 mol of N2 produces 2 mol of NO2, hence mol of NO2 produced =

Part (a): Step (1) N2 + 2 O2 2 NO2 Step (2) No conversion is needed here, since the amount of starting material, N2, is given in moles. Step (3) From the balanced equation (step (1)), 1 mol of N2 produces 2 mol of NO2, hence mol of NO2 produced = Step (4) Not required.

Part (b) Step (1) N2 + 2 O2 2 NO2

Part (b) Step (1) N2 + 2 O2 2 NO2 Step (2) The molar mass of O2 is 31 Part (b) Step (1) N2 + 2 O2 2 NO2 Step (2) The molar mass of O2 is 31.9988 g/mol, therefore the number of moles of O2 is

Part (b) Step (1) N2 + 2 O2 2 NO2 Step (2) The molar mass of O2 is 31 Part (b) Step (1) N2 + 2 O2 2 NO2 Step (2) The molar mass of O2 is 31.9988 g/mol, therefore the number of moles of O2 is Step (3) Since 2 moles of O2 produce 2 moles of NO2, the number of moles of NO2 obtained =

Part (b) Step (1) N2 + 2 O2 2 NO2 Step (2) The molar mass of O2 is 31 Part (b) Step (1) N2 + 2 O2 2 NO2 Step (2) The molar mass of O2 is 31.9988 g/mol, therefore the number of moles of O2 is Step (3) Since 2 moles of O2 produce 2 moles of NO2, the number of moles of NO2 obtained =

Step (4) The molar mass of NO2 is 46 Step (4) The molar mass of NO2 is 46.0055 g/mol, hence mass of NO2 formed =

After some practice it is convenient to combine steps 2, 3, and 4 into a single factor label equation.

After some practice it is convenient to combine steps 2, 3, and 4 into a single factor label equation. Example: From the following balanced equation, calculate the number of grams of sodium phosphate needed to prepare 45.97 g of sodium chloride.

After some practice it is convenient to combine steps 2, 3, and 4 into a single factor label equation. Example: From the following balanced equation, calculate the number of grams of sodium phosphate needed to prepare 45.97 g of sodium chloride. 3 CaCl2 + 2 Na3PO4 Ca3(PO4)2 + 6 NaCl Step (1) given above. Steps (2), (3), and (4)

Mass of sodium phosphate produced = = 42.98 g

Limiting reagents When a reaction is carried out, the reactants are usually not present in the exact stoichiometric amounts indicated by the balanced equation.

Limiting reagents When a reaction is carried out, the reactants are usually not present in the exact stoichiometric amounts indicated by the balanced equation. Often, one or more reactants will be present in excess of the theoretical amount needed. The reactant used up first in a reaction is called the limiting reagent.

Limiting reagents When a reaction is carried out, the reactants are usually not present in the exact stoichiometric amounts indicated by the balanced equation. Often, one or more reactants will be present in excess of the theoretical amount needed. The reactant used up first in a reaction is called the limiting reagent. The maximum amount of product produced depends solely on how much of the limiting reagent is present.

The concept of limiting reagent plays an important role in gravimetric analysis.

The concept of limiting reagent plays an important role in gravimetric analysis. Example: Consider the determination of sodium chloride in an unknown sample using the reaction: AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

The concept of limiting reagent plays an important role in gravimetric analysis. Example: Consider the determination of sodium chloride in an unknown sample using the reaction: AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) We must add sufficient AgNO3 (i.e. an excess) to the unknown sample of NaCl. The NaCl is the limiting reagent in this case.

Example: The chief ingredient of milk of magnesia is magnesium hydroxide, Mg(OH)2. Being a base, it neutralizes excess acid – largely hydrochloric acid – in our stomachs. The reaction is Mg(OH)2 + 2 HCl MgCl2 + 2 H2O.

Example: The chief ingredient of milk of magnesia is magnesium hydroxide, Mg(OH)2. Being a base, it neutralizes excess acid – largely hydrochloric acid – in our stomachs. The reaction is Mg(OH)2 + 2 HCl MgCl2 + 2 H2O. If 16.14 g of Mg(OH)2 is combined with 10.97 g of HCl, how many grams of MgCl2 could be formed?

Approach 1: Since we cannot tell by inspection which of the two reactants is the limiting reagent, we must proceed by first converting the reactant masses into moles.

The molar masses are: for Mg(OH)2 58.32 g/mol for HCl 36.46 g/mol Approach 1: Since we cannot tell by inspection which of the two reactants is the limiting reagent, we must proceed by first converting the reactant masses into moles. The molar masses are: for Mg(OH)2 58.32 g/mol for HCl 36.46 g/mol

The molar masses are: for Mg(OH)2 58.32 g/mol for HCl 36.46 g/mol Approach 1: Since we cannot tell by inspection which of the two reactants is the limiting reagent, we must proceed by first converting the reactant masses into moles. The molar masses are: for Mg(OH)2 58.32 g/mol for HCl 36.46 g/mol Moles of Mg(OH)2 = = 0.2768 mol

Moles of HCl =

Moles of HCl = Now 0.2768 moles of Mg(OH)2 would require

Moles of HCl = Now 0.2768 moles of Mg(OH)2 would require But there are only 0.3009 moles of HCl present, hence HCl is the limiting reagent. The rest of the calculation is based on the amount of limiting reagent.

Therefore grams of MgCl2 formed =

Approach 2: Determine the mass of MgCl2 formed from the amount of each reactant, then take the smaller of the two results, since this arises from the limiting reagent.

Approach 2: Determine the mass of MgCl2 formed from the amount of each reactant, then take the smaller of the two results, since this arises from the limiting reagent. g of MgCl2 possible from the available Mg(OH)2 =

Approach 2: Determine the mass of MgCl2 formed from the amount of each reactant, then take the smaller of the two results, since this arises from the limiting reagent. g of MgCl2 possible from the available Mg(OH)2 =

Approach 2: Determine the mass of MgCl2 formed from the amount of each reactant, then take the smaller of the two results, since this arises from the limiting reagent. g of MgCl2 possible from the available Mg(OH)2 = = 26.35 g MgCl2

g of MgCl2 possible from the available HCl = = 14.32 g MgCl2

g of MgCl2 possible from the available HCl = = 14.32 g MgCl2 Since the smaller result is 14.32 g, then 14.32 g of MgCl2 is formed, and HCl is the limiting reagent.

Theoretical, actual, and percentage yields

Theoretical, actual, and percentage yields Theoretical yield: The amount of product obtained when all the limiting reagent has reacted.

Theoretical, actual, and percentage yields Theoretical yield: The amount of product obtained when all the limiting reagent has reacted. Actual yield: Amount of product actually obtained by experiment from a chemical reaction.