Chapter 10

Calculate the molar mass of the following compounds: a)NaOH - 40g/mol b)AuCl 3 – g/mol c)(NH 4 ) 2 SO 4 – g/mol Q1

Convert 3.4 moles of NH 4 into grams of NH 4. Q2 = 3.4 mole of NH 4 61g 18.04g of NH 4 1 mol of NH 4

Convert 4.2 grams of Ca(NO 3 ) 2 into moles of Ca(NO 3 ) 2. Q3 = 4.2g of Ca(NO 3 ) mol 1mol of Ca(NO 3 ) g of Ca(NO 3 ) 2

Convert 7.8 L of N 2 into moles. Q4 = 7.8L of N mol 1mol of N L of N 2

Convert 3.5 mol of H 2 into liters Q5 = 3.5 mol of H 2 78 L of H L of H 2 1 mol of H 2

Convert 1.4 moles of CO 2 into molecules of CO 2. Q6 = 1.4 mol of CO x x molecules of CO 2 1 mol of CO 2

Convert 2.3 X atoms of Mg(NO 3 ) 2 into grams of Mg(NO 3 ) 2. Q7 = 2.3 x molecules of Mg(NO 3 ) X 10 3 g 1 mol of Mg(NO 3 ) x molecules of Mg(NO 3 ) g of Mg(NO 3 ) 2 1 mol of Mg(NO 3 ) 2

Q8 What is the percent composition of sulfur in Al 2 (SO 4 ) 3 ? = % Comp Part (3 x molar mass of S) Whole (molar mass of Al 2 (SO 4 ) 3 ) X 100% 96.18g g X 100% = 28.1% S

What is the percent composition of carbon in pentane, C 5 H 12 ? Q9 = % Comp Part (5 x molar mass of C) Whole (molar mass of C 5 H 12 ) X 100% 60.05g 72.15g X 100% = 83.23% C

q10

Chapter 12

(Ch 12) Q1 Mole-Mole Problem: According to the following equation, how many moles of Al react with 5.2 moles of H 2 SO 4 ? 2Al + 3H 2 SO 4  Al 2 (SO 4 ) 3 + 3H 2 = 5.2 moles of H 2 SO mole 2 mol of Al 3 moles of H 2 SO 4

(Ch 12) Q2 Mole-Mole Problem: According to the following equation, how many moles of oxygen are needed to form 3.7 mol Al 2 O 3 ? 4Al + 3O 2  2Al 2 O 3 = 3.7 moles of Al 2 O mole 3 mol of O 2 2 moles of Al 2 O 3

(Ch 12) Q3 Mass-Mole Problem: According to the equation in problem #2, how many moles of Al 2 O 3 are formed when 7.8 grams of aluminum completely reacts with oxygen? 4Al + 3O 2  2Al 2 O 3 = 7.8g of Al 0.14mole 1 mol of Al 26.98g of Al 4 mol of Al 2 mol of Al 2 O 3

(Ch 12) Q4 Mass-Mole-Mole-Mass Problem: According to the following equation, how many grams of CO 2 would be produced if 45g of C 2 H 2 completely reacted with oxygen? 2C 2 H 2 + 5O 2  4CO 2 + 2H 2 O = 45g of C 2 H g CO 2 1 mol C 2 H g of C 2 H 2 2 mol C 2 H 2 4 mol CO 2 1 mol CO g of CO 2

(Ch 12) Q5a/b/c In the equation below 3.4g of Al reacts with 4.2g of O 2 to form aluminum oxide. a.) What is the limiting reactant? 4Al + 3O 2  2Al 2 O 3 = 3.4g of Al 6.4g Al 2 O 3 1 mol Al 27.0g of Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O g Al 2 O 3 = 4.2g of O 2 8.9g Al 2 O 3 1 mol O g of O 2 3 mol O 2 2 mol Al 2 O 3 1 mol Al 2 O g Al 2 O 3 Theoretical Yield Limiting Reagent Excess Reagent

(Ch 12) Q5d d.) How much of the excess reactant is left over? 4Al + 3O 2  2Al 2 O 3 = 6.4g of Al2O3 3.02g O 2 4 mol Al 3 mol O 2 1 mol O g O 2 Excess Reactant (O 2 ) = g = 1.18g 101.9g Al 2 O 3 1 mol Al 2 O 3

(Ch 12) Q5e e) If the actual yield of aluminum oxide is 1.24g Al 2 O 3 what is the percent yield? 4Al + 3O 2  2Al 2 O 3 = % Yield Actual yield (grams) Theoretical yield (grams) X 100% 1.24g of Al 2 O 3 6.4g of Al 2 O 3 X 100% = 19.3%

(Ch 12) Q6 Sodium chloride decomposes through electrolysis by the below. What is the percent yield of sodium produced if 45g of NaCl reacts and produces an actual yield of 6.2g of Na? 2NaCl  2Na + Cl 2 = 45g of NaCl 17.67g of Na 1 mol NaCl 58.44g of NaCl 2 mol NaCl 2 mol Na 1 mol Na 23.0g Na Theoretical Yield

(Ch 12) Q6 Sodium chloride decomposes through electrolysis by the below. What is the percent yield of sodium produced if 45g of NaCl reacts and produces an actual yield of 6.2g of Na? 2NaCl  2Na + Cl 2 6.2g of Na (actual yield) 17.67g Na (Theoretical yield) X 100% = 35%

(Ch 12) Q7 The combustion of propane, C 3 H 8 is shown in the equation below. Identify the limiting reactant and the volume of CO 2 formed when 6 L of C 3 H 8 reacts with 12 L of O 2 to produce CO 2 gas and H 2 O vapor at STP. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g) = 6L of C 3 H 8 18L of CO 2 3L of CO 2 1L of C 3 H 8 = 12L of O 2 7.2L of CO 2 3L of CO 2 5L of O 2 Limiting Reactant

(Ch 12) Q8 From problem #7, how much of excess reactant is left unreacted? C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g) Excess Reactant ( C 3 H 8 ) = = 3.6L = 2.4L of C 3 H 8 needed 1L of C 3 H 8 3L of CO 2 7.2L of CO 2

(Ch 12) Q9 From problem #7, if the actual yield was 6.9 L of CO 2, what is the percent yield? C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g) 6.9L of CO 2 (actual yield) 7.2L CO 2 (Theoretical yield) X 100% = 95.8%

(Ch 12) Q10a In the reaction below, 5.50g of Mg reacts with 3.25g of O 2 to produce MgO. a) Balance this equation: 2 Mg + 1 O 2  2 MgO

(Ch 12) Q10b/c/d In the reaction below, 5.50g of Mg reacts with 3.25g of O 2 to produce MgO. b) Which is the theoretical yield of MgO? 2 Mg + 1 O 2  2 MgO = 5.5g of Mg 9.12g MgO 1 mol Mg 24.3g of Mg 2 mol Mg 2 mol MgO 1 mol MgO 40.3g MgO = 3.25g of O g MgO 1 mol O g of O 2 1 mol O 2 2 mol MgO 1 mol MgO 40.3g MgO Limiting Reactant Excess Reactant

(Ch 12) Q10e In the reaction below, 5.50g of Mg reacts with 3.25g of O 2 to produce MgO. e) How much of the excess reactant is left unreacted? 2 Mg + 1 O 2  2 MgO = 8.19g MgO 4.94g Mg 1 mol MgO 40.3g MgO2 mol MgO 2 mol Mg 1 mol Mg 24.3g Mg Excess Reactant (Mg) = = 0.56g

(Ch 12) Q10f In the reaction below, 5.50g of Mg reacts with 3.25g of O 2 to produce MgO. f) Calculate the percent yield of MgO if the actual yield was 6.80g MgO. 2 Mg + 1 O 2  2 MgO 6.8g of MgO (actual yield) 8.19g MgO (Theoretical yield) X 100% = 83.0%

Chapter 14

Ch14 Q5 P 1 V 1 = P 2 V 2 A gas has a volume of 50cm 3 at a pressure of 200 mm Hg. If the pressure is changed to 190 mm Hg, what is the new volume of that sample of gas? P2P2 P1P1 V1V1 P 1 V 1 = P 2 V 2 V2V2 = P1V1P1V1 P2P2 = 200 x V 2 = 53cm 3

Ch14 WP Q6 V 1 /T 1 = V 2 /T 2 If 400mL of nitrogen is collected at 30 o C, what will the new volume be if the temperature is increased to 50 o C? T2T2 T1T1 V1V1 V1V1 V2V2 = V1T2V1T2 T1T1 = 400 x V 1 = 426mL T1T1 = V2V2 T2T2

Ch14 Q7 P 1 /T 1 = P 2 /T 2 A cylinder of carbon dioxide left outside has a pressure of 3.5 atm and a temperature of 50 o C. If the cylinder is taken indoors and cooled to 25 o C, what will the new pressure be? T2T2 P1P1 P2P2 = P1T2P1T2 T1 = 3.5 x P 2 = 3.2 atm T1T1 = P2P2 T2T2 P1P1 T1T1

Ch14 Q8 P 1 V 1 = P 2 V 2 A gas has a volume of 125 mL at a pressure of 2 atm. If the pressure is changed to 7.2 atm, what is the new volume of that sample of gas? P2P2 P1P1 V1V1 P 1 V 1 = P 2 V 2 V2V2 = P1V1P1V1 P2P2 = 2 x V 2 = 35mL

Ch14 Q9 P 1 /T 1 = P 2 /T 2 To what temperature must a sample of oxygen gas at 350 K and pressure of 4.2 atm be cooled to so the final pressure is 1.25 atm? P2P2 P1P1 T2T2 = P2T1P2T1 P1P1 = 1.25 x T 2 = 104K T1T1 = P2P2 T2T2 P1P1 T1T1

Ch14 Q10 V 1 /T 1 = V 2 /T 2 A gas has a volume of 20 L at a temperature of 350 K. What will the volume of the gas occupy if the Kelvin temperature is increased to 423 K? T2T2 T1T1 V1V1 V1V1 V2V2 = V1T2V1T2 T1T1 = 20 x V 2 = 24 L T1T1 = V2V2 T2T2

Chapter 16

Ch16 Q1 M = mol/L Calculate the molarity of a 3.4 mol of Na 2 CO 3 dissolved in a 100mL solution? 3.4 mole 0.100L = 34M

Ch16 Q2 M = mol/L Calculate the molarity of a solution prepared by dissolving 14.5 g of BaCl 2 in 250 mL of solution. = 14.5g of BaCl mole of BaCl 2 1 mol BaCl g of BaCl mole 0.250L = 0.028M

Ch16 Q3 ML = mol Calculate the number of grams of solute needed to make a 150 mL solution of 0.40M H 2 SO 4. = 0.4M x 0.150L0.06mole of H 2 SO mole x 89.08g/mole=5.88g of H 2 SO 4

Ch16 Q4 L = mol/M If you have 5.2 g of CaCl 2 and you want to make a 2.5 M solution, what volume in liter should CaCl 2 be diluted to? = 5.2g of CaCl mole of CaCl 2 1 mol CaCl g of CaCl mole 2.5M = 0.019L

Ch16 Q5 M 1 V 1 = M 2 V 2 You want to make a solution with a final concentration of 2.25M HCl from a stock solution of 6.0M HCl. How many milliliters of the stock solution are required to make a solution with a final volume of 50mL? M1M1 M2M2 V2V2 M 1 V 1 = M 2 V 2 M2V2M2V2 V1V1 = M1M1 = 2.25 x V 1 = 18.75mL

Ch16 Q6 M 1 V 1 = M 2 V 2 You want to make a 2.50 L solution of 3.50 M HNO 3 from a 9.0M HNO 3 solution. How many mL of the 9.0M HNO 3 should you use? (1L = 1,000mL) M1M1 v2v2 M2M2 M 1 V 1 = M 2 V 2 M2V2M2V2 V1V1 = M1M1 = 3.50 x V 1 = 972mL

Ch16 Q7 M 1 V 1 = M 2 V 2 You want to make a 250 mL solution of 0.75 M H 2 SO 4 from a 6.0M H 2 SO 4 solution. How many mL of the 6.0M H 2 SO 4 should you use? M1M1 v2v2 M2M2 M 1 V 1 = M 2 V 2 M2V2M2V2 V1V1 = M1M1 = 0.75 x V 1 = 31.25mL

Ch16 Q8 M 1 V 1 = M 2 V 2 How many mL of a 15.8M HNO 3 stock solution are required to make a 3.0M HNO 3 solution with a final volume of 250mL? M2M2 M1M1 V2V2 M 1 V 1 = M 2 V 2 M2V2M2V2 V1V1 = M1M1 = 3.0 x V 1 = 47.5mL