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301 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.

302 2 K+(aq) + SO42-(aq) + Ba2+(aq) + 2 Cl-(aq) 2 K+(aq) + 2 Cl-(aq) + BaSO4(s)

303 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.

304 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)

305 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.

306 Quantitative Aspects of Chemical Reactions

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

308 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.

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

310 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.

311 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:

312 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.

313 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.

314 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.

315 Solving Stoichiometry Problems using the mole method

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

317 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.

318 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.

319 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).

320 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.

321 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?

322 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 g of dioxygen with sufficient dinitrogen?

323 Part (a): Step (1) N O NO2

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

325 Part (a): Step (1) N O 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 =

326 Part (a): Step (1) N O 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 =

327 Part (a): Step (1) N O 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.

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

329 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 g/mol, therefore the number of moles of O2 is

330 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 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 =

331 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 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 =

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

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

334 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 g of sodium chloride.

335 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 g of sodium chloride. 3 CaCl2 + 2 Na3PO Ca3(PO4) NaCl Step (1) given above. Steps (2), (3), and (4)

336 Mass of sodium phosphate produced =
= g

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

338 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.

339 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.

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

341 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)

342 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.

343 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) HCl MgCl H2O.

344 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) HCl MgCl H2O. If g of Mg(OH)2 is combined with g of HCl, how many grams of MgCl2 could be formed?

345 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.

346 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) g/mol for HCl g/mol

347 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) g/mol for HCl g/mol Moles of Mg(OH)2 = = mol

348 Moles of HCl =

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

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

351 Therefore grams of MgCl2 formed =

352 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.

353 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 =

354 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 =

355 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 = = g MgCl2

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

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

358 Theoretical, actual, and percentage yields

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

360 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.


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