Chem I Homework

Chem I Homework Page

| Matter, Conversion Factors and Dimensional Analysis | Atoms, Molecules, and Ions | Stoichiometry | Aqueous Reactions | Nuclear Physics | Optics |

Homework Page with Visable Answers

About This Page

This page has all of the required homework for the first semester of General Chemistry. You can go directly to each section using the links above. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12^th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition will do for this course.
These are bare-bones questions. You are expected to go to the end of chapter problems in your textbook and find similar questions.
The questions in the textbook will have additional information that may be useful and that connects the questions to real life applications, many of them in biology. You should also study the Exercises within the chapters as well. The exercises are worked out examples of the questions at the back of the chapter.

Matter, Conversion Factors and Dimensional Analysis (Chapter One)

Label each of the following as either a chemical or a physical process:
- Burning paper
- A nail that is rusting.
- Dissolving sugar in water.
- Acid rain reacting with limestone headstones.
- Dissolving table salt in water.
Convert to or from exponential notation as needed.
- 23,465
- 0.005316
- 6.74 x 10⁵
- 4.54 x 10^-4
- 78.632 x 10⁴
Answer the following density questions.
- Under normal conditions aluminum has a density of 2.70 g/ml. What is the volume of 1.85 g of Aluminum?
- One way to liquify carbon dioxide is to put it under a lot of pressure. At a certain pressure the density of the liquid is 0.466 g/cm³. What is the mass of a 15 mL sample of this liquid CO₂?
- A student was given the task of determining if the pure metal she had was nickel (d = 8.90 g/mL) or zinc (d = 7.14 g/mL). She used a balance to determine a mass of 9.7 g. She then used a displacement technique to determine a volume of 1.36 mL. What would she report?
A reported value with its uncertainty (+/- or ±) should encompass the correct value for the measurement. A group of students reported a value of 2.64 ± 0.1 g/mL for the density of aluminum (known value is 2.70 g/mL). What is the range of their reported value and does it encompass the known value for the density of aluminum?
Answer
The range goes from 2.54 g/mL to 2.74 g/mL. The known value is between these two numbers.

Use units and dimensional analysis to do the following problems.

The periodic table provides conversion factors between grams and moles. We call these numbers the atomic weight or the molar mass. Atomic nitrogen has a molar mass of 14 g/mole. How many grams are in 1.5 moles of atomic nitrogen?

Answer

(

1.5 mole N

)

(

14 g N

1 mole N

)

= 21 g N

How many grams are in 1.5 moles of gaseous nitrogen (N₂)? See the last question for data.

Answer

(

1.5 mole N₂

)

(

2 mole N

1 mole N₂

)

(

14 g N

1 mole N

)

= 42 g N

How many liters (L) are there in 500 mL?

Answer

(

500 mL

)

(

1 L

1000 mL

)

= 0.5 L

How many centimeters (cm) are there in 1.2 m?

Answer

(

1.2 m

)

(

100 cm

1 m

)

= 120 cm

Atoms, Molecules, and Ions (Chapter Two)

What is the difference between an atom and an element?
Answer
An atom is the smallest piece of an element that retains the properties of that element.
Define a molecule two different ways.
Answer
A molecule is two or more atoms that are hooked together, are stuck together, and act as one unit. A molecule is the smallest piece of a substance that retains the properties of that substance.
Do ionic compounds have molecules? Explain.
Answer
Ionic compounds do not have molecules. There is no single unit that retains the properties of a compound.
Use the periodic table to predict the charge of an ion created from each of the following elements:
- K
- Ca
- Br
- Ba
- F
Define the word isotope.
Answer
Two atoms are isotopes if they have the same number of protons, but a different number of neutrons. Two atoms of the same element that have a different number of neutrons.
Silver has two naturally occurring isotopes: ¹⁰⁷Ag (mass = 106.905095 g/mole) has an abundance of 51.83% and ¹⁰⁹Ag (mass = 108.904754 g/mole) has an abundance of 48.17. Calculate the molar mass that is reported on the periodic table.
Answer
0.5183(106.905095) + 0.4817(108.904754) = 107.868 g/mole
Lithium has two naturally occurring isotopes: ⁶Li (mass = 6.015123 g/mole) has an abundance of 7.5% and ⁷Li has an abundance of 92.5%. Calculate the molar mass of ⁷Li. The periodic table reports 6.941 g/mole for the molar mass of lithium.
Answer
0.075(6.015123) + 0.925(x) = 6.941
x = [6.941 - (0.075)(6.015123)]/0.925 = 7.016 g/mole
Boron has two naturally occurring isotopes: ¹⁰B (mass = 10.012938 g/mole) and ¹¹B (mass = 11.009305 g/mole). The atomic weight on the periodic table for boron is 10.811 g/mole. Calculate the percent abundance for each isotope.
Answer
x(10.012938) + (1 - x)(11.009305) = 10.811
x = [10.811 - 11.009305]/[10.012938 - 11.009305] = 0.199
¹⁰B is 19.9% and ¹¹B is 80.1%
Identify the following elements and the number of neutrons in the nucleus of the isotope:
- ³⁷₁₇X
- ²⁵₁₂X
- ⁷⁰₃₁X
- ²⁴³₉₅X
How many protons, neutrons, and electrons are in the following isotopes:
- ³⁵Cl
- ⁸⁵Sr²⁺
- ¹³¹I^-
- ¹²⁹Cs
Fill in the following table:

Symbol ⁵⁶Fe³⁺

Protons 29 55

Neutrons 34 78 7

Electrons 28 10

Net Charge 0 2-

Answer

Symbol ⁵⁶Fe³⁺ ⁶³Cu⁺ ¹³³Cs ¹⁵O^2-

Protons 26 29 55 8

Neutrons 30 34 78 7

Electrons 23 28 55 10

Net Charge 3+ 1+ 0 2-

Predict the compounds formed by combining the cations across the top with the anions down the side.

Ion	Zn²⁺	Na⁺	Ba²⁺	Al³⁺
SO₄^2-
NO₃^-
CO₃^2-
OH^-
Cl^-
PO₄^3-

Answer

Ion	Zn²⁺	Na⁺	Ba²⁺	Al³⁺
SO₄^2-	ZnSO₄	Na₂SO₄	BaSO₄	Al₂(SO₄)₃
NO₃^-	Zn(NO₃)₂	NaNO₃	Ba(NO₃)₂	Al(NO₃)₃
CO₃^2-	ZnCO₃	Na₂CO₃	BaCO₃	Al₂(CO₃)₃
OH^-	Zn(OH)₂	NaOH	Ba(OH)₂	Al(OH)₃
Cl^-	ZnCl₂	NaCl	BaCl₂	AlCl₃
PO₄^3-	Zn₃(PO₄)₂	Na₃PO₄	Ba₃(PO₄)₂	AlPO₄

Give the name or formula for each of the following compounds.
- CuCl
- HgCO₃
- FeBr₃
- Chromium(III)acetate
- Lead(II)hydroxide
- Manganese(II)nitrate
Give the name or formula for each of the following molecules.
- Cl₂O
- N₂O₄
- NF₃
- chlorine pentaflouride
- Dinitrogen oxide
- Phosphorous trichloride

Stoichiometry, Etc. (Chapter Three)

Balance the following chemical equations.
- CH₄ + O₂ → H₂O + CO₂
- HgCO₃ + HCl → H₂O + CO₂ + HgCl₂
- FeBr₃ + AgNO₃ → Fe(NO₃)₃ + AgBr
- H₂ + O₂ → H₂O
- K + N₂ → K₃N
- Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
Caculate the formula weights (also called molecular weight or molar mass) of the following substances. See Chapter Three problems for interesting examples.
- Ammonium hydroxide
- HgCO₃
- H₂SO₄
- Copper(II)chloride
Caculate the percentage by mass of the indicated element in the following substances.
- Oxygen in ammonium hydroxide
- Hg in HgCO₃
- N in Fe(NO₃)₃
- Chlorine in copper(II)chloride
What is a mole? Where are the conversion factors between grams and moles found?
Answer
1 mole = 6.02 x 10²³ objects. The atomic mass on the periodic table is the number of grams associated with one mole of that element.

Calculate the following:

How many grams are in 1.5 moles of atomic nitrogen?

Answer

(

1.5 mole N

)

(

14 g N

1 mole N

)

= 21 g N

How many grams of nitrogen are in 1.5 moles of gaseous nitrogen (N₂)?

Answer

(

1.5 mole N₂

)

(

2 mole N

1 mole N₂

)

(

14 g N

1 mole N

)

= 42 g N

How many moles of Br are in 79.9 g of Br₂?

Answer

(

79.9 g Br₂

)

(

2 mole Br

1 mole Br₂

)

(

79.9 g Br

1 mole Br

)

= 0.5 mole Br

How many moles of O₂ are there in 32 g O₂?

Answer

(

32 g O₂

)

(

1 mole O₂

32 g O₂

)

= 1 mole O₂

Calculate the empirical formulas.

What is the empirical formula for acetone, a common solvent, if it is 62.07% carbon, 27.59% oxygen, and 10.34% hydrogen?

Answer

(

62.07 g C

)

(

1 mole C

12.01 g C

)

= 5.17 mole C

(

27.59 g O

)

(

1 mole O

16 g O

)

= 1.72 mole O

(

10.34 g H

)

(

1 mole H

1 g H

)

= 10.34 mole H

(

5.17 mole C

1.72 mole O

)

= 3

(

1.72 mole O

)

= 1

(

10.34 mole H

1.72 mole O

)

= 6

⇒ C, 3; O, 1; H, 6 ⇒ C₃H₆O

What is the empirical formula for phenol if a sample analysis yielded 0.32 g hydrogen, 0.85 g oxygen, and 3.83 g carbon?

Answer

(

0.32 g H

)

(

1 mole H

1 g H

)

= 0.32 mole H

(

0.85 g O

)

(

1 mole O

16 g O

)

= 0.053 mole O

(

3.83 g C

)

(

1 mole C

12 g C

)

= 0.319 mole C

(

0.320 mole H

0.053 mole O

)

= 6

(

0.053 mole O

)

= 1

(

0.319 mole C

0.053 mole O

)

= 6

⇒ C, 6; H, 6; O, 1 ⇒ C₆H₆O

What is the empirical formula of an aluminum oxide sample if it is 52.92% aluminum and 47.08% oxygen?

Answer

(

52.92 g Al

)

(

1 mole Al

26.98 g Al

)

= 1.96 mole Al

(

47.08 g O

)

(

1 mole O

16 g O

)

= 2.94 mole O

(

1.96 mole Al

)

= 1

(

2.94 mole O

1.96 mole Al

)

= 1.5

⇒ Al, 1; O, 1.5 ⇒ Al, 2; O, 3 ⇒ Al₂O₃

A common organic solvent only contains carbon, hydrogen, and oxygen. Combustion analysis of 1.000 g of this solvent gives 1.913 g CO₂ and 1.174 g H₂O. What is the empirical formula?

Answer

(

1.913 g CO₂

)

(

1 mole CO₂

44 g CO₂

)

(

1 mole C

1 mole CO₂

)

= 0.0435 mole C

(

1.174 g H₂O

)

(

1 mole H₂O

18 g H₂O

)

(

2 mole H

1 mole H₂O

)

= 0.130 mole H

(

0.0435 mole C

)

(

12 g C

1 mole C

)

= 0.522 g C

(

0.130 mole H

)

(

1 g H

1 mole H

)

= 0.13 g H

1.000 g sample - 0.522 g C - 0.130 g H = 0.348 g O

(

0.348 g O

)

(

1 mole O

16 g O

)

= 0.0218 mole O

(

0.0435 mole C

0.0218 mole O

)

= 2

(

0.130 mole H

0.0218 mole O

)

= 6

(

0.218 mole O

0.0218 mole O

)

= 1

⇒ C, 2; H, 6; O, 1 ⇒ C₂H₆O (ethanol or ethyl alcohol)

In combustion reactions carbon-containing substances react with oxygen to form carbon dioxide and water. Burning ethanol is an example:

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

How many moles of O₂ are needed to react with 1.5 moles of C₂H₅OH?

Answer

(

1.5 mole C₂H₅OH

)

(

3 mole O₂

1 mole C₂H₅OH

)

= 4.5 mole O₂

How many moles of CO₂ are produced when when 1.5 moles of C₂H₅OH are burned?

Answer

(

1.5 mole C₂H₅OH

)

(

2 mole CO₂

1 mole C₂H₅OH

)

= 3 mole CO₂

How many moles of oxygen are needed to produce 3.3 moles of carbon dioxide?

Answer

(

3.3 mole CO₂

)

(

2 mole O₂

3 mole CO₂

)

= 2.2 mole O₂

How many grams of C₂H₅OH are needed to produce 36 g H₂O?

Answer

(

36 g H₂O

)

(

1 mole H₂O

18 g H₂O

)

= 2 mole H₂O

(

2 mole H₂O

)

(

1 mole C₂H₅OH

3 mole H₂O

)

(

46 g C₂H₅OH

1 mole C₂H₅OH

)

= 30.67 g C₂H₅OH

How many grams of CO₂ are produced when 15 g O₂ are reacted?

Answer

(

15 g O₂

)

(

1 mole O₂

32 g O₂

)

(

2 mole CO₂

3 mole O₂

)

(

44 g CO₂

1 mole CO₂

)

= 13.75 g CO₂

How many mole of CO₂ are produced when 4 moles of O₂ are reacted with 1.5 moles of C₂H₅OH?

Answer

Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen O₂ below.

(

4 mole O₂

)

(

1 mole C₂H₅OH

3 mole O₂

)

= 1.33 mole C₂H₅OH needed

We need 1.33 moles, but we have 1.5 moles according to the problem statement. We have extra C₂H₅OH which means that O₂ is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

(

4 mole O₂

)

(

2 mole CO₂

3 mole O₂

)

= 2.67 mole CO₂

How many grams of H₂O are produced when 256 g of O₂ are reacted with 138 g of C₂H₅OH?

Answer

Determine the moles of each reactant that we are given:

(

256 g O₂

)

(

1 mole O₂

32 g O₂

)

= 8 mole O₂

(

138 g C₂H₅OH

)

(

1 mole C₂H₅OH

46 g C₂H₅OH

)

= 3 mole C₂H₅OH

Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen C₂H₅OH below. Get the mole ratios from the balanced equation.

(

3 mole C₂H₅OH

)

(

3 mole O₂

1 mole C₂H₅OH

)

= 9 mole O₂ needed

We need 9 mole O₂, but have 8 moles. We don't have enough O₂ which means that O₂ is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

(

8 mole O₂

)

(

3 mole H₂O

3 mole O₂

)

(

18 g H₂O

1 mole H₂O

)

= 144 g H₂O

How many grams of Fe₂O₃ are produced when 56 g of Fe are reacted with 32 g of O₂? How many grams of the excess reactant will remain after the reaction is complete?

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

Answer

Determine the moles of each of the reactants that we are given.

(

32 g O₂

)

(

1 mole O₂

32 g O₂

)

= 1 mole O₂

(

56 g Fe

)

(

1 mole Fe

55.845 g Fe

)

= 1 mole Fe

Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen Fe below. Get the mole ratios from the balanced chemical equation.

(

1 mole Fe

)

(

3 mole O₂

4 mole Fe

)

= 0.752 mole O₂ needed

We need 0.752 mole O₂, but have 1 mole. We have extra O₂ which means that Fe is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

(

56 g Fe

)

(

1 mole Fe

55.845 g Fe

)

(

2 mole Fe₂O₃

4 mole Fe

)

(

159.69 g Fe₂O₃

1 mole Fe₂O₃

)

= 80.07 g Fe₂O₃

We have 1 mole - 0.752 mole = 0.248 moles of extra O₂

(

0.248 mole O₂

)

(

32 g O₂

1 mole O₂

)

= 7.94 g O₂ left over

Aqueous Reactions (Chapter Four)

What ions will be present when each of the following substances is placed in water?
- BaCl₂
- ZnSO₄
- HNO₃
- Na₂CO₃
- NaOH
What will be dissolved in the solution when each of the following weak acids are placed in water?
- Hypochlorous acid, HClO
- Propionic acid, C₂H₅COOH or HC₃H₅O₂
Write the molecular, ionic and net ionic equations when the following substances are placed together in an aqueous solution.
- CaBr₂ and K₂CO₃
- CdSO₄ and BaC₂H₃O₂
- HNO₃ and PbCO₃
- (NH₄)₃PO₄ and CuCl₂
- MgBr₂ and CaSO₄
A solution may contain any or all of the following ions: Ba²⁺, Ni²⁺, or Pb²⁺. From the following information, which ions are present? Show reasoning.
When HCl is added to the solution a precipitate is formed. The precipitate is filtered from the solution. When K₂SO₄ is added to the filtered solution no precipitate is formed. When NaOH is then added to the filtered solution a precipitate is formed.

Answer
A reaction table showing what precipitates are possible would look like the following:

Ba²⁺ Ni²⁺ Pb²⁺

HCl BaCl₂(aq) NiCl₂(aq) PbCl₂(s)

K₂SO₄ BaSO₄(s) NiSO₄(aq) --

NaOH -- Ni(OH)₂(s) --

A precipitate with HCl means that Pb²⁺ must be present. No precipitate with K₂SO₄ means that Ba²⁺ can not be present. A precipitate with NaOH means that Ni²⁺ must be present.
Answer: Pb²⁺ and Ni²⁺ are in the solution.
A solution may contain any or all of the following ions: Sr²⁺, Fe²⁺, or Ca²⁺. From the following information, which ions are present? Show reasoning.
When KOH is added to the solution a precipitate is formed. The precipitate is filtered from the solution. When Na₂SO₄ is added to the filtered solution a precipitate is again formed. This new precipitate is then filtered from the solution. When (NH₄)₂CO₃ is then added to the filtered solution a precipitate is not formed.

Answer
A reaction table showing what precipitates are possible would look like the following:

Sr²⁺ Fe²⁺ Ca²⁺

KOH Sr(OH)₂(aq) Fe(OH)₂(s) Ca(OH)₂(aq)

Na₂SO₄ SrSO₄(s) -- CaSO₄(aq)

(NH₄)₂CO₃ -- -- CaCO₃(s)

A precipitate with KOH means that Fe²⁺ must be present. A precipitate with Na₂SO₄ means that Sr²⁺ is present. No precipitate with NaOH means that Ca²⁺ is not present.
Answer: Sr²⁺ and Fe²⁺ are in the solution.
Answer the following oxidation/reduction related questions:
- What element has the strongest attraction for electrons within a bond?
- What is the attraction for electrons within a bond called?
- Can oxidation occur without reduction?
- What do we know about the number of electrons lost compared to the number of electrons gained?
- What is the trend for electronegativity across and down the periodic table?
- True or False: The assignment of oxidation numbers assumes that the most electronegative element will get the electrons, or in other words that the bonds will be like ionic bonds.
- What is the oxidation number of the alkali metals?
- What is the oxidation number of flourine?
- What is the oxidation number of a single element ion?
- What is the oxidation number of elements in their standard state?
- What is the oxidation number rule for polyatomic ions?
- What is the oxidation number rule for molecules that don't have any charge?
- What is the normal oxidation number of oxygen?
What is the oxidation number of each element in each of the following substances?
- HClO
- C₂H₅OH
- HSO₄^-
- PO₄^3-
- Cl₂(g)
- COCl₂
- KMnO₄
- Al(NO₃)₃
For each of the following reactions: (a) Use the oxidation number method to balance the reaction (show work, i.e., oxidation numbers, electrons transferred, etc.). (b) Identify the element that is reduced and the element that is oxidized. (c) Identify the oxidizing agent and the reducing agent.
- Cd(s) + Ag⁺(aq) → Cd²⁺(aq) + Ag(s)
- Fe(NO₃)₂(aq) + 2Al(s) → Fe(s) + Al(NO₃)₃(aq)
- Cr²⁺(aq) + MnO₄^-(aq) + H⁺(aq) → Cr³⁺(aq) + Mn²⁺(aq) + H₂O(l)
- P₄(s) + HClO(aq) + H₂O → H₃PO₄(aq) + HCl(aq)
- C₂H₅OH(l) + O₂(g) → H₂O(l) + CO₂(g)
Use the half reaction method to complete and balance the following reactions.
- SO₂ + NO₃^- → SO₄^2- + NO (acidic solution)
- ClO₃^- + Br^- → Cl₂ + Br₂ (acidic solution)
- Cr²⁺ + MnO₄^- → Cr³⁺ + Mn²⁺ (acidic solution)
- Sn(OH)₄^2- + BrO^- → SnO₂ + Br^- (basic solution)
- S^- + NO₃^- → S + NO (basic solution)
- I^- + OCl^- → IO₃^- + Cl^-
- NO₂^- + Ga → NH₄⁺ + GaO₂^-

Answer each of these molarity problems.

What is the definition of molarity?

Answer

molarity is defined as moles of solute divided by liters of solution.

What is the difference between 0.4 mol and 0.4 M?

Answer

0.4 mol is the number of moles while 0.4 M is the number of moles per liter of solution.

What is the molarity when 1.2 moles of NaCl is placed in a solution where the total volume is 2.4 L?

Answer

molarity =

1.2 mole NaCl

2.4 L soln

0.5 mole NaCl

1 L soln

= 0.5 M

What is the molarity when 5.0 g of KMnO₄ are placed in water to make a total solution having a volume of 500 mL?

Answer

(

5 g KMnO₄

)

(

1 mole KMnO₄

158.04 g KMnO₄

)

= 0.0316 mole KMnO₄

molarity =

0.0316 mole KMnO₄

0.5 L soln

0.0633 mole KMnO₄

1 L soln

= 0.0633 M KMnO₄

How many moles of KCl are in 250 mL of a 1.2 M KCl solution?

Answer

(

0.25 L soln

)

(

1.2 mole KCl

1 L soln

)

= 0.3 moles KCl

What volume of 2.2 M LiBr is needed to have 1.1 mole of LiBr?

Answer

(

1.1 mole LiBr

)

(

1 L soln

2.2 mole LiBr

)

= 0.5 L soln

How many grams of NaOH are needed to make 500 mL of a 1.5 M NaOH solution?

Answer

(

0.5 L soln

)

(

1.5 mole NaOH

1 L soln

)

(

39.99 g NaOH

1 mole NaOH

)

= 29.99 g NaOH

What concentration should a stock solution be in order to use 10 mL of it to prepare 250 mL of a 0.2 M solution?

Answer

(

0.250 L soln

)

(

0.2 mole solute

1 L soln

)

= 0.05 mole solute

0.05 mole solute

0.010 L soln

= 5 M

How much of a 1.5 M stock solution should you use to prepare 100 mL of a 0.5 M solution?

Answer

(

0.1 L soln

)

(

0.5 mole solute

1 L soln

)

(

1 L stock soln

1.5 mole solute

)

= 0.033 L stock soln

0.033 L or 33 mL of the stock soln are needed

A condition called hyponatremia occurs when the sodium concentration in the blood falls below 0.130 M. A typical adult has 4.7 L of blood. (a) How many grams of sodium are in the blood of a typical adult? (b) How much sodium would a person need to get into their blood if the sodium concentration was 0.12 M?

Answer

(

4.7 L blood

)

(

0.130 mole Na

1 L blood

)

(

22.99 g Na

1 mole Na

)

= 14.05 g Na (a)

(

4.7 L blood

)

(

0.120 mole Na

1 L blood

)

(

22.99 g Na

1 mole Na

)

= 12.97 g Na

Amount needed = 14.05 g - 12.97 g = 1.08 g Na needed (b)

Many places have laws that say a person is legally drunk when they have a concentration of more than 0.013 M in ethanol, C₂H₅OH. A typical adult has 4.7 L of blood. (a) How many grams of ethanol are in the blood of a typical adult when legally drunk? (b) A can of beer typically has about 20 g of ethanol. How many cans of beer will provide the legal limit?

Answer

(

4.7 L blood

)

(

0.013 mole C₂H₅OH

1 L blood

)

(

62 g C₂H₅OH

1 mole C₂H₅OH

)

= 3.79 g C₂H₅OH (a)

(

3.79 g C₂H₅OH

)

(

1 can

20 g C₂H₅OH

)

= 0.19 can Only one fifth of a can! (b)

Assume a silver acetate solution that is 1.4 M AgC₃H₃O₂ when doing the following problems.

Write the balanced chemical equation for the reaction of silver acetate with sodium chloride in an aqueous solution.

Answer

AgC₃H₃O₂(aq) + NaCl(aq) → AgCl(s) + NaC₃H₃O₂(aq)

How many grams of NaCl are needed to completely react with 200 mL of the AgC₃H₃O₂ solution?

Answer

(

0.2 L AgC₃H₃O₂ solution

)

(

1.4 mole AgC₃H₃O₂

1 L AgC₃H₃O₂ solution

)

= 0.28 mole AgC₃H₃O₂

(

0.28 mole AgC₃H₃O₂

)

(

1 mole NaCl

1 mole AgC₃H₃O₂

)

(

58.5 g NaCl

1 mole NaCl

)

= 16.4 g NaCl

How much silver chloride is created when 100 mL of the AgC₃H₃O₂ solution is combined with 200 mL of 0.6 M NaCl?

Answer

Determine the moles of each reactant. We have:

(

0.1 L AgC₃H₃O₂ soln

)

(

1.4 mole AgC₃H₃O₂

1 L AgC₃H₃O₂ soln

)

= 0.14 mole AgC₃H₃O₂

(

0.2 L NaCl soln

)

(

0.6 mole NaCl

1 L NaCl soln

)

= 0.12 mole NaCl

Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen NaCl below.

(

0.12 mole NaCl

)

(

1 mole AgC₃H₃O₂

1 mole NaCl

)

= 0.12 mole AgC₃H₃O₂ needed

We need 0.12 mole AgC₃H₃O₂, but have 0.14 mole. We have extra AgC₃H₃O₂ and so NaCl is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

(

0.12 mole NaCl

)

(

1 mole AgCl

1 mole NaCl

)

(

143.4 g AgCl

1 mole AgCl

)

= 17.2 g AgCl

What are the concentrations of each of the ions in the final solution when 100 mL of the AgC₃H₃O₂ solution is combined with 200 mL of 0.6 M NaCl? Use information calculated in the last problem and assume the volumes are additive (the total volume is the sum of the volumes that are added together).

Answer

From the last problem we see that the NaCl was the limiting reactant. That means that all of the Cl will be in AgCl(s) and the concentration of the Cl^- is zero: [Cl^-] = 0.

There was extra Ag⁺. The amount used was 0.12 moles, but we had 0.14 moles. That means that there was 0.02 moles extra that would remain in the solution after the reaction was complete. Those 0.02 moles would be in a total of 300 mL, so

[Ag⁺] =

0.02 mole Ag⁺

0.3 L soln

0.067 mole Ag⁺

1 L soln

= 0.067 M Ag⁺

Na⁺ doesn't make a precipitate, so all of the original amount would be in the final solution and

[Na⁺] =

0.12 mole Na⁺

0.3 L soln

0.4 mole Na⁺

1 L soln

= 0.4 M Na⁺

C₃H₃O₂^- doesn't make a precipitate, so all of the original amount would be in the final solution and

[C₃H₃O₂^-] =

0.14 mole C₃H₃O₂^-

0.3 L soln

0.467 mole C₃H₃O₂^-

1 L soln

= 0.467 M C₃H₃O₂^-

Symbol	⁵⁶Fe³⁺
Protons		29	55
Neutrons		34	78	7
Electrons		28		10
Net Charge			0	2-

Symbol	⁵⁶Fe³⁺	⁶³Cu⁺	¹³³Cs	¹⁵O^2-
Protons	26	29	55	8
Neutrons	30	34	78	7
Electrons	23	28	55	10
Net Charge	3+	1+	0	2-

	Ba²⁺	Ni²⁺	Pb²⁺
HCl	BaCl₂(aq)	NiCl₂(aq)	PbCl₂(s)
K₂SO₄	BaSO₄(s)	NiSO₄(aq)	--
NaOH	--	Ni(OH)₂(s)	--

	Sr²⁺	Fe²⁺	Ca²⁺
KOH	Sr(OH)₂(aq)	Fe(OH)₂(s)	Ca(OH)₂(aq)
Na₂SO₄	SrSO₄(s)	--	CaSO₄(aq)
(NH₄)₂CO₃	--	--	CaCO₃(s)