Unit VI Stoichiometry

Lesson Day Date Topic Worksheet

1. Stoichiometry 1

2. Molar Volume of a Gas at STP Lab 2

3. Stoichiometry Percentage Yield Energy 3

4. STP Gas Problems 4

5. Review of moles to STP 5

6. Limiting and Excess Reactant 1 6

7. Limiting and Excess Reactant 2 7

8. Practice Test 1 8

9. Practice Test 2 9

10. Test

Worksheet #1 Stoichiometry

1. Calculate the number of grams water produced by the complete reaction of 100. g of hydrogen with excess oxygen (theoretical yield). 2H₂ + O₂ → 2H₂O

100. g H₂ x 1 mole x 2 mole H₂O x 18.02 g = 892 g H₂O

2.02 g 2 mole H₂ 1 mole

2. Calculate the mass of carbon required to consume 5.67 g of iron III oxide

2Fe₂O₃ + 3C → 4Fe +3CO₂

5.67 g Fe₂O₃ x 1 mole x 3 mole C x 12.0 g = 0.639 g C

159.6 g 2 mole Fe₂O₃ 1 mole

3. Calculate the amount of oxygen in grams produced by the reaction of 69.0 g of water.

2H₂O → 2H₂ + O₂

69.0 g H₂ O x 1 mole x 1 mole O₂ x 32.0 g = 61.3 g O₂

18.02 g 2 mole H₂O 1 mole

4. Calculate the theoretical yield in grams of Fe produced by the reaction of 5.67 g of

iron III oxide.

2Fe₂O₃ + 3C → 4Fe +3CO₂

5.67 g Fe₂O₃ x 1 mole x 4 mole Fe x 55.8 g = 3.96 g Fe

159.6 g 2 mole Fe₂O₃ 1 mole

5. Calculate the number of moles CO₂ produced by the reaction of 8.45 g of C.

2Fe₂O₃ + 3C → 4Fe +3CO₂

8.45 g C x 1 mole x 3 moles CO₂ = 0.704 moles CO₂

12.0 g 3 mole C

6. Calculate the number of Fe atoms consumed in the reaction if 100. g of Fe₂O₃ react.

2Fe₂O₃ + 3C + 235 KJ → 4Fe +3CO₂

100. g Fe₂O₃ x 1 mole x 4 mole Fe x 6.02 x 10²³ at = 7.54 x 10²³ at Fe

159.6 g 2 mole Fe₂O₃ 1 mole

7. Calculate the number of grams water produced by the complete reaction of 14.5 g of oxygen (theoretical yield).

2H₂ + O₂ → 2H₂O

14.5 g O₂ x 1 mole x 2 mole H₂O x 18.02 g = 16.3 g H₂O

32.0 g 1 mole O₂ 1 mole

8. Calculate the mass of carbon required to produce 5.67 g of iron.

2Fe₂O₃ + 3C → 4Fe +3CO₂

5.67 g Fe x 1 mole x 3 mole C x 12.0 g = 0.915 g C

55.8 g 4 mole Fe 1 mole

9. Calculate the amount of oxygen in grams produced by the reaction of 125 g of water.

2H₂O → 2H₂ + O₂

125 g H₂O x 1 mole x 1 moleO₂ x 32.0 g = 111 g O₂

18.02 g 2 mole H₂O 1 mole

10. Calculate the theoretical yield in grams of CO₂ produced by the reaction of 15.6 g of iron III oxide.

2Fe₂O₃ + 3C → 4Fe +3CO₂

15.6 g Fe₂O₃ x 1 mole x 3 mole CO₂ x 44.0 g = 6.45 g CO₂

159.6 g 2 mole Fe₂O₃ 1 mole

11. Calculate the number of molecules of CO₂ produced by the reaction of 2.45g of

iron III oxide.

2Fe₂O₃ + 3C → 4Fe +3CO₂

2.45 g Fe₂O₃ x 1 mole x 3 mole CO₂ x 6.02 x 10²³ molecules = 1.39 x 10²² molecules CO₂

159.6 g 2 mole Fe₂O₃ 1 mole

Worksheet #2 Molar Volume of a Gas at STP Lab

Purpose To determine the molar volume of H₂ gas at STP.

Theory The molar volume of a gas is the volume in litres per 1 mole of gas. STP is standard temperature and pressure, which is, 0^oC and 101 Kpa. To calculate the molar mass, divide the number of litres by the number of moles.

Procedure

1. Weigh a 3.9 cm piece of magnesium strip.

2. Add about 10 mL of concentrated HCl to a gas collecting tube. Caution: HCl is very corrosive! Carefully pour water over the HCl to fill the gas collecting tube.

3. Fill a 500 mL beaker to 400 mL with water.

4. Roll the magnesium into a coil that is small to fit into the gas collecting tube. Place it into the tube and allow it to sink down to the acid. When it reaches the acid it will begin to bubble and rise. As soon as it begins to rise, place your thumb on the top of the tube to completely seal the opening, invert the tube and place under the water in the 500 ml beaker. Remove your thumb as soon as the tube is under the surface of the water. Clamp the tube to a ring stand to secure it.

5. After the reaction is over allow five minutes for the hydrogen gas to reach room temperature. Record the volume of the hydrogen gas.

6. Clean and put away all of your equipment. Wash your lab bench.

Data

Mass of Mg 0.470 g

Volume of H₂ 48.6 mL = 0.0486 L

Calculations

1. Write a balanced single replacement equation for the reaction between HCl and Mg.

Mg + 2HCl ® H₂ + MgCl₂

2. Covert grams Mg into moles H₂ at room pressure and temperature.

0.0470 g Mg x 1 mol x 1 mol H₂ = 1.93 x 10^-3 mol H₂

24.3 g    1 mol Mg

3. Convert the volume of H₂ in litres to STP conditions by multiplying by the conversion factor 0.894. This lowers the volume to what it would be at 0 ⁰C and 101 Kpa.

0.0486 L H₂ x 0.894 = 0.04345 L H₂

4. Calculate the molar volume by dividing the volume of H₂ in litres by the number of mole H₂.

0.04345 L = 22.5 L/mol

0.00193 mol

5. Determine an accepted value from your textbook for the molar volume of a gas at STP.

 Molar volume of gas = 22.4 L/mol

6. Calculate the percentage difference between the accepted value and the experimental value of the molar volume. Take the difference and divide it by the accepted value and then multiple by 100 %.

(22.5 –22.4) x 100 % = 0.446 %

 ( 22.4 )

STP Calculations

1. Calculate the volume of F₂ gas at STP produced by the electrolysis of 8.2 g of KF.

2KF → 2K + F₂

8.2 g ? L

8.2 g x 1 mole x 1 mole F₂ x 22.4 L = 1.6 L

58.1 g 2 mole KF 1 mole

2. Calculate the mass of KF required to produce 100. L of F₂ gas at STP.

2KF → 2K + F₂

100. L x 1 mole x 2 mole KF x 58.1 g = 519 g

22.4 L 1 mole F₂ 1 mole

3. 5.31 g of a common gas at STP occupies 4.25 L. Calculate the molar mass of the gas. Determine the gas.

4.25 L x 1 mole = 0.1888 mole Molar Mass = 5.31 g = 28.1 g/mole

22.4 L 0.18888 mole

4. 2.00 g of a common gas at STP occupies 1.018 L. Calculate the molar mass of the gas. Determine the gas.

1.018 L x I mole = 0.04545 mole Molar Mass = 2.00 g = 44.0 g/moleCO₂

22.4 L 0.04545 mole

5. Calculate the volume of 10. lb of CO₂ gas at STP (2.21 lb = 1.00Kg).

10 lb x 1.00 Kg x 1000 g x 1 mole x 22.4 L = 2.3 x 10³ L

2.21 lb 1 Kg 44.0 g 1 mole

6. Calculate the volume of H₂ gas at STP produced by the complete reaction of

10.0 g Na with water.

2Na + 2HOH ® H₂ + 2NaOH

10.0 g ? L

10.0 g Na x 1 mole x 1 mole H₂ x 22.4 L = 4.87 L

23.0 g 2 mole Na 1 mole

7. How many litres of 0₂, measured at STP, will be released on the decomposition of 8.56 g of mercury (II) oxide: 2HgO → 2Hg + O₂

0.443 L

8. Calculate the number of molecules in 10.0 lb of CO₂ gas at STP

(2.21 lb = 1.00Kg).

6.19 x 10²⁵ molecules

9. Calculate the molar mass of Co(NH₃)₆Cl₃.

267.58 g/mole

10. Calculate the number of Cl atoms in 50.0 g Co(NH₃)₆Cl₃.

50.0 g x 1 mole x 6.02 x 10²³ FU x 3 at Cl = 3.37 x 10²³ atoms

267.58 g 1mole 1 FU

11. Covert 2.0 x 10²⁶ FU of MgCl₂ to Kg.

2.0 x 10²⁶ FU x 1 mole x 95.3 g x 1 kg = 32 kg

6.02 x 10²³ 1 mole 1000 g

Name the following:

12. Co₂O₃^.6H₂O Cobalt III oxide hexahydrate

13. P₂O₅ Diphosphorous Pentoxide

14. H₂SO_4(aq) Sulphuric Acid

15. H₂SO_4(l) Hydrogen Sulphate

14. HI _(aq) Hydriodic Acid

15. HI _(l) Hydrogen Iodide

16. (NH₄)Cr₂O₇ Ammonium Dichromate

Write the quantum electron configuration for:

17. Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²

18. Sr²⁺ 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶

19. I 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁵

20. I^- 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶

21. Give an example of a formula unit NaCl

22. Give an example of a molecule CO₂

23. Give an example of an atom Mg

24. Write the formula eqaution.

H₂SO_4(aq) + 2NaOH_(aq)® Na₂SO_4(aq) + 2H₂O_(l)

Worksheet #3 Stoichiometry and Percentage Yield and Energy

1. Calculate the energy consumed in the reaction if 150. g of C react.

2Fe₂O₃ + 3C + 235 KJ → 4Fe +3CO₂

150. g C x 1mole x 235 KJ = 979 KJ

12.0 g 3 mole C

2. Calculate the energy consumed in the reaction if 250. Kg of C react.

2Fe₂O₃ + 3C + 235 KJ → 4Fe + 3CO₂

250. Kg C x 1000 g x 1mole x 235 KJ = 1.63 x 10⁶ KJ

1 Kg 12.0 g 3 mole C

3. Calculate the actual yield in grams of Fe produced by the reaction of 0.22 Kg of C assuming a 85.0% yield.

2Fe₂O₃ + 3C → 4Fe + 3CO₂

0.22 Kg C x 1000 g x 1mole x 4 mole Fe x 55.8 g x 0.850 = 1.2 x 10³ g Fe

1 Kg 12.0 g 3 mole C 1 mole

4. Calculate the number of grams water produced by the complete reaction of 95.5 g of hydrogen with excess oxygen. If the percentage yield is 65.0% calculate the actual yield.

2H₂ + O₂ → 2H₂O

95.5 g H₂ x 1 mole x 2 mole H₂O x 18.02 g x 0.650 = 554 g H₂O

2.02 g 2 mole H₂ 1 mole

5. Calculate the mass of carbon dioxide produced when 5.67 g of iron III oxide reacts with excess C. If the percentage yield is 85.0 % calculate the actual yield.

2Fe₂O₃ + 3C → 4Fe + 3CO₂

5.67 g Fe₂O₃ x 1 mole x 3 mole CO₂ x 44.0 g x 0.850 = 1.99 g CO₂

159.6 g 2 mole Fe₂O₃ 1 mole

6. Calculate the energy in KJ produced by the reaction of 10.0 Kg of hydrogen with excess oxygen.

2H₂ + O₂ → 2H₂O + 213KJ

10.0 Kg H₂ x 1000 g x 1 mole x 213 KJ = 5.27 x 10⁵ KJ

1 Kg 2.02 g 2 mole H₂

7. Calculate the amount of CO₂ produced by the reaction of 8.45 g of iron III oxide. If the percentage yield is 75 %, calculate the actual yield.

2Fe₂O₃ + 3C → 4Fe +3CO₂

8.45 g Fe₂O₃ x 1 mole x 3 mole CO₂ x 44.0 g x 0.75 = 2.6 g CO₂

159.6 g 2 mole Fe₂O₃ 1 mole

8. Calculate the number of grams water produced by the complete reaction of 100. g of hydrogen with excess oxygen (theoretical yield). If the percentage yield is 65.0 % calculate the actual yield. 2H₂ + O₂ → 2H₂O

100. g H₂ x 1 mole x 2 mole H₂O x 18.02 g x 0.650 = 580 g H₂O

2.02 g 2 mole H₂ 1 mole

9. Calculate the theorectical mass of iron produced from the reaction of 5.67 g of iron III oxide. If the percentage yield is 85.0 % calculate the actual yield. 2Fe₂O₃ + 3C → 4Fe +3CO₂

5.67 g Fe₂O₃ x 1 mole x 3 mole C     x 12.0 g =     3.96 g Fe      x 0.850 = 0.544 g C

159.6 g     2 mole Fe₂O₃    1 mole

10. Calculate the theoretical yield of oxygen in grams produced by the reaction of 69.0 g of water. If the actual yield of O₂ is 50.0 g, calculate the percentage yield. 2H₂O → 2H₂ + O₂

69.0 g H₂O x 1 mole x 1 mole O₂ x 32.0 g = 61.3 g

18.02 g 2 mole H₂O 1 mole

% Yield = 50.0 g x 100 % = 81.6 %

61.1 g

11. Calculate the energy in KJ produced by the reaction of 10.0 g of hydrogen with excess oxygen.

2H₂ + O₂ → 2H₂O + 213 KJ

10.0 g H₂ x 1 mole x 213 KJ = 527 KJ

2.02 g 2 mole

12. Calculate the theoretical yield in grams of Fe produced by the reaction of 5.67 g of iron III oxide. If the actual yield is 2.00 g calculate the percentage yield. 2Fe₂O₃ + 3C → 4Fe +3CO₂

5.67 g Fe₂O₃ x 1 mole x 4 mole Fe x 55.8 g = 3.96 g Fe

159.6 g 2 mole Fe₂O₃ 1 mole

% Yield = 2.00 g x 100 % = 50.4 %

3.96 g

13. Calculate the amount of CO₂ produced by the reaction of 8.45 g of iron III oxide. If the percentage yield is 75 %, calculate the actual yield. 2Fe₂O₃ + 3C → 4Fe +3CO₂

8.45 g Fe₂O₃ x 1 mole x 3 mole CO₂ x 44.0 g = 3.49 g CO₂ x 0.75      = 2.6 g CO₂

159.6 g 2 mole Fe₂O₃ 1 mole

14. Calculate the number of moles CO₂ produced by the reaction of 8.45 g of C.

2Fe₂O₃ + 3C → 4Fe +3CO₂

8.45 g C x 1mole x 3 mole CO₂ = 0.704 mole CO₂

12.0 g 3 mole C

15. Calculate the number of Fe atoms consumed in the reaction if 500. J of energy are absorbed.

2Fe₂O₃ + 3C + 235 KJ → 4Fe +3CO₂

500. J x 1 KJ x 4 mole Fe x 6.02 x 10²³ at = 5.12 x 10²¹ at Fe

1000 J 235 KJ 1 mole

Worksheet # 4 STP Gas Problems

1. Calculate the STP volume of 150. g NH₃ gas.

150. g x 1 mole x 22.4 L = 197 L

17.03 g 1 mole

2. Calculate the mass of 350. mL of CO₂ at STP.

350. mL x 1 L x 1 mole x 44.0 g = 0.688 g

1000 mL 22.4 L 1 mole

3. Calculate the mass of carbon required to consume 5.67 g of iron III oxide.

2Fe₂O₃ + 3C → 4Fe +3CO₂

0.639 g C

4. Calculate the volume of oxygen at STP produced by the reaction of 69.0 g of water.

2H₂O → 2H₂ + O₂

42.9 L

5. If the actual volume of oxygen is 40.0 L, what is the % yield?

93.2 %

6. Calculate the energy produced by the reaction of 100. L of hydrogen at STP with excess oxygen.

2H₂ + O₂ → 2H₂O + 213KJ

475 KJ

7. Calculate the volume of CO₂ produced at STP by the reaction of 5.67 g of iron III oxide if the percentage yield is 70.0 %. 2Fe₂O₃ + 3C → 4Fe +3CO₂

0.836 L

8. Calculate the density of CO₂ gas at STP.

1.96 g/L

9. Calculate the density of NH₃ gas at STP

0.760 g/L

Stoichiometry Review

10. Chloroform or trichloromethane, CHCl₃, may be prepared in the laboratory by the reaction between Chlorine and methane. Calculate the number of grams of chlorine that are required to produce 50.0 g of trichloromethane.

3Cl_2(g) + CH_4(g) → CHCl_3(l) + 3HCl_(g)

50.0 g CHCl₃ x 1 mole x 3 mole Cl₂ x 71.0 g = 89.1 g Cl₂

119.51 g 1 mole CHCl₃ 1 mole

11. The purification ability and bleaching action of bleaching activity of chlorine both come from the effective release of oxygen by the reaction below. How many grams of oxygen will be liberated by 255 g of chlorine?

2H₂O_(l) + 2Cl_2(g) → 4HCl_(aq) + O_2(g)

255 g Cl₂ x 1 mole x 1 mole O₂ x 32.0 g = 57.5 g O₂

71.0 g 2 mole Cl₂ 1 mole

12. Photosynthesis is the process by which carbon dioxide is converted into sugar, C₆H₁₂O₆, with the help of sunlight and the chlorophyll of green plants acting as a catalyst. It may be summarized by the equation below. How many grams of sugar may be produced in the reaction that consumes 6.90 grams of carbon dioxide?

6CO_{2 (g)} + 6 H₂O _(g) →     C₆H₁₂O₆ + 6 0_{2 (g)}

6.90 g CO₂ x 1 mole x 1 mole C₆H₁₂O₆ x 180.12 g = 4.71 g C₆H₁₂O₆

44.0 g   6 mole CO₂ 1 mole

13. Dinitrogen oxide N₂O- Known also as nitrous oxide and laughing gas –was one of early anesthetics, and has recently regained popularity in this role in the dental field. It is made by the decomposition of ammonium nitrate which equations shown below. How many grams of NH₄NO_{3 (S)} are required to prepare 12.8 grams of N₂O?

NH₄NO_3(s) → N₂O_(g) + 2H₂O_(l)

12.8 g N₂O x 1 mole x 1 mole NH₄NO₃ x 80.04 g = 23.3 g NH₄NO₃

44.0 g 1 mole N₂O 1 mole

14. In the extraction of copper from its sulfide ore, the overall process may be summarized by the equation shown below. If the percentage yield is 61.2%, how much copper will result from treatment of 7.00 x 10⁶ grams of Cu₂S?

Cu₂S_(s) + O_2(g) → 2Cu_(s) + SO_2(g)

7.00 x 10⁶ g Cu₂S x 1 mole x 2 mole Cu x 63.5 g x 0.612 = 3.42 x 10⁶ g Cu

159.1 g 1 mole Cu₂S 1 mole

15. One step of the Ostwald process for manufacturing nitric acid involves making nitrogen monoxide by oxidizing ammonia in the presence of platinum catalyst, which is shown by the equation below. If the percentage yield is 80.3%, how many grams of NO can be produced from 4.00 x 10³ grams of NH₃?

Pt

4 NH_3(g) + 50_2(g) → 4 NO_(g) + 6 H₂O_(g)

4.00 x 10³ g NH₃ x 1 mole x 4 moles NO x 30.0 g x 0.803 = 5.66 x 10³ g NO

17.03 g 4 moles NH₃ 1 mole

16. The direct combination of powdered zinc with powdered sulfur is a spectacular reaction, though not one to be tried by students, with bright light, flame and smoke. The equation is shown below. Calculate the energy released by the reaction of 9.63 grams of zinc.

Zn_(s) + S_(s) → ZnS_(s) + 148.5 KJ

9.63 g Zn x 1 mole x 148.5 KJ = 21.9 KJ

65.4 g 1 mole

17. Carbon monoxide is used as a fuel in many industrial processes. How much carbon monoxide must be burned in a process requiring 183 KJ?

2CO_(g) + O_2(g) → 2CO_2(g) + 135 KJ

183 KJ x 2 moles CO x 28 g = 75.9 g CO

135 KJ 1 mole

Worksheet # 5 Review of moles to STP

1. Convert 1.65 Kg CO₂ to molecules.

2.26 x 10²⁵ molecules CO₂

2. Covert 2.0 x 10²⁶ FU of MgCl₂ to Kg.

32 Kg MgCl₂

3. In a propane tank there are 9.0 Kg of C₃H₈, calculate the number of H atoms. (First calculate molecules and then H atoms).

9.8 x 10²⁶ at H

4. A certain mass of complex Co(NH₃)₆Cl₃ was found to contain 2.65 x 10²¹ atoms of H, calculate the mass of cobalt III chloride hexammine (change atoms to FU’s 1st).

0.0654 g Co(NH₃)₆Cl₃

5. Calculate the percentage composition of Co(NH₃)₆Cl₃ to three significant digits.

22.0% Co, 31.4% N, 6.79% H, and 39.8% Cl

6. A compound is 71.6 % C, 6.03 % H, 10.4 % N, and 11.9 % O. If the molecular mass of the compound is 268.16 g/mol, calculate the empirical and the molecular formula.

C₁₆H₁₆N₂O₂

7. A compound is 63.55% Ag, 8.23% N, and 28.24% O, calculate the empirical formula.

AgNO₃

8. A volume of a common gas used for welding has a mass of 0.856 g. An equal volume of H₂ gas has a mass of 0.0658 g. Calculate the molar mass of the gas and determine its formula.

26.3g/mol

9. Calculate the energy produced by the complete reaction of 150. g H₂.

2H₂ + O₂ → 2H₂O + 250 KJ

9.28 x 10³ KJ

10. Calculate the number of grams H₂ required to produce 500 J of energy.

2H₂ + O₂ → 2H₂O + 250 KJ

0.00808 g

11. The fermentation of sugar in the presence of zymase, an enzyme in yeast, can be described by the equation below. If 500. g of sugar is fermented and 200 g of alcohol are produced, calculate the theoretical and percentage yield of alcohol.

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

255g 78.3%

12. Calculate the mass of Fe produced by the reaction of 8.56 g of Fe₂O₃ if the percentage yield

is 75.0 %.

2Fe₂O₃ + 3C → 4 Fe + 3CO₂

4.49 g

13. What is the mass of 3.25 litres of neon at STP?

2.93 g

14. Find the STP volume of 4.62 grams of ammonia (NH₃).

6.08 L

15. How many moles of ethane, C₂H₆, are in 16.9 liters at STP?

0.754 moles

16. How many liters of O₂, measured at STP, will be released on the decomposition of 2.65 g of mercury (II) oxide:

2HgO → 2Hg + O₂

0.137 L

17. What mass of potassium chlorate must be decomposed to produce 1.50 liters of oxygen at STP?

2KClO_3(s) → 2KCl_(s) + 3O_2(g)

5.47g

18. When sodium hydrogen carbonate is treated with sulfuric acid, carbon dioxide bubbles off:

H₂SO_4(aq) + 2NaHCO_3(s) → Na₂SO_4(aq) + 2H₂O_(l) + 2CO_2(g)

What volume of CO₂, measured at STP, is available from 8.58 g NaHCO₃?

2.29L

19. What quantity of magnesium must a student react with excess hydrochloric acid to produce

85.0 ml hydrogen, measured at STP?

Mg_(s) + 2HCl_(aq) → H_2(g) + MgCl_2(aq)

0.0922g

Worksheet # 6 Limiting and Excess Reactant

Fill in the ICE charts- all units are in moles.

1.          1 A                +             2 B                   →                 1 C                +            2 D

I           8.00                               6.00                                        0.00                            0.00

C         3.00                               6.00                                        3.00                             6.00

E         5.00                               0.00                                        3.00                             6.00

2.          3 A                +             4 B                   →                4 C                +            3 D

I           9.00                             14.00                                      0.00                             0.00

C         9.00                              12.00                                     12.00                            9.00

E         0.00                               2.00                                       12.00                            9.00

3.          3 A                +             2 B                   →                 5 C                +            4 D

I           10.00                             8.00                                 0.00                                 0.00

C          10.00                            6.667                                16.67                              13.33

E          0.00                               1.33                                 16.67                              13.33

4           3 A                +           5 B                   →                 5 C                +            4 D

I           8.00                            16.00                                     0.00                             0.00

C          8.00                            13.33                                   13.33                            10.67

E          0.00                              2.67                                    13.3                              10.7

5. 2 Al + 3 I₂ → 2 AlI₃

I 9.0 mol 9.0 mol 0

C 6 mol 9 mol 6 mol

E 3 mol 0 6 mol

6. C + 2Cl₂ → CCl₄

I 8.0 mol 6.0 mol 0

C 3 mol 6 mol 3 mol

E 5 mol 0 3 mol

7. 4 Fe + 3 O₂ → 2 Fe₂O₃

I 10.0 mol 6.0 mol 0

C 8 mol 6 mol 4 mol

E 2 mol 0 4 mol

8. 2 NO + O₂ → 2 NO₂

100.0 g 100.0g 0

I 3.3333 mol 3.1250 mol 0

C 3.3333 mol 1.6667 mol 3.3333 mol

E 0 1.4584 mol 3.3333 mol

Grams 0 46.7 g 153 g     Note the molar masses had 3 sig figs!

9. 2 Fe + 3 Cl₂ → 2 FeCl₃

100.0 g 120.0 g 0

I 1.792 mol 1.690 mol 0

C 1.127 mol 1.690 mol 1.127 mol

E 0.665 mol 0 1.127 mol

Grams 37.11 g 0 183.4 g

10. Mg + 2HCl → MgCl₂           + H₂

45.8 g 65.4 g 0 0

I 1.885 mol 1.791 mol 0   0

C 0.8955 mol 1.791 mol 0.8956 mol 0.8956 mol

E 0.9895 mol 0 0.8956 mol 0.8956 mol

Grams 24.0 g 85.4 g 1.81 g

11. If 75.4 g of Cu is reacted with 189.7 g of I₂, then CuI₂ is formed.

1 Cu + 1 I₂ → 1 CuI₂

75.4 g 189.7 g 0

I 1.187 mol 0.7474 mol 0

C 0.7474 mol 0.7474 mol 0.7474 mol

E 0.4396 g 0 mol 0.7474 mol

Grams 27.9 g 0 g 237g

(a) Which reactant is in excess?

Cu

(b) How many grams are in excess?

27.9 g

(c) Calculate the amount of CuI₂ produced.

237 g

12. If 15.5 g of Al are reacted with 46.7 g of Cl₂, then AlCl₃ is formed.

2Al + 3Cl₂ → 2AlCl₃

15.5 g 46.7 g 0

I 0.5741 mol   0.6577 mol 0

C 0.4385 mol 0.6577 mol 0.4385 mol

E 0.1356 mol 0 mol 0.4385 mol

Grams 3.66 g 0 g 58.5 g

(a) Which reactant is in excess?

Al

(b) How many grams are in excess?

3.66 g

(c) Calculate the amount of AlCl₃ produced.

58.5 g

13. If 5.45g of KClO₃ are decomposed to form KCl and 1.95g of O₂ are collected. Calculate the theoretical and percentage yield.

2KClO₃ → 2KCl + 3O₂

5.45 g ?g

5.45 g KClO₃ x 1 mol x 3 mol O₂ x 32.0g = 2.13 g

122.6 g 2 mol KClO₃ 1 mol

% Yield = 1.95 x 100% = 91.4%

2.13

Worksheet # 7 Limiting Reactants

1. 2 Al + 3 I₂ → 2 AlI₃

I 12 mol 15 mol 0

C 10 mol 15 mol 10 mol

E 2 mol 0 10 mol

2. C + 2Cl₂ → CCl₄

I 13.5 mol 28.6 mol 0

C 13.5 mol 27.0 mol 13.5 mol

E 0 1.6 mol 13.5 mol

3. 4 Fe + 3 O₂ → 2 Fe₂O₃

I 15.0 mol 13.2 mol 0

C 15.0 mol 11.3 mol 7.50 mol

E 0 1.9 mol 7.50 mol

4. 2 NO + O₂ → 2 NO₂

80 g x 1 mol 50.0g x 1 mol 0

30.0 32.00

I 2.667 mol 1.563 mol 0

C 2.667 mol 1.334 mol 2.667 mol

E 0 0.230 mol 2.667 mol

Grams 0 7.34 g 123 g

5. 2 Fe + 3 Cl₂ → 2 FeCl₃

45.5g 55.6g 0

I 0.8154 mol 0.7831 mol 0

C 0.5221 mol 0.7831 mol 0.5221 mol

E: 0.2933 mol 0 0.5221 mol

Grams 16.4 g 0 84.7 g

6. Mg + 2HCl → MgCl₂ + H₂

22.3 g 15.6 g 0 0

I 0.9177 mol 0.4273 mol 0 0

C 0.2136 mol 0.4273 mol 0.2136 mol 0.2136 mol

E 0.7041 mol 0 0.2136 mol 0.2136 mol

Grams 17.1 g 0 g 20.4 g 0.431 g

7. If 100.0 g of Cu is reacted with 150.0 g of I₂, then CuI₂ is formed.

1 Cu + 1 I₂ → 1 CuI₂

100.0 g 150.0 g 0

I 1.575 mol 0.5910 mol 0

C 0.5910 mol 0.5910 mol 0.5910 mol

E 0.984 mol 0 mol 0.5910 mol

Grams 62.5 g 0 g 188 g

(a) Which reactant is in excess?

Cu

(b) How many grams are in excess?

62.5 g

(c) Calculate the amount of CuI₂ produced.

188 g

8. If 25.0 g of Al are reacted with 40.0 g of Cl₂, then AlCl₃ is formed.

2Al + 3Cl₂ → 2AlCl₃

25.0 g x 1 mole 40.0 g x 1mole 0

27.0 g 71.0 g

I 0.9259 mol 0.5634 mol 0

C 0.3756 mol 0.5634 mol 0.3756 mol

E 0.5503 mol 0 mol 0.3756 mol

Grams 14.9 g 0 g 50.1 g

(a) Which reactant is in excess?

Al

(b) How many grams are in excess?

14.9 g

(c) Calculate the amount of AlCl₃ produced.

50.1 g

9. If 100.0 g of Ca are reacted with 100.0 g of N₂, then Ca₃N₂ is formed.

Equation:

3Ca + N₂ ® Ca₃N₂

100.0 g x 1 mole 100.0 g x 1mole 0

40.1 g 28.0 g

I 2.4938 mol 3.5714 mol 0

C 2.4938mol 0.8313 mol 0.8313 mol

E 0 mol 2.7401 mol            0.8313 mol

Grams 0 g 76.7 g 123 g

                                                                                      Note that the molar masses had 3 sig figs

(a) Which reactant is in excess? N₂

(b) How many grams are in excess? 76.72 g

(c) Calculate the amount of Ca₃N₂ produced. 123.3 g

10. If 100.0 g of Ca are reacted with 100.0 g of Cl₂, then CaCl₂ is formed.

Equation:

Ca + Cl₂ ® CaCl₂

100.0 g x 1 mole 100.0 g x 1mole 0

40.1 g 71.0 g

I 2.4938 mol 1.4085 mol 0

C 1.4085 mol 1.4085 mol 1.4085 mol

E 1.0853 mol 0 mol 1.4085 mol

Grams 43.52 g 0g 156.5 g

(a) Which reactant is in excess? Ca

(b) How many grams are in excess? 43.52 g

(c) Calculate the amount of CaCl₂ produced. 156.5 g

11. If 25.0 g of Na are reacted with 40.0 g of O₂, and one ionic compound is formed.

Equation:

4Na + O₂ ® 2Na₂O

          25.0 g   x 1 mole 40.0 g x 1mole 0

23.0 g   32.0 g

I 1.0870 mol 1.2500 mol 0

C 1.0870 mol 0.2718 mol 0.5435 mol

E 0 mol 0.9782 mol 0.5435 mol

Grams 0 g 31.3g 33.7 g

(a) Which reactant is in excess? O₂

(b) How many grams are in excess? 31.3g

(c) Calculate the amount of AlCl₃ produced. 33.7 g

12. If 32.7 g of KClO₃ are decomposed to form KCl and 11.7 g of O₂ are collected.

Calculate the theoretical and percentage yield.

2KClO₃ → 2KCl + 3O₂

32.7 g ?g

32.7 g KClO₃ x 1 mol x 3 mol O₂ x 32.0g = 12.8 g

122.6 g 2 mol KClO₃ 1 mol

% Yield = 11.7 x 100% = 91.4%

12.8

13. Equal volumes of gas "X" and O₂ have weights of 8.75g and 10.0g.

Calculate the molar mass of gas "X".

Gas x = 8.75 x 32.0 g/ mol = 28.0 g/mol N₂

O₂ 10.0 g

Worksheet # 8 Practice Test #1

1. Change 8.75 moles of H₂O to molecules.

5.27 x 10²⁴ molecules H₂O

2. Change 9.7x10¹⁹ atoms Fe to moles.

1.6 x10^-4 mol Fe

3. Convert 800. g AgNO₃ to formula units and then to atoms of O.

2.83 x10²⁴ FU AgNO₃

8.50 x 10²⁴ atoms O

4. Convert 3.8 x 10²⁵ H₂ molecules to grams.

1.3x10²g H

B Percentage Composition

5. Calculate the percentage composition of Na₂SO₄.

32.4% Na 22.6%S 45.0%O

C Molecular Formula and Empirical Formula

6. Calculate the empirical formula of a compound that is 62.2 % Pb,

8.454 % N, and 28.8 % O. Is this compound ionic or covalent?

Pb (NO₃) ₂ ionic

7. A compound is 42.3 % C, 5.94 % H, 32.9 % N, and 18.8 % O and has a molecular mass of 425.25 g/mol. Calculate the empirical and molecular formula.

C₃H₅N₂O C₁₅H₂₅N₁₀O₅

D Stoichiometry

8. How many grams of O₂ are required to consume 100. g Al?

4Al + 302 → 2Al₂O₃

88.9g O

9. How many moles of Al₂O₃ are produced by the reaction 200. g Al?

4Al + 30₂ → 2Al₂O₃

3.70 moles Al₂O₃

10. How many moles Al are required to produce 300. g Al₂O₃?

4Al + 30₂ → 2Al₂O₃

5.88 moles Al

11. How many litres of O₂ gas are required to produce 100. g Al₂O₃?

4Al + 30₂ → 2Al₂O₃

32.9 L

E Percentage Yield

12. 100. g Al reacts with excess O₂ to produce 150. g Al₂O₃ according to

Calculate the theoretical and percentage yield. 4Al + 30₂ → 2 Al₂O₃.

Theoretical Yield = 189g Al₂O₃

F Energy Calculations

13. Calculate the energy produced by the complete reaction of 150.g H₂.

2H₂ + O₂ → 2H₂O + 130KJ

4.83 x 10³ KJ

14. How many grams of H₂ would be needed to produce 260. KJ of energy?

2H₂ + O₂ → 2H₂O + 130KJ

8.08 g of H₂

G Limiting Reactants

15. 20. mol H₂ reacts with 8.0 mol O₂ to produce H₂O. Determine the number of grams reactant in excess and number of grams H₂O produced. Identify the limiting reactant.

8 g excess H₂ 2.9 x 10² g H₂O in excess O₂ is limiting

16. 128g of Al reacts with 128g of O₂ to produce Al₂O₃. How many grams of Al₂O₃ are produced? Determine the mass of the reactant in excess and the limiting reactant.

Al is limiting 14g O₂ are in excess 242g Al₂O₃

17. 13.6 g of Al reacts with 8.33 g O₂ to produce Al₂O₃.How many grams of Al₂O₃ are produced? Determine the mass of the reactant in excess and the limiting reactant.

O₂ is limiting 4.23 g Al are in excess 17.7 g of Al₂O₃ are produced

18. A gas weighs 21.9 g. An equal volume of He weighs 1.991 g. Determine the molar mass and formula of the gas (hint- the gas can cook a steak).

44.0 g/mole propane

19. Calculate the molar mass of a gas that weighs 19.43 g and has a STP volume of 9.894 L. If the gas is a very funny one containing nitrogen and used by the dentist, determine the molar mass and molecular formula for the gas.

44.0 g/mole N₂O

20. A 45.0 g sample of bronze, a Cu and Sn alloy, is completely reacted with concentrated HNO₃ according to the following equations. A total of 59.978 g of NO₂ was isolated. Assuming a 100 % yield of NO₂, determine the mass of Cu and Sn in the alloy.

Cu_(s)     +    4HNO_3(aq)        →     Cu(NO₃)_2(aq)                  +          2NO_2(g)      +     2H₂O_(l)

X g Cu   x    1 mole   x    2 mole NO₂   x   46.0 g    =    1.4488 X

                    63.5 g           1 mole Cu          1 mole

Sn_(s)     +    4HNO_3(aq)        →     Sn(NO₃)_2(aq)                  +          2NO_2(g)      +     2H₂O_(l)

(45.0  -  X) g Sn   x    1 mole   x    2 mole NO₂   x   46.0 g    =   34.878  -  0.77506 X

                                   118.7 g       1 mole Sn           1 mol

1.4488 X   +   34.878  -  0.7751 X =   59.978

0.6737 X  =   25.1

X   =   Cu  =  37.3 g

Sn  =  45.0 - 37.3 = 7.7 g        

Sn_(s)     +    4HNO_3(aq)        →     Sn(NO₃)_2(aq)                   +          2NO_2(g)      +     2H₂O_(l)

21. A 80.0 g sample of brass, a Cu/Zn alloy, is completely reacted with concentrated HNO₃ according to the following equations. A total of 115.232 g of NO₂ was isolated. Assuming a 100 % yield of NO₂, determine the mass of Cu and Zn in the alloy.

Cu_(s)     +    4HNO_3(aq)        →     Cu(NO₃)_2(aq)                  +          2NO_2(g)      +     2H₂O_(l)

X g Cu   x    1 mole   x    2 mole NO₂   x   46.0 g    =    1.4488 X

                    63.5 g           1 mole Cu          1 mole

Zn_(s)     +    4HNO_3(aq)        →     Zn(NO₃)_2(aq)                   +          2NO_2(g)      +     2H₂O_(l)

(80.0  -  X) g Zn   x    1 mole   x    2 mole NO₂   x    46.0 g    =   112.54  -  1.4067 X

                                   65.4 g           1 mole Cu          1 mol

1.4488 X   +   112.54  -  1.4067 X =   115.232

0.042072 X  =   2.692

X   =   Cu  =  64.0 g

Zn  =  16.0 g

With Correct Sig Figs:  Cu = 60 g and Zn = 20 g

Worksheet # 9 Practice Test 2

1. Covert 200 g NaCl to formula units.

200 g x 1 mole x 6.02 x 10²³ FU = 2 x 10²⁴ FU

58.5 g 1 mole

2. Convert 5.99 x 10²⁵ H₂O molecules to grams.

5.99 x 10²⁵ H₂O molecules x 1 mole x 18.02 g = 1.79 x 10³ g H₂O

6.02 x 10²³ molecules 1 mole

3. A certain mass of CH₄ contains 2.00 x 10²⁵ atoms of H. Calculate the mass of CH₄.

2.00 x 10²⁵ atoms H x 1 molecule CH₄ x   1 mole x 16.04 g = 133 g CH₄

4 at H 6.02 x 10²³ molecules 1 mole

4. Calculate the number of grams SO₂ in 250. L @ STP.

250. L SO₂ x 1 mole x 64.1 g = 715 g SO₂

  22.4 L 1 mole

5. The mass of a diatomic gas is 15.13 g. The mass of an equal volume of He is

1.593 g. What is the molar mass of the diatomic gas and what is the gas?

Gas X = 15.13 g x 4.0 g/mole = 38 g/mole F₂

1.593 g

6. Calculate the volume of F₂ gas at STP produced by the reaction of 15.6 g of KF.

2KF → 2K + F₂

15.6 g KF x 1 mole x 1 mole F₂ x 22.4 L = 3.01 L F₂

58.1 g 2 mole KF 1 mole

7. A compound was found to be composed of 6.89 g Si and 5.89 g O. If the molecular mass of the compound is 417 g/mol, what is the molecular formula of the compound?

6.89 g Si x 1 mole = 0.2452 mole = 1 x 2 = 2

28.1 g 0.2452 mole

5.89 g O x 1 mole = 0.3681 mole = 1.501 x 2 = 3

16.0 g 0.2452 mole

Si₂O₃ 104.2 g/mole

Si₈O₁₂ 417 g/mole

8. Calculate the energy produced by the complete reaction of 25.0 g of C₄H₁₀ .

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O + 256 KJ

25.0 g C₄H₁₀ x 1 mole x 256 KJ = 55.1 KJ

58.1 g 2 mole C₄H₁₀

9. Calculate the mass of C₄H₁₀ required to produce 2.06 x 10⁶ J of energy.

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O + 256KJ

2.06 x 10⁶ J x 1 KJ x 2 mole C₄H₁₀ x 58.1 g = 935 g

1000 J 256 KJ 1 mole

10. 500. g of Fe₂O₃ are refined to produce 200. g of Fe. Calculate the percentage yield of Fe.

2Fe₂O₃ + 3C → 4Fe + 3CO₂

500. g Fe₂O₃ x 1 mole x 4 mole Fe x 55.8 g = 350 g

159.6 g 2 mole Fe₂O₃ 1 mole

Percentage Yield = 200 g x 100% = 57.1 %

350 g

11. 500. g of Fe₂O₃ are refined to produce Fe at a 78.0 % yield, calculate the actual yield of Fe.

2Fe₂O₃ + 3C → 4Fe + 3CO₂

500 g Fe₂O₃ x 1 mole x 4 mole Fe x 55.8 g x 0.780 = 273 g

159.6 g 2 mole Fe₂O₃ 1 mole

12. Calculate the molar mass of Cr₂(H₂0)₆(SO₄)₃.

500.42 g/mole

13. Calculate the mass of 3.25 L of SO₂ gas at STP.

3.25 L x 1 mole x 64.1 g = 9.30 g

22.4 L 1 mole

14. What mass of magnesium must react with excess hydrochloric acid to produce 85.0 ml hydrogen, measured at STP?

Mg_(s) + 2HCl_(aq) → H₂(g) + MgCl_2(aq)

85.0 ml H₂ x 1L x 1 mole x 1 mole Mg x 24.3 g = 0.0922 g

1000ml 22.4 L 1 mole H₂ 1 mole

15. 2 Al + 3 I₂ → 2 AlI₃

Initial: 12 mol 15 mol 0

Changes: 10mol 15mol 10mol

End: 2 mol 0 10 mol

16. 12.8 g of Al reacts with 12.8 g of O₂ to produce Al₂O₃. How many grams of Al₂O₃ are produced? Determine the mass of the reactant in excess and the limiting reactant.

27.0 g 32.0g 102.0g

4Al + 3O₂ → 2Al₂O₃

12.8g 12.8g 0

I 0.4741mol 0.4000mol 0

C 0.4741 0.3555 mol 0.2371mol

E 0 0.0445 mol 0.237mol

Grams 24.2 g

Limiting Reactant Al= 0g

Excess Reactant O₂=1.42g

17. Calculate the molar mass of a gas that weighes 38.87 g and has a STP molar volume of 19.789 L. If the gas is one used in cars in the Fast and Furious determine the molar mass of the gas and the formula.

                          44.0 g/mole       N₂O

18. 100. g of an aqueous compound that is 62.2 % Pb, 8.454 % N, and 28.8 % O reacts with another compound that is 28.16 % N, 8.13 % H, 20.79 % P, and 42.91 % O. If the actual yield of the product containing lead is 60.0 g, calculate the percentage yield. And you thought the last challenge question was tough!!!!

3Pb(NO₃)_2(aq) + 2(NH₄)₃PO_4(aq) → Pb₃(PO₄)_2(s) + 6NH₄NO_3(aq)

100. g

? g

Theoretical Yield = 81.68 g Percentage Yield = 60.0 g x 100 % = 73.5 %

81.7 g

19.     A 10.0 g sample of a Cu/Ag alloy reacted with concentrated HNO₃ according to the following equations.

          A total of 7.3307g of NO₂ was isolated. Assuming a 100 % yield of NO₂, determine the mass of Cu

and Ag in the alloy.

Cu_(s)     +    4HNO_3(aq)        →     Cu(NO₃)_2(aq)              +       2NO_2(g)      +     2H₂O_(l)

2Ag_(s)      +    4HNO_3(aq)       →       2AgNO_3(aq)              +       2NO_2(g)       +     2H₂O_(l)

7.00 g Ag and 3.00 g Cu