Chemical Reactions and Quantities Practice …ion.chem.usu.edu/~scheiner/LundellChemistry/practice...7 – Chemical Reactions and Quantities – Practice Problems I’m trying a different

7 – Chemical Reactions and Quantities – Practice Problems

I’m trying a different set up for the practice problems. This still contains the practice problems you need to

master for the test. I’ve organized the problems by sections in the chapter and provided a summary that may

prove useful. For some sections, there’s also a part called Understanding the Concepts. It’s comprised of a few

problems that use visual guides rather than equations to solve the problems which might prove useful as you

first begin working through the problems. I won’t use the pictures on the test, I’ll use the wordings you find in

the Practice Problems section. Understanding the Concepts is to give you another way of looking at the

problems that may help the concepts click better.

Outline:

Section heading

Goal: what you should learn from the section.

Summary: a brief summary of the section’s material and an example.

Understanding the Concepts – A visual representation of the problems that may prove helpful.

Practice Problems – the normal problems you need to master for the test

Challenge Problems – some sections will have a few select problems that are harder than the others, mainly

because they require you to combine knowledge from multiple sections. I will choose one challenge

problem from each chapter to put on the test. It will either be identical or very slightly different than

the wording written here. (I’ve always added a couple difficult questions to the test so this isn’t new.

It will simply give you more of a guide as to what to expect.)

Section 7.1 – Equations for Chemical Reactions

Goal: Write a balanced chemical equation from the formulas of the reactants and products for a reaction;

determine the number of atoms in the reactants and products.

Summary

In a balanced equation, there are the same number and type of atoms on each side of the arrow. To achieve this,

whole numbers called coefficients written to the left of the formulas increase the amount of that formula. For

example: 2H2O means there are 2 water molecules for a total of 4 hydrogen atoms and 2 oxygen atoms.

Example: Balance the following chemical equation:

Ag(s) + S Ag2S(s)

reactants products

Ag 1 2

S 1 1

Need 2 Ag on left hand side. Put a 2 in front of Ag(s)

2Ag(s) + S(s) Ag2S(s)

reactants products

Ag 1 2 2

S 1 1

Understanding the Concepts

Balance each of the following by adding coefficients:

Balance each of the following by adding coefficients:

If red spheres represent oxygen atoms and blue spheres represent nitrogen atoms

a. Write the formula for each of the reactants and products.

b. Write a balanced equation for the reaction.

If blue spheres represent nitrogen atoms and purple spheres represent iodine,



If green spheres represent chlorine atoms, yellow-green spheres represent fluorine atoms, and white spheres

represent hydrogen atoms,



If green spheres represent chlorine atoms and red spheres represent oxygen atoms



If blue spheres represent nitrogen atoms and purple spheres represent iodine atoms



Practice Problems

1. Balance the following chemical reaction:

P4(s) + O2(g) P4O10(s)

a. P4(s) + O10(g) P4O10(s)

b. 2P4(s) + 4O2(g) 2P4O10(s)

c. P4(s) + 5O2(g) P4O10(s)

d. P4(s) + 4O2(g) P4O10(s)

2. Balance the following chemical reaction:

C4H8(g) + O2(g)

∆→ CO2(g) + H2O(g)

a. C4H8(g) + 3O2(g)

∆→ 4CO2(g) + 4H2O(g)

b. C4H8(g) + 6O2(g)

∆→ 4CO2(g) + 4H2O(g)

c. 2C4H8(g) + 6O2(g)

∆→ 4CO2(g) + 6H2O(g)

d. 2C4H8(g) + 3O2(g)

∆→ 3CO2(g) + 3H2O(g)

3. Balance the following chemical equation:

Zn(s) + HNO3(aq) Zn(NO3)2(aq) + H2(g)

a. Zn(s) + 2HNO3(aq) Zn(NO3)2(aq) + H2(g)

b. 2Zn(s) + 2HNO3(aq) 2Zn(NO3)2(aq) + H2(g)

c. Zn(s) + 2HNO3(aq) 2Zn(NO3)2(aq) + H2(g)

d. 2Zn(s) + HNO3(aq) Zn(NO3)2(aq) + 2H2(g)


K2SO4(aq) + BaCl2(aq) BaSO4(s) + KCl(aq)

a. 2K2SO4(aq) + 2BaCl2(aq) BaSO4(s) + 4KCl(aq)

b. 2K2SO4(aq) + BaCl2(aq) 2BaSO4(s) + KCl(aq)

c. K2SO4(aq) + 2BaCl2(aq) BaSO4(s) + 2KCl(aq)

d. K2SO4(aq) + BaCl2(aq) BaSO4(s) + 2KCl(aq)


Ca(OH)2(aq) + HNO3(aq) Ca(NO3)2(aq) + H2O(l)

a. Ca(OH)2(aq) + 2HNO3(aq) Ca(NO3)2(aq) + 2H2O(l)

b. 2Ca(OH)2(aq) + 2HNO3(aq) Ca(NO3)2(aq) + H2O(l)

c. 2Ca(OH)2(aq) + HNO3(aq) Ca(NO3)2(aq) + 2H2O(l)

d. Ca(OH)2(aq) + 3HNO3(aq) Ca(NO3)2(aq) + 2H2O(l)


AlCl3(aq) + KOH(aq) Al(OH)3(s) + KCl(aq)

a. AlCl3(aq) + 3KOH(aq) Al(OH)3(s) + 3KCl(aq)

b. AlCl3(aq) + 2KOH(aq) Al(OH)3(s) + 2KCl(aq)

c. 2AlCl3(aq) + KOH(aq) Al(OH)3(s) + KCl(aq)

d. AlCl3(aq) + KOH(aq) Al(OH)3(s) + 2KCl(aq)


H3PO4(aq) + KOH(aq) H2O(l) + K3PO4(aq)

a. H3PO4(aq) + KOH(aq) 2H2O(l) + K3PO4(aq)

b. H3PO4(aq) + 2KOH(aq) 3H2O(l) + K3PO4(aq)

c. H3PO4(aq) + 3KOH(aq) 3H2O(l) + K3PO4(aq)

d. H3PO4(aq) + 3KOH(aq) H2O(l) + K3PO4(aq)

Section 7.2 – Types of Reactions

Goal: Identify a reaction as a combination, decomposition, single replacement, double replacement, or

combustion reaction.

Summary


Identify each of the following as: combination, decomposition, single replacement, double replacement, or

combustion. (If it helps, use the equations you wrote for Understanding the Concepts for section 7.1. These are

the same reactions.)

Practice Problems

8. Classify as a combination, decomposition, single replacement, double replacement, or combustion reaction:

__________________ H2(g) + Br2(g) 2HBr(g)

__________________ AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

__________________ 2H2O2(aq) 2H2O(l) + O2(g)

__________________ Zn(s) + CuCl2(aq) Cu(s) + ZnCl2(aq)

__________________ C5H8(g) + 7O2(g) 5CO2(g) + 4H2O(g)

9. Classify as a combination, decomposition, single replacement, double replacement, or combustion reaction:

__________________ CuO(s) + 2HCl(aq) CuCl2(aq) + H2O(l)

__________________ 2Al(s) + 3Br3(g) 2AlBr3(s)

__________________ C6H12(l) + 9O2(g) 6CO2(g) + 9H2O(g)

__________________ Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g)

__________________ C6H12O6(aq) 2C6H6O(aq) + 2CO2(g)

Challenge Problems (from sections 7.1 and 7.2)

For problems 10-11: Balance each of the following chemical equations and (ii) identify the type of reaction.

10. C4H8(g) + O2(g) CO2(g) + H2O(g)

(i) (ii)

a. C4H8(g) + 3O2(g) 4CO2(g) + 4H2O(g) a. single replacement

b. C4H8(g) + 6O2(g) 4CO2(g) + 4H2O(g) b. double replacement

c. 2C4H8(g) + 6O2(g) 4CO2(g) + 6H2O(g) c. combination

d. 2C4H8(g) + 3O2(g) 3CO2(g) + 3H2O(g) d. combustion

11. Sb(s) + Cl2(g) SbCl3(s)

(i) (ii)

a. Sb(s) + 3Cl2(g) SbCl3(s) a. combination

b. 2Sb(s) + 2Cl2(g) 2SbCl3(s) b. decomposition

c. 2Sb(s) + 3Cl2(g) 2SbCl3(s) c. single replacement

d. Sb(s) + 2Cl2(g) SbCl3(s) d. double replacement

12. NI3(s) N2(g) + I2(g)

(i) (ii)

a. 2NI3(s) N2(g) + 3I2(g) combination

b. 3NI3(s) N2(g) + 2I2(g) decomposition

c. NI3(s) N2(g) + 3I2(g) single replacement

d. NI3(s) N2(g) + I3(g) double replacement

Section 7.3 – Oxidation-Reduction Reactions

Goal: Define the terms oxidation and reduction; identify reactants as oxidized or reduced.

Summary

In an oxidation-reduction reaction (abbreviated redox), one reactant is oxidized when it loses electrons, and

another reactant is reduced when it gains electrons.

• OIL – oxidation is loss (of electrons)

• RIG – reduction is gain (of electrons)

Example: For the following redox reaction, identify the reactant that is oxidized, and the reactant that is

reduced:

Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)

Answer: Fe0(s) Fe2+(aq) + 2e-

Fe starts with 0 charge, then changes to +2 (0 +2, Fe must have lose 2 electrons)

Fe loses electrons; it is oxidized

Cu2+(aq) + 2e- Cu0(s)

Cu starts with +2 charge, then changes to 0 (+2 0, Cu must have gained 2 electrons)

Cu2+ gain electrons; it is reduced

Practice Problems

13. Identify each of the following as oxidation or reduction:

O2(g) + 4e- 2O2-(aq)

Ag(s) Ag+(aq) + e-

Fe3+(aq) + e- Fe2+(aq)

2Br-(aq) Br2(l) +2e-

14. In the following reaction ___(i)___ is oxidized and ___(ii)___ is reduced.

2Li(s) + F2(g) 2LiF(s)

a. (i) Li (ii) F2

b. (i) Fe2 (ii) Li

c. (i) Li (ii) Li+

d. (i) Li (ii) F-

e. (i) Li+ (ii) F-


Cl2(g) + 2KI(aq) 2KCl(aq) + I2(s)

a. (i) Cl2 (ii) K+

b. (i) K+ (ii) Cl2

c. (i) I- (ii) Cl2

d. (i) I (ii) Cl-

e. (i) Cl- (ii) K+


2Al(s) + 3Sn2+(aq) 2Al3+(aq) + 3Sn(s)

a. (i) Sn (ii) Al

b. (i) Al (ii) Sn

c. (i) Al3+ (ii) Sn

d. (i) Sn2+ (ii) Al

e. (i) Al (ii) Sn2+


Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

a. (i) Fe (ii) Cu

b. (i) Fe (ii) Cu2+

c. (i) Cu2+ (ii) Fe

d. (i) Cu (ii) Fe

e. (i) SO42- (ii) Cu2+

Section 7.4 – The Mole

Goal: Use Avogadro’s number (6.02 x 1023) to determine the number of particles in a given number of moles.

Summary:

• In chemistry, atoms, molecules, and ions are counted by the mole, a unit that contains 6.02 x 1023 items,

which is Avogadro’s number.

• For example, 1 mole of carbon contains 6.02 x 1023 atoms of carbon; 1 mole of H2O contains 6.02 x 1023

molecules of H2O

• Avogadro’s number is used to convert between particles and moles.

Example: How many moles of nickel contain 2.45 x 1024 Ni atoms?

Answer: 2.45 x 1024 Ni atoms 1 mole Ni

6.02 x 1023 Ni atoms= 4.07 moles Ni

Practice Problems:

18. What is a mole?

19. What is Avogadro’s number and what is it used for?

20. Calculate the number of Li atoms in 4.5 moles of Li.

a. 2.7 x 1021 atoms

b. 1.3 x 1023 atoms

c. 2.7 x 1024 atoms

d. 2.7 x 10-24 atoms

e. 1.3 x 10-23 atoms

21. Calculate the number of CO2 molecules in 0.0180 moles CO2.

a. 1.1 x 1022 molecules

b. 3.3 x 1025 molecules

c. 7.2 x 1026 molecules

d. 3.3 x 10-25 molecules

e. 1.1 x 10-22 molecules

22. Calculate the number of moles of Cu in 7.8 x 1021 atoms of Cu

a. 1.3 x 1044 moles

b. 77 moles

c. 0.15 moles

d. 0.013 moles

e. 2.6 moles

23. Calculate the moles of C2H6 in 3.75 x 1023 molecules of C2H6.

a. 0.011 moles

b. 0.62 moles

c. 1.6 moles

d. 2.3 x 1047 moles

e. 6.2 x 1045 moles

24. Calculate the number of moles of C in 0.185 moles of C6H14O.

a. 1.11 moles

b. 0.031 moles

c. 3.1 x 10-25 moles

d. 1.11 x 1023 moles

e. 2.61 moles

25. Calculate the number of atoms of H in 0.185 moles of C6H14O.

a. 2.59 atoms

b. 3.07 x 10-25 atoms

c. 1.11 x 1023 atoms

d. 7.96 x 1021 atoms

e. 1.56 x 1024 atoms

26. How many moles of S are present in 3.0 moles of Al2(SO4)3?

a. 5.0 x 10-24 moles

b. 1.8 x 1024 moles

c. 6.0 moles

d. 9.0 moles

e. 3.0 moles

27. How many moles of aluminum ion (Al3+) are present in 0.40 moles of Al2(SO4)3?

a. 0.4 moles

b. 0.8 moles

c. 2.0 moles

d. 2.4 x 1023 moles

e. 6.6 x 10-24 moles

28. How many moles of the sulfate ions (SO42-) are present in 1.5 moles of Al2(SO4)3?

a. 9.0 x 1023 moles

b. 4.0 x 1023 moles

c. 1.5 moles

d. 4.5 moles

e. 3.0 moles

Section 7.5 – Molar Mass and Calculations

Goal: Calculate the molar mass for a substance given its chemical formula; use molar mass to convert between

grams and moles.

Summary:

Calculating Molar Mass:

The molar mass of a compound is the sum of the molar mass of each element in it chemical formula multiplied

by it subscript in the formula. The molar masses of individual elements are found on the periodic table as the

atomic mass (the decimal number under the symbol).

Example:

Pinene, C10H16, which is found in pine tree sap and essential oils, has anti-inflammatory properties. Calculate

the molar mass for pinene.

Answer:

Using Molar Mass as a Conversion Factor

• The molar mass of an element is its mass in grams per mole equal numerically to its atomic mass.

• The molar mass of a compound is its mass in grams per mole equal numerically to the sum of the masses

of its elements.

• Molar mass is used as a conversion factor to convert between the moles and grams of a substance.

Example:

The frame of a bicycle contains 6500 g of aluminum. How many moles of aluminum are in the bicycle frame?

Answer:

\

Understanding the Concepts:

Using the models of the molecules (black = C, white = H, yellow =S, green = Cl) determine each of the

following for models of the compounds.

a. molecular formula

b. molar mass

c. number of moles in 10.0 g

Using the models of the molecules (black = C, white = H, yellow = S, Red = O) determine each of the

following for the models of the compounds

a. molecular formula

b. molar mass

c. number of moles in 10.0 g

Practice Problems:

29. Calculate the molar mass of the following:

a. FeSO4 _______ g/mol

b. C7H5NO3S _______ g/mol

c. (NH4)2CO3 _______ g/mol

d. O2 _______ g/mol

e. Fe(ClO4)3 _______ g/mol

30. Calculate the mass in grams for 1.50 moles K

a. 11.2 g

b. 39.1 g

c. 26.1 g

d. 58.6 g

e. 28.5 g

31. Calculate the mass in grams of 2.5 moles of C.

a. 24.0 g

b. 30.0 g

c. 4.80 g

d. 15.0 g

e. 16.2 g

32. Calculate the number of grams in 5.00 moles of C2H6O.

a. 212 g

b. 29.0 g

c. 460 g

d. 230 g

e. 46.1 g

33. Calculate the number of grams in 0.488 mole of C3H6O3.

a. 29.0 g

b. 90.1 g

c. 44.0 g

d. 57.3 g

e. 185 g

34. How many moles are contained on 25.0 g of Ca?

a. 1.25 moles

b. 1.60 moles

c. 0.624 moles

d. 6.42 moles

e. 0.500 moles

35. How many moles are contained in 5.00 g of S.

a. 0.116 moles

b. 11.6 moles

c. 0.761 moles

d. 160 moles

e. 4.70 moles

36. Calculate the number of moles of Cr(OH)3 in 4.00 g of Cr(OH)3.

a. 42.1 moles

b. 0.721 moles

c. 412 moles

d. 0.0388 moles

e. 103 moles

37. Calculate the number of moles of Ca3N2 in 4.00 g of Ca3N2.

a. 1.01 moles

b. 148 moles

c. 37.1 moles

d. 0.0270 moles

e. 593 moles

Section 7.6 – Mole Relationships in Chemical Equations

Goal: Given the quantity in moles of reactants or products, use a mole – mole factor from the balanced

chemical equation to calculate the number of moles of another substance in the reaction.

Summary:

Using Mole – Mole Factors

Consider the balanced chemical equation:

• The coefficients in a balanced chemical equation represent the moles of the reactants and the moles of

products. Thus, 4 moles of Na react with 1 mole of O2 to form 2 moles of Na2O.

• From the coefficients, mole-mole factors can be written for any two substances as follows:

• A mole – mole factor is used to convert the number of moles of one substance in the reaction to the

number of moles of another substance in the reaction.

Example:

How many moles of sodium are needed to produce 3.5 moles of sodium oxide?

Answer:

Need to convert from moles of Na2O to moles of Na

Practice Problems

38. Write all the mole – mole factors for 2Al(s) + 3Cl2(g) 2AlCl3(s)

39. Write all the mole – mole factors for 4HCl(g) + O2(g) 2Cl2(g) + 2H2O(g)

For problems 40-42 use the following:

Ammonia is produced by the chemical reaction of nitrogen and hydrogen.

40. How many moles of H2 are needed to react with 1.0 mole of N2?

a. 1 mole

b. 2 moles

c. 2.5 moles

d. 3 moles

e. 4.6 moles

41. How many moles of N2 reacted if 0.60 mole of NH3 is produced?

a. 0.6 moles

b. 0.3 moles

c. 1 moles

d. 2 moles

e. 3 moles

42. How many moles of NH3 are produced when 1.4 moles of H2 react?

a. 2 moles

b. 5 moles

c. 4 moles

d. 2.5 moles

e. 0.5 moles

For problems 43-46, use the following:

In an acetylene torch, acetylene gas (C2H2) burns in oxygen to produce carbon dioxide and water.

43. How many moles of O2 are needed to react with 2.00 moles of C2H2?

a. 2 moles

b. 5 moles

c. 4 moles

d. 2.5 moles

e. 0.5 moles

44. How many moles of CO2 are produced when 3.5 moles of C2H2 reacts?

a. 2 moles

b. 7 moles

c. 5 moles

d. 4 moles

e. 2.5 moles

45. How many moles of C2H2 are needed to produce 0.50 mole of H2O?

a. 0.5 moles

b. 2 moles

c. 5 moles

d. 4 moles

e. 1 mole

46. How many moles of CO2 are produced from 0.100 mole of O2?

a. 4 moles

b. 5 moles

c. 2 moles

d. 1.25 moles

e. 0.08 moles

Section 7.7 – Mass Calculations for Reactions

Goal: Given the mass in grams of a substance in a reaction, calculate the mass of another substance in the

reaction.

Summary:

Converting grams to grams

When we have a balanced chemical equation for a reaction, we can use the mass of substance A and then

calculate the mass of substance B.

A B

The process is as follows:

1. Use the molar mass of A to convert the mass (in grams) of A to moles of A

2. Use the mole-mole factor that converts moles of A to moles of B

3. Use the molar mass of B to convert the moles of B to mass (in grams) of B.

grams A moles A moles B grams B

Example: How many grams of O2 are needed to completely react with 14.6 g of Na?

4Na(s) + O2(g) 2Na2O(s)

Answer:

Practice Problems

47. Nitrogen gas reacts with hydrogen gas to produce ammonia:

N2(g) + 3H2(g) 2NH3(g)

If you have 3.64 g of H2, how many grams of NH3 can be produced?

a. 10.3 g

b. 46.1 g

c. 83.3 g

d. 0.646 g

e. 187 g

48. Iron (III) oxide reacts with carbon to give iron and carbon monoxide.

Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g)

How many grams of CO are produced when 36.0 g of C reacts?

a. 0.107 g

b. 84.0 g

c. 252 g

d. 1.21 x 104 g

e. 28.0 g

molar mass A

molar mass B

mole-mole factor

49. Calcium cyanamide reacts with water to form calcium carbonate.

CaCN2(s) + 3H2O(l) CaCO3(s) + 2NH3(g)

How many grams of H2O are needed to react with 75.0 grams of CaCN2?

a. 79.1 g

b. 5.62 g

c. 111 g

d. 1.48 g

e. 50.6 g

Challenge Questions

50. When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3),

which causes acid rain and nitrogen oxide (Sections 7.1, 7.5, 7.6, 7.7)

a. Write the balanced chemical equation.

b. How many moles of each product are produced from 0.230 mole of H2O?

c. How many grams of HNO3 are produced when 60.0 g of NO2 completely reacts?

d. How many grams of NO2 are needed to form 75.0 g of HNO3?

51. Propane gas (C3H8) reacts with oxygen to produce carbon dioxide and water (Sections 7.5, 7.6, 7.7)

C3H8(g) + 5O2(g) 3CO2(g) + 5H2O(l)

a. How many moles of H2O form when 5.00 moles of C3H8 completely reacts?

b. How many grams of CO2 are produced from 18.5 g of oxygen gas?

c. How many grams of H2O can be produced when 56.4 g of C3H8 reacts?

52. Gasohol is a fuel containing ethanol (C2H6O) that burns in oxygen (O2) to give carbon dioxide and water

(Sections 7.1, 7.6, 7.7)

a. Write the balanced chemical equation.

b. How many moles of O2 are needed to completely react with 8.0 moles of C2H6O?

c. If a car produces 4.4g of CO2, how many grams of O2 are used up in the reaction?

d. If you burn 125 g of C2H6O, how many grams of CO2 and H2O can be produced?

Section 7.8 – Limiting Reactants and Percent Yield

Goal: Identify a limiting reactant and calculate the amount of product formed from the limiting reactant. Given

the actual quantity of product, determine the percent yield for a reaction.

Summary:

Calculating quantity of product from a limiting reactant

Often in reactions, the reactants are not consumed at exactly the same time. Then one of the reactants, called the

limiting reactant, determines the maximum amount of product that can for.

• To determine the limiting reactant, we calculate the amount of product that is possible from each

reactant.

• The limiting reactant is the one that produces the smaller amount of product.

Example: If 12.5 g of S reacts with 17.2 g of O2, what is the limiting reactant and the mass, in grams, of SO3

produced?

2S(s) + 2O2(g) 2SO3(g)

Answer:

Mass of SO3 from S:

Mass of SO3 from O2:

=

Conclusion: The limiting reactant is O2 because it produced less SO3 28.7 g vs. 31.2 g. So the

reaction will stop once all the O2 is used up and it will produce 28.7 g of SO3

Calculating Percent Yield

• The theoretical yield or a reaction is the amount of product (100%) formed if all the reactants were

converted to desired product.

• The actual yield for the reaction is the mass, in grams, of the product obtained at the end of the

experiment. Because some product is usually lost, the actual yield is less than the theoretical yield.

• The percent yield is calculated from the actual yield divided by the theoretical yield and multiplied by

100%.

Percent (%) yield = actual yield

theoretical yield x 100

Example: If 22.6 g of Al reacts completely with O2 and 37.8 g of Al2O3 is obtained, what is the percent yield of

Al2O3 for the reaction:

4Al(s) + 3O2(g) 2Al2O3(s)

Answer:

Calculation of theoretical yield:

Calculation of percent yield:


If green spheres represent chlorine atoms, yellow-green spheres represent fluorine atoms, and white spheres represent

hydrogen atoms, (7.1, 7.8)

a. Write a balanced equation for the reaction.

b. Identify the limiting reactant.

If blue spheres represent nitrogen atoms and white spheres represent hydrogen atoms, (7.1, 7.8)

a. Write a balanced equation.

b. Identify the diagram that shows the products.

Practice Problems

For problems 53-55, use the following:

Iron and oxygen react to form iron (III) oxide.

4Fe(s) + 3O2(g) 2Fe2O3(s) Determine the limiting reactant in each of the following mixture of reactants:

53. 2.0 moles of Fe and 6.0 moles of O2: __________ 54. 5.0 moles of Fe and 4.0 moles of O2:___________

55. 16.0 moles of e and 20.0 moles of O2:____________

56. 20.0g of each reactant are present initially. Determine the limiting reactant, and calculate the grams of product of

Al2O3 that would be produced.

4Al(s) + 3O2(g) 2Al2O3(s)

a. Al, 37.8 g

b. Al, 42.5 g

c. O2, 37.8 g d. O2, 42.5 g

e. Al2O3, 37.8 g

57. 20.0g of each reactant are present initially. Determine the limiting reactant, and calculate the grams of product of H2O

that would be produced.

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)

a. C2H5OH, 23.5 g

b. C2H5OH, 11.3 g

c. O2, 23.5 g d. O2, 11.3 g

e. CO2, 23.5 g

58. Iron (III) oxide reacts with carbon monoxide to produce iron and carbon dioxide: Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

What is the percent yield of iron if the reaction of 65.0g of iron(III) oxide produces 15.0 g of iron? a. 12.3 %

b. 51.3 %

c. 24.8 % d. 71.3 %

e. 403 %

59. Iron (III) oxide reacts with carbon monoxide to produce iron and carbon dioxide: Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

What is the percent yield of carbon dioxide of the reaction of 75.0 g of carbon monoxide produces 85.0 g of carbon dioxide?

a. 40.3 %

b. 72.0 % c. 53.1 %

d. 64.0 %

e. 139 %

60. The equation for the decomposition of potassium chlorate is written as: 2KClO3(s) 2KCl(s) + 3O2(g)

When 46.0 g of KClO3 is completely decomposed, what is the theoretical yield (in grams) of O2?

(This is the same as in section 7.7, calculated grams of B from grams of A.)

a. 18.0 g b. 117 g

c. 72.3 g

d. 8.01 g e. 36.0 g

61. Continuing with Problem 60, If 12.1 g of O2 is produced, what is the percent yield of O2? a. 1.80 %

b. 149 %

c. 51.3 %

d. 73.2 % e. 67.2 %

Challenge Problems

62. Aluminum and chlorine combine to form aluminum chloride. 2Al(s) + 3Cl2(g) 2AlCl3(g)

If 45.0 g of Al and 62.0 of Cl2 are mixed, and 66.5 g of AlCl3 is actually obtained what is the percent yield of AlCl3 for the

reaction?

a. 67.7 % b. 23.0 %

c. 95.4 %

d. 85.6 % e. 104 %

Section 7.9 – Energy in Chemical Reactions

Goal: Given the heat of reaction, calculate the loss or gain of heat for an exothermic or endothermic reaction.

Summary:

Using the heat of reaction

• The heat of reaction is the amount of heat, in kJ or kcal, that is absorbed or released during a reaction.

• The heat of reaction, symbol ΔH, is the difference in the energy of the products and the reactants.

ΔH = ΔHproducts – ΔHreactants

• In an exothermic reaction (exo mean “out”) the energy of the products is lower than that of the reactants. This means that heat is released along with the products that form. Then the sign for the heat of reaction, ΔH, is

negative.

• In an endothermic reaction (endo means ”within”) the energy of the products is higher than that of the reactants.

Thus heat is required to convert the reactants to products. Then the sign for the heat of reaction, ΔH, is positive.

Example: How many kilojoules are released when 3.50 g of CH4 undergoes combustion?

Answer:

Practice Problems

63. Classify each of the following as endothermic or exothermic: a. The energy level of the products is lower than that of the reactants. ______________

b. In the body, the synthesis of proteins requires energy. ______________

c. A reaction absorbs 125kJ. ______________

64. Classify each of the following as exothermic or endothermic and give the ΔH for each:

a. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) + 2220 kJ

b. 2Na(s) + Cl2(g) 2NaCl(s) + 819 kJ c. PCl5(g) + 67 kJ PCl3(g) + Cl2(g)

65. How many kilojoules are released when 75.0 g of methanol (CH3OH) reacts? 2CH3OH(l) + 3O2(g) CO2(g) + 4H2O(l) ΔH = -726 kJ

a. 726 kJ

b. 624 kJ

c. 423 kJ d. 845 kJ

e. 764 kJ

66. How many kilojoules are absorbed when 315 g of Ca(OH)2 reacts?

Ca(OH)2(s) CaO(s) + H2O(l) ΔH = +65.3 kJ

a. 278 kJ b. 357 kJ

c. 65.3 kJ

d. 2.06 x 104 kJ

e. 76.1 kJ

67. How many kilojoules are released when 125 g of Cl2 reacts with silicon? Si(s) + 2Cl2(g) SiCl4(g) ΔH = -657 kJ

a. 986 kJ

b. 321 kJ

c. 657 kJ d. 1158 kJ

e. 579 kJ

Challenge Problems

68. The equation for the reaction of iron and oxygen gas to form rust (Fe2O3) is written as (7.5, 7.6, 7.7, 7.9):

4Fe(s) + 3O2(g) 2Fe2O3(s) Δ = -1.7 x 103 kJ

(i) How many kilojoules are released when 2.00 g of Fe reacts? (ii) How many grams of rust form when 150 kJ are released?

a. (i) 60.9 kJ (ii) 14.1 g b. (i) 15.2 kJ (ii) 14.1 g

c. (i) 60.9 kJ (ii) 28.2 g

d. (i) 15.2 kJ (ii) 28.2 g e. (i) 0.263 kJ (ii) 28.2 g

69. When hydrogen peroxide (H2O2) is used in rocket fuels, it produces water, oxygen, and heat. (7.6, 7.7. 7.9):

2H2O2(l) 2H2O(l) + O2(g) ΔH = -196 kJ (i) Is the reaction exothermic or endothermic?

(ii) How many kilojoules are released when 2.50 moles of H2O2 reacts?

(iii) How many kilojoules are released wen 275 g of O2 is produced?

a. (i) exothermic (ii) 490 kJ (iii) 1684 kJ

b. (i) exothermic (ii) 245 kJ (iii) 1684 kJ

c. (i) endothermic (ii) 245 kJ (iii) 3369 kJ d. (i) endothermic (ii) 490 kJ (iii) 3369 kJ

e. (i) exothermic (ii) 490 kJ (iii) 3369 kJ

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