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Tro IC3 1. Reacts with 3 mol of O 2 2. Produces 1 mol of Al 2 O 3 3. Produces 0.5 mol Al 2 O 3 4. Reacts with 2 mol of O 2 5. Cannot react with O 2 8.1 The balanced equation of 4Al + 3O 2 2Al 2 O 3 can be interpreted by saying that 1 mol of Al:

The balanced equation of 4Al + 3O 2 → 2Al 2 O 3 can be interpreted by saying that 1 mol of Al:

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The balanced equation of 4Al + 3O 2 → 2Al 2 O 3 can be interpreted by saying that 1 mol of Al:. 8.1. Reacts with 3 mol of O 2 Produces 1 mol of Al 2 O 3 Produces 0 . 5 mol Al 2 O 3 Reacts with 2 mol of O 2 Cannot react with O 2. - PowerPoint PPT Presentation

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Page 1: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. Reacts with 3 mol of O2

2. Produces 1 mol of Al2O3

3. Produces 0.5 mol Al2O3

4. Reacts with 2 mol of O2

5. Cannot react with O2

8.1 The balanced equation of 4Al + 3O2 → 2Al2O3 can be interpreted by saying that 1 mol of Al:

Page 2: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. Reacts with 3 mol of O2

2. Produces 1 mol of Al2O3

3. Produces 0.5 mol Al2O3

4. Reacts with 2 mol of O2

5. Cannot react with O2

8.1 The balanced equation of 4Al + 3O2 → 2Al2O3 can be interpreted by saying that 1 mol of Al:

Page 3: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 3

2. 1

3. 6

4. 9

5. 18

8.2 How many moles of oxygen are needed to produce 6 mol of Fe2O3 in the following equation:

4Fe + 3O2 → 2Fe2O3?

Page 4: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 3

2. 1

3. 6

4. 9

5. 18

8.2 How many moles of oxygen are needed to produce 6 mol of Fe2O3 in the following equation:

4Fe + 3O2 → 2Fe2O3?

Page 5: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 4.0

2. 8.0

3. 13

4. 26

5. 52

8.3 How many moles of gaseous oxygen are required for the complete combustion of 4.0 mol of butane (C4H10)?

Page 6: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 4.0

2. 8.0

3. 13

4. 26

5. 52

8.3 How many moles of gaseous oxygen are required for the complete combustion of 4.0 mol of butane (C4H10)?

Page 7: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 5.90 g

2. 21.4 g

3. 4.28 g

4. 10.7 g

5. 29.5 g

8.4 How many grams of oxygen are required to completely burn 5.90 g of C3H8 in the following unbalanced equation:

C3H8(g) + O2(g) → CO2(g) + H2O(g)?

Page 8: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 5.90 g

2. 21.4 g

3. 4.28 g

4. 10.7 g

5. 29.5 g

8.4 How many grams of oxygen are required to completely burn 5.90 g of C3H8 in the following unbalanced equation:

C3H8(g) + O2(g) → CO2(g) + H2O(g)?

Page 9: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 305 g

2. 457 g

3. 272 g

4. 914 g

5. 342 g

8.5 How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid in the following unbalanced equation:

Al2O3(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2O(l)?

Page 10: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 305 g

2. 457 g

3. 272 g

4. 914 g

5. 342 g

8.5 How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid in the following unbalanced equation:

Al2O3(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2O(l)?

Page 11: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 34.9 mol

2. 0.0703 mol

3. 4.00 mol

4. 2.18 mol

5. 0.282 mol

8.6 How many moles of phosphorus are required to produce 15.5 g P4O6 in the following equation:

4P(s) + 3O2(g) → P4O6(s)?

Page 12: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 34.9 mol

2. 0.0703 mol

3. 4.00 mol

4. 2.18 mol

5. 0.282 mol

8.6 How many moles of phosphorus are required to produce 15.5 g P4O6 in the following equation:

4P(s) + 3O2(g) → P4O6(s)?

Page 13: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 4.5 mol

2. 3.8 mol

3. 5.3 mol

4. 3.0 mol

5. 4.2 mol

8.7 How many moles of oxygen gas are produced from the following decomposition of 3.0 mol of Al2O3,

assuming a 85% yield: 2Al2O3 → 4Al + 3O2?

Page 14: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 4.5 mol

2. 3.8 mol

3. 5.3 mol

4. 3.0 mol

5. 4.2 mol

8.7 How many moles of oxygen gas are produced from the following decomposition of 3.0 mol of Al2O3,

assuming a 85% yield: 2Al2O3 → 4Al + 3O2 ?

Page 15: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 115 kg

2. 74.3 kg

3. 134 kg

4. 25.0 kg

5. 66.2 kg

8.8 How many kg of methane (CH4) are produced from 25.0 kg of hydrogen in the following catalytic

methanation: 3H2 + CO → CH4 + H2O?

Page 16: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 115 kg

2. 74.3 kg

3. 134 kg

4. 25.0 kg

5. 66.2 kg

8.8 How many kg of methane (CH4) are produced from 25.0 kg of hydrogen in the following catalytic

methanation: 3H2 + CO → CH4 + H2O?

Page 17: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 7.0 mol

2. 5.2 mol

3. 3.5 mol

4. 10. mol

5. 1.8 mol

8.9 How many moles of water are produced from the combustion of 3.5 mol of hydrogen and 5.2 mol of oxygen?

Page 18: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 7.0 mol

2. 5.2 mol

3. 3.5 mol

4. 10. mol

5. 1.8 mol

8.9 How many moles of water are produced from the combustion of 3.5 mol of hydrogen and 5.2 mol of oxygen?

Page 19: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 0.305 g

2. 25.5 g

3. 18.3 g

4. 9.15 g

5. 0.851 g

8.10 How many grams of NO are produced from the oxidation of 14.5 g of NH3 by 12.2 g of O2 in the

following equation: 4NH3 + 5O2 → 4NO + 6H2O?

Page 20: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 0.305 g

2. 25.5 g

3. 18.3 g

4. 9.15 g

5. 0.851 g

8.10 How many grams of NO are produced from the oxidation of 14.5 g of NH3 by 12.2 g of O2 in the

following equation: 4NH3 + 5O2 → 4NO + 6H2O?

Page 21: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 2.26 g

2. 4.14 g

3. 2.3 g

4. 21.5 g

5. 10.4 g

8.11 How many grams of the excess reagent are left over after the reaction of 4.36 g of Al and 8.50 g HCl in the following unbalanced equation:

Al + HCl → AlCl3 + H2?

Page 22: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 2.26 g

2. 4.14 g

3. 2.3 g

4. 21.5 g

5. 10.4 g

8.11 How many grams of the excess reagent are left over after the reaction of 4.36 g of Al and 8.50 g HCl in the following unbalanced equation:

Al + HCl → AlCl3 + H2?

Page 23: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 1.08 x 1025

2. 2.77 x 1025

3. 2.56 x 1025

4. 1.39 x 1025

5. 6.02 x 1023

8.12 How many molecules of water are produced from the combustion of 2.56 mol of octane in the following equation:

2C8H18 + 25O2 → 16CO2 + 18H2O?

Page 24: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 1.08 x 1025

2. 2.77 x 1025

3. 2.56 x 1025

4. 1.39 x 1025

5. 6.02 x 1023

8.12 How many molecules of water are produced from the combustion of 2.56 mol of octane in the following equation:

2C8H18 + 25O2 → 16CO2 + 18H2O?

Page 25: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 8.75 x 1024

2. 7.74 x 1022

3. 1.55 x 1022

4. 5.27 x 1024

5. 6.02 x 1023

8.13 How many molecules of oxygen are produced from the decomposition of 8.75 g of hydrogen peroxide

in the following equation: 2H2O2 → 2H2 + O2?

Page 26: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 8.75 x 1024

2. 7.74 x 1022

3. 1.55 x 1022

4. 5.27 x 1024

5. 6.02 x 1023

8.13 How many molecules of oxygen are produced from the decomposition of 8.75 g of hydrogen peroxide

in the following equation: 2H2O2 → 2H2 + O2?

Page 27: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 2 molecules 2. 7.54 x 1025 molecules3. 5.03 x 1025 molecules 4. 1.51 x 1026 molecules5. 6.02 x 1023 molecules

8.14 How many molecules of ammonia are produced from 7.54 x 1025 molecules of hydrogen in the

following equation: 3H2 + N2 → 2NH3?

Page 28: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 2 molecules 2. 7.54 x 1025 molecules3. 5.03 x 1025 molecules 4. 1.51 x 1026 molecules5. 6.02 x 1023 molecules

8.14 How many molecules of ammonia are produced from 7.54 x 1025 molecules of hydrogen in the

following equation: 3H2 + N2 → 2NH3?

Page 29: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 1.93 x 1024 molecules2. 16 molecules3. 9.63 x 1024 molecules4. 3 molecules5. 3.2 x 1024 molecules

8.15 How many molecules of CO2 are produced from the combustion of 2 molecules of C8H18 and 5 molecules of O2? (Hint: This is a tricky question.)

Page 30: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 1.93 x 1024 molecules2. 16 molecules3. 9.63 x 1024 molecules4. 3 molecules5. 3.2 x 1024 molecules

8.15 How many molecules of CO2 are produced from the combustion of 2 molecules of C8H18 and 5 molecules of O2? (Hint: This is a tricky question.)

Page 31: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 24.2 g2. 32.0 g3. 42.2 g4. 68.6 g5. 84.4 g

8.16 What is the theoretical yield of acetic acid when 22.5 g of methanol reacts with 32.0 g of carbon monoxide in the following equation:

CH3OH + CO → HC2H3O2?

Page 32: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 24.2 g2. 32.0 g3. 42.2 g4. 68.6 g5. 84.4 g

8.16 What is the theoretical yield of acetic acid when 22.5 g of methanol reacts with 32.0 g of carbon monoxide in the following equation:

CH3OH + CO → HC2H3O2?

Page 33: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 3.00 g2. 320. g3. 2.00 g4. 4.00 g5. 240. g

8.17 What is the theoretical yield of SO3 from the combustion of 3.00 mol of S and 6.00 mol of O2?

Page 34: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 3.00 g2. 320. g3. 2.00 g4. 4.00 g5. 240. g

8.17 What is the theoretical yield of SO3 from the combustion of 3.00 mol of S and 6.00 mol of O2?

Page 35: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 35.1 g2. 2.17 x 1025 g3. 31.4 g4. 3.78 x 1025 g5. 62.8 g

8.18 What is the theoretical yield of MgO from the combustion of 5.25 x 1023 atoms of Mg and 4.69 x 1023 molecules of O2?

Page 36: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 35.1 g2. 2.17 x 1025 g3. 31.4 g4. 3.78 x 1025 g5. 62.8 g

8.18 What is the theoretical yield of MgO from the combustion of 5.25 x 1023 atoms of Mg and 4.69 x 1023 molecules of O2?

Page 37: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. H2. O3. O2

4. H2

5. H2O

8.19 Which reactant is in excess from the reaction of 35.0 g of hydrogen with 276 g of oxygen to form water?

Page 38: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. H2. O3. O2

4. H2

5. H2O

8.19 Which reactant is in excess from the reaction of 35.0 g of hydrogen with 276 g of oxygen to form water?

Page 39: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. It has the lowest mass.2. Its coefficient in the balanced equation is the

smallest one.3. It has the highest mass.4. It doesn’t react completely.5. All of the above

8.20 Which of the following statements is NOT correct about the limiting reactant in a reaction?

Page 40: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. It has the lowest mass.2. Its coefficient in the balanced equation is the

smallest one.3. It has the highest mass.4. It doesn’t react completely.5. All of the above

8.20 Which of the following statements is NOT correct about the limiting reactant in a reaction?

Page 41: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 100. %2. 91.3 % 3. 109 %4. 85.2 %5. 55.9 %

8.21 In the combustion of a sample of methane, the theoretical yield is 28.9 g CO2. When a chemistry student performed the experiment, the yield was 26.4 g CO2. What is the percent yield of their reaction by the student?

Page 42: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 100. %2. 91.3 % 3. 109 %4. 85.2 %5. 55.9 %

8.21 In the combustion of a sample of methane, the theoretical yield is 28.9 g CO2. When a chemistry student performed the experiment, the yield was 26.4 g CO2. What is the percent yield of their reaction by the student?

Page 43: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 25.9 g2. 2.50 g3. 92.0 g4. 28.5 g5. 30.7 g

8.22 An experiment had a percent yield of 91.8% and a calculated theoretical yield of 28.2 g. How many grams were actually produced?

Page 44: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 25.9 g2. 2.50 g3. 92.0 g4. 28.5 g5. 30.7 g

8.22 An experiment had a percent yield of 91.8% and a calculated theoretical yield of 28.2 g. How many grams were actually produced?

Page 45: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. K2. H2O

3. KOH4. H2

5. Both are used up at the same time.

8.23 What is the limiting reactant for the reaction between 18.2 g of K and 8.38 g of water in the following unbalanced reaction:

K + H2O → KOH + H2?

Page 46: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. K2. H2O

3. KOH4. H2

5. Both are used up at the same time.

8.23 What is the limiting reactant for the reaction between 18.2 g of K and 8.38 g of water in the following unbalanced reaction:

K + H2O → KOH + H2?

Page 47: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 40. %2. 63 %3. 80. %4. 90. %5. 161%

8.24 5.0 kg of hydrogen is reacted with 31 kg of oxygen. The experiment yields 28 kg of water. Calculate the percent yield of the reaction.

Page 48: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 40. %2. 63 %3. 80. %4. 90. %5. 161%

8.24 5.0 kg of hydrogen is reacted with 31 kg of oxygen. The experiment yields 28 kg of water. Calculate the percent yield of the reaction.

Page 49: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 28.1%2. 47.3%3. 79.9%4. 82.8%5. 91.2%

8.25 4.25 mol of Cl2 are reacted with 1.49 mol NaOH.

12.5 g of NaClO3 are produced. Calculate the

percent yield of the following reaction: 3Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3H2O.

Page 50: The balanced equation of 4Al + 3O 2 →  2Al 2 O 3  can be interpreted by saying that 1 mol of Al:

Tro IC3

1. 28.1%2. 47.3%3. 79.9%4. 82.8%5. 91.2%

8.25 4.25 mol of Cl2 are reacted with 1.49 mol NaOH.

12.5 g of NaClO3 are produced. Calculate the

percent yield of the following reaction: 3Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3H2O.