Example Problems. PV nt =R so P i. V i n i T i. = P f V f n f T f. (since V and n are constant) = P f T f. T i

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1 The temperature of 2.5 L of a gas initially at STP is raised to 250 o C at constant volume. Calculate the final pressure of the gas in atm. PV nt =R so P i V i n i T i = P f V f n f T f or P i T i = P f T f (since V and n are constant)

2 The temperature of 2.5 L of a gas initially at STP is raised to 250 o C at constant volume. Calculate the final pressure of the gas in atm. P f = P i T f T i = 1atm 523 K 298 K = 1.75atm

3 At STP, L of a gas weighs 0.400g. Calculate the molar mass of the gas. M= mrt PV = 0.400g Latm/molK 298 K 1 atm 0.280l =35.0 g/mol

4 A compound has the empirical formula SF 4. At 20 o C, 0.100g of the gaseous compound occupies a volume of 22.1 ml and exerts a pressure of 1.02 atm. What is the molecular formula of the compound? Empirical formula mass = g/mol M = mrt/pv = x x 293/ 1.02 x = 107 g/mol ratio molar mass to empirical formula mass: 107 : 108 1:1 molecular formula: SF 4

5 Methane, the principal component of natural gas, is used for heating and cooking. The combustion process is: CH 4 + 2O 2 CO 2 + 2H 2 O If 15.0 moles of CH 4 are reacted what is the volume of CO 2 in liters produced at 23.0 o C and atm? Mols CO 2 produced = 15.0 mol CH 4 x 1 mol CO 2 /1 mol CH 4 = 15.0 mols V = nrt/p = 15.0 mol x latm/molk x 296 K/ atm = 370 L

6 A piece of sodium metal reacts completely with water as follows: 2Na + 2H 2 O 2NaOH + H 2 The H 2 gas is collected over water at 25.0 o C. The volume of the gas is 246 ml measured at 1.00 atm. Calculate the number of grams of Na used in the reaction (Vapor pressure of water at 25 o C = atm). P T = P H2 + P H2O P H2 = P T - P H2O = = 0.97 atm

7 A piece of sodium metal reacts completely with water as follows: 2Na + 2H 2 O 2NaOH + H 2 The H 2 gas is collected over water at 25.0 o C. The volume of the gas is 246 ml measured at 1.00 atm. Calculate the number of grams of Na used in the reaction (Vapor pressure of water at 25 o C = atm). n H2 = PV/RT = 0.97 atm x L / Latm/molK x 298 K = mol mols Na = mol H 2 x 2 mol Na/ 1 mol H 2 = mol mass Na = mol x g/mol = 0.45g

8 A mixture of Ar and N 2 gases has a density of g/l at STP. What is the mole fraction of the gas? Recalling the equation M = drt/p that can be used to calculate a molar mass from a density: M = g/l x Latm/molK x 298 K / 1 atm = 34.6 g/mol This is intermediate between the molar masses of N 2, g/mol, and Ar, g/mol and is the weighted average of the two... in exactly the same way that the average atomic mass is the weighted average of isotopes.

9 A mixture of Ar and N 2 gases has a density of g/l at STP. What is the mole fraction of the gas? 34.6 = X N2 x (1-X N2 ) x X N2 = ( )/( ) = therefore, X Ar = = 0.55

10 A 36.4 L volume of methane gas is heated from 25 o C to 88 o C at constant pressure. What is the final volume of the gas? P f = P i T f T i = 1atm 523 K 298 K =1.75atm

11 Iodine, I 2, reacts with aqueous thiosulfate ion in neutral solution according to the equation: I 2 + S 2 O 3 2- S 4 O I - How many grams of I 2 are present in a solution if ml of M Na 2 S 2 O 3 solution is needed to titrate the I 2 solution? Balance the equation: 1. I 2 + 2e - 2I - 2S 2 O 3 2- S 4 O e - 2. I 2 + 2S 2 O 3 2- S 4 O I -

12 Iodine, I 2, reacts with aqueous thiosulfate ion in neutral solution according to the equation: I 2 + 2S 2 O 3 2- S 4 O I - How many grams of I 2 are present in a solution if ml of M Na 2 S 2 O 3 solution is needed to titrate the I 2 solution? Mols Na 2 S 2 O 3 = L x mol/l = mol mols I 2 = mols Na 2 S 2 O 3 x 1 mol I 2 / 2 mols Na 2 S 2 O 3 = mol mass I 2 = mol x g/mol = g

13 Iron in a sample can be determined by titrating with cerium ion: Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ What is the mass percentage of iron in a sample if titration of g of the sample requires ml of M Ce 4+? Mols Ce 4+ = L x mol/l = mol mols Fe 2+ = mol Ce 4+ x 1 mol Fe 2+ /1 mol Ce 4+ = mol mass Fe = mol x g/mol = g mass %Fe = (0.3292g/1.2284g) x 100 = 26.80%

14 If 1.87 g of acetic acid, C 2 H 4 O 2, reacts with 2.31 g of isopentyl alcohol, C 5 H 12 O to give 2.96 g of isopentyl acetate, C 7 H 14 O 2, what is the percent yield? Balanced equation: C 2 H 4 O 2 + C 5 H 12 O C 7 H 14 O 2 + H 2 O limiting reagent approach 1. mols acetic acid = 1.87 g / g/mol = mol mols C 7 H 14 O 2 = mol 2. mols C 5 H 12 O = 2.31 g / g/mol = mol mols C 7 H 14 O 2 = mol So, C 5 H 12 O is the limiting reagent.

15 If 1.87 g of acetic acid, C 2 H 4 O 2, reacts with 2.31 g of isopentyl alcohol, C 5 H 12 O to give 2.96 g of isopentyl acetate, C 7 H 14 O 2, what is the percent yield? Mass C 7 H 14 O 2 = mol x g/mol = 3.41 g (theoretical yield) Percent yield = (actual yield/theoretical yield) x 100 = (2.96/3.41) x 100 = 78.9%

16 An average cup of coffee contains about 125 mg of caffeine, C 8 H 10 N 4 O 2. How many moles of caffeine are in a cup? How many molecules of caffeine? Mols caffeine = 125 mg x (1 g/1000 mg) / g/mol = 6.44 x 10-4 mol Molecules caffeine = 6.44 x 10-4 mol x x molecule/mol = 3.88 x molecules

17 A g sample of a metal carbonate MCO 3 was reacted with ml of M H 2 SO 4 yielding CO 2 gas and an aqueous solution of the metal sulfate. The CO 2 was removed by boiling and the remaining H 2 SO 4 was titrated with M NaOH, requiring ml to reach the endpoint. What is the metal? 2NaOH + H 2 SO 4 Na 2 SO 4 + 2H 2 O starting mols H 2 SO 4 = L x mol/l = mol mols H 2 SO 4 left = mols NaOH x 1 mol H 2 SO 4 / 2 mol NaOH = L x mol/l x ½ = mol mols H 2 SO 4 used = = mol

18 A g sample of a metal carbonate MCO 3 was reacted with ml of M H 2 SO 4 yielding CO 2 gas and an aqueous solution of the metal sulfate. The CO 2 was removed by boiling and the remaining H 2 SO 4 was titrated with M NaOH, requiring ml to reach the endpoint. What is the metal? MCO 3 + H 2 SO 4 MSO 4 + H 2 O mols H 2 SO 4 used = mols MCO 3 = mol molar mass MCO 3 = g/ mol = g/mol mass M in 1 mol = mass MCO 3 mass CO 3 = = g/mol Barium

19 The fictional isotope Xium-60 has an atomic mass of units while Xium 70 has an atomic mass of units. The average mass of Xium is units. What is the fractional abundance of Xium 70? Average atomic mass = fraction A x atomic mass A + fraction B x atomic mass B amu = fraction 60 x amu + fraction 70 x = (1 - fraction 70) x fraction 70 x fraction 70 =

20 Calculate the number of neutrons in an atom of 239 Pu. Atomic number of plutonium = 94 mass number = 239 = number of neutrons + number of protons number of neutrons = mass number number of protons = = 145

21 Permanganate, MnO 4 -, reacts with nitrite, NO 2 -, to produce manganese (IV) oxide and nitrate. Balance the equation. 1. half reactions: MnO e - MnO 2 NO 2 - NO e - 2. balance oxygens: MnO e - MnO 2 + 2H 2 O NO H 2 O NO e - 3. balance hydrogens: MnO H + + 3e - MnO 2 + 2H 2 O NO H 2 O NO H + + 2e -

22 Permanganate, MnO 4 -, reacts with nitrite, NO 2 -, to produce manganese (IV) oxide and nitrate. Balance the equation. 4. add half reactions: 2MnO H + + 6e - 2MnO 2 + 4H 2 O 3NO H 2 O 3NO H + + 6e - 2MnO H + + 3NO 2-2MnO 2 + 3NO H 2 O 5. adjust for basic solution if necessary: 2MnO H + + 2OH - + 3NO 2-2MnO 2 + 3NO H 2 O + 2OH - 2MnO H 2 O + 3NO 2-2MnO 2 + 3NO OH -

23 Acetylsalicylic acid (C 9 H 8 O 4 ) commonly known as "aspirin" is an acid that can ionize to produce one H + ion. A typical aspirin tablet, however, contains only a small amount of the acid. In an experiment to determine its composition, an aspirin tablet was crushed and dissolved in water. It took ml of M NaOH to neutralize the solution. Calculate the number of grams of aspirin in the tablet. HA + NaOH NaA + H 2 O mols NaOH = L x mol/l = mol mols HA = mols NaOH x 1 mol HA / 1 mol NaOH = mol mass HA = mol x g/mol = g

24 Titration with solutions of potassium bromate, KBrO 3, can be used to determine the concentration of As(III). What is the molar concentration of As(III) in a solution if ml of M KBrO3 is needed to titrate ml of the As(III) solution? The (unbalanced) equation is: H 3 AsO 3 + BrO 3 - Br - + H 3 AsO 4 1. half reactions: H 3 AsO 3 H 3 AsO 4 + 2e - BrO e - Br - 2. balance oxygens: H 3 AsO 3 + H 2 O H 3 AsO 4 + 2e - BrO e - Br - + 3H 2 O

25 Titration with solutions of potassium bromate, KBrO 3, can be used to determine the concentration of As(III). What is the molar concentration of As(III) in a solution if ml of M KBrO3 is needed to titrate ml of the As(III) solution? The (unbalanced) equation is: H 3 AsO 3 + BrO 3 - Br - + H 3 AsO 4 3. balance hydrogens: H 3 AsO 3 + H 2 O H 3 AsO 4 + 2H + + 2e - BrO H + + 6e - Br - + 3H 2 O 4. add half reactions: 3H 3 AsO 3 + 3H 2 O 3H 3 AsO 4 + 6H + + 6e - BrO H + + 6e - Br - + 3H 2 O 3H 3 AsO 3 + BrO 3-3H 3 AsO 4 + Br -

26 Titration with solutions of potassium bromate, KBrO 3, can be used to determine the concentration of As(III). What is the molar concentration of As(III) in a solution if ml of M KBrO3 is needed to titrate ml of the As(III) solution? The (unbalanced) equation is: H 3 AsO 3 + BrO 3 - Br - + H 3 AsO 4 Mols BrO 3 - = L x mol/l = mol mols H 3 AsO 3 = mols BrO 3 - x 3 mol H 3 AsO 3 / 1 mol BrO 3 - = mol [H 3 AsO 3 ] = mol / L = mol/l

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