EXAM 4 REVIEW CHEM 110 Last Name Professor BEAMER 4 Review Practice Exam... · EXAM 4 REVIEW CHEM 110 Page 1 of 13 Last Name Professor BEAMER First Name Which lab section are you

EXAM 4 REVIEW CHEM 110

Page 1 of 13

Last Name Professor BEAMER

First Name Which lab section are you in?

M W R (Circle One)

Note: As Part 1 is supposed to be memorized material, you can cross-check your answers

with the information in your textbook and/or from your notes.

PART 2: MATH

15a) Calculate the molar mass of Na2Cr2O7. Rewrite final answer 261.98 g/mol Na2Cr2O7

(Time Range: 3 to 5 min)

g/mol g/mol

2 × Na 2 × 22.99 45.98

2 × Cr 2 × 52.00 104.00

7 × O 7 × 16.00 + 112.00

261.98 g/mol Na2Cr2O7

15b) Rewrite the molar mass of the substance above in conversion factor form:

261.98 g Na2Cr2O7 = 1 mol Na2Cr2O7

(Time Range: 1 min − all levels)


Page 2 of 13

16) Calculate the number of molecules of NH3 in a 1.400 × 10−5 mole sample of NH3.


Solution/Explanation:

You are converting between particles (molecules) and moles (mol). Therefore, you need to use Avogadro’s Number.

You need to devise a conversion factor: 1 mol NH3 = 6.022 × 1023 NH3 molec

Always start with your initial value: 1.400 × 10−5 NH3 molec

1 mol NH3 = 6.022 × 1023 NH3 molec

initial value Avogadro’s Number Conversion

(

1.400 × 10-5 NH3 mol

1) (

6.022 × 1023 NH3 molec

1 mol NH3) = 8.431 × 1018 NH3 molec

Identify initial value!


Page 3 of 13

17) A sample of CaCl2 contains 3.8 × 1025 formula units of CaCl2. Calculate the mass (in grams) of CaCl2 in the sample. The molar mass of CaCl2 is 110.98 g/mol.



You are converting between particles (molecules) and mass (grams). Therefore, you need to use Avogadro’s Number.

There is no direct pathway from particles to moles: mass moles particles

You must go through moles first. You need two conversion factors:

110.98 g CaCl2 = 1 mol CaCl2

6.022 × 1023 CaCl2 form = 1 mol CaCl2

initial value Avogadro’s Number Conversion

(

3.8 × 1025 CaCl2 form

1) (

1 mol CaCl2

6.022 × 1023 CaCl2 form ) = 63 mol CaCl2

molar mass

(

63 mol CaCl21

) (110.98 g CaCl21 mol CaCl2

) = 7003 g CaCl2 x7.0 × 103 g CaCl2x

Also acceptable:

initial value Avogadro’s Number Conversion molar mass

(

3.8 × 1025 CaCl2 form

1) (

1 mol CaCl2

6.022 × 1023 CaCl2 form ) (

110.98 g CaCl21 mol CaCl2

) = 7003 g CaCl2 x7.0 × 103 g CaCl2x


Identify final unit: mass (grams)


Page 4 of 13

Reaction for Question 18

283.88 g/mol 63.01 g/mol 98.00 g/mol 108.01 g/mol

P4O10(s) + 12 HNO3(aq) 4 H3PO4(aq) + 6 N2O5(s)

Notice that I have highlighted the coefficients of the known (HNO3) and the unknown (N2O5)

18) Calculate the theoretical yield of dinitrogen pentoxide, N2O5, that can be prepared from

500.0 grams of nitric acid, HNO3(aq).



Steps to Theoretical Yield Calculations

Step 1: Convert the initial value (the KNOWN) to moles

Step 2: Use mole ratios to convert moles KNOWN moles UNK

Step 3: Convert moles UNK to mass UNK

STEP 1:

initial value molar mass

(500.0 g HNO3

1) (

1 mol HNO3

63.01 g HNO3 ) = 7.935 mol HNO3

* Use molar mass of the balanced equation to convert mass of HNO3 to moles of HNO3.

STEP 2:

mole ratio

(7.935 mol HNO3

1) (

6 mol N2O5

12 mol HNO3 ) = 3.968 mol N2O5

* Use the coefficients of the balanced equation to convert moles of HNO3 to moles of N2O5.

STEP 3:

molar mass

(3.968 mol N2O5

1) (

108.05 g N2O5

1 mol N2O5 ) = x428.7 g N2O5 OR 4.287 × 102 g N2O5x

* Convert moles of N2O5 to mass of N2O5 using molar mass.

* You can always put answers to calculations in standard SCINOT. (Make sure the SCINOT is

correct.)

See next page to see “all-in-one-step” calculation:



Page 5 of 13

Reaction for Question 18

283.88 g/mol 63.01 g/mol 98.00 g/mol 108.01 g/mol

P4O10(s) + 12 HNO3(aq) 4 H3PO4(aq) + 6 N2O5(s)

Notice that I have highlighted the coefficients of the known (HNO3) and the unknown (N2O5)

18) Calculate the theoretical yield of dinitrogen pentoxide, N2O5, that can be prepared from 500.0 grams of nitric acid, HNO3(aq).



Steps to Theoretical Yield Calculations

Step 1: Convert the initial value (the KNOWN) to moles

Step 2: Use mole ratios to convert moles KNOWN moles UNK

Step 3: Convert moles UNK to mass UNK

INITIAL VALUE STEP 1 STEP 2 STEP 3

initial value molar mass mole ratio molar mass

(500.0 g HNO3

1) (

1 mol HNO3

63.01 g HNO3 ) (

6 mol N2O5

12 mol HNO3 ) (

108.05 g N2O5

1 mol N2O5 ) = x428.7 g N2O5 OR 4.287 × 102 g N2O5x

* Again, I have no preference as to whether you break down this solution into three steps or choose the “all-in-one-step” method.



Page 6 of 13

19) Assume that a scientist perform this experiment and obtains a yield of 411.7 grams of N2O5. Calculate the %-yield value for this

experiment. Use the value obtained in Question 18 for the theoretical yield. Note: You must rewrite the %-yield equation first with variables only as your first step.


Solution

%-yield = experimental yield

theoretical yield × 100 =

experimental yield

theoretical yield × 100

= 411.7 g N2O5

428.7 g N2O5 × 100

= 96.03 % yield N2O5


Page 7 of 13

PART 3: NON-CALCULATOR

Resources:

You will be given a periodic table (symbols only)

You will be given the memorized table of polyatomic ions chart

You will be given the Activity Series Table (Top Dog)

You will be given the Solubility Rules (all of them)

No calculators for this section

Timing: 10 min (mastery) 25 min (competence)

20) Indicate whether the following elements are solid (s), liquid (l), or gas (g).

20a) Cu (#29) (s) 20b) Hg (#80) (l)

(Time Range: 1 min maximum – all levels)

21) List the charges of the following elements when they become ions:

21a) nitrogen (#7) −3 or 3− 21c) potassium (#19) +1 or 1+

21b) sulfur (#16) −2 or 2− 21d) iodine (#53) −1 or 1−

(Time Range: 1 min 30 s maximum – all levels)

IMPORTANT: For positive ions, you must write the “+” symbol!

Question 21a has been corrected. It should be −3 (not +3). Sorry.


Page 8 of 13

22) Circle the metals that require a Roman Numeral in their names when they are present in ionic compounds.

Na (#11) Fe (#26) Sn (#50)

Al (#13) Zn (#30) Ba (#56)


Na Group 1A (alkali metals) always have a charge of 1+. Therefore, Roman numerals are not

required when naming salts that contain Group 1A metals.

Fe Transition metal. With three exceptions, transitions metals always require a Roman

numeral in their names to indicate their charge.

Sn Post-Transition metal. All post-transition metals always require a Roman numeral in

their names to indicate their charge.

Al Group 3A. This is a little tricky. Even though there are transition metals in Group 3A,

aluminum itself is not a transition metal. Therefore, it does not require a Roman numeral

to indicate its charge.

Zn Zinc is one of the three exceptions of the transition metals. Zinc always has a charge

2+ in ionic compounds. Therefore, a Roman numeral is not required. Here are the three

exceptions again:

Ag (silver) is always 1+ When present in ionic compounds,

these three metals do not require a

Roman numeral when naming them. Cd (cadmium) is always 2+ in ionic compounds

Zn (zinc) is always 2+ in ionic compounds

Ba Group 2A (alkali earth metals) always have a charge of 2+. Therefore, Roman numerals

are not required when naming salts that contain Group 2A metals.


Page 9 of 13

23) Circle the compounds that are soluble in water.

potassium sulfate V2O5 ammonium phosphate

CuCl2 chromium(III)

hydroxide Au2(Cr2O7)3


potassium sulfate

K2SO4

potassium K+ Rule 1 = soluble

Rule 1 beats Rule 6

sulfate SO4 2−

Rule 6 = soluble

V2O5

vanadium(V) oxide

vanadium(V) V5+ Rule 7 = insoluble Rule 5 beats Rule 7

oxide O 2− Rule 5 = insoluble

ammonium phosphate

(NH4)3PO4

ammonium NH4 + Rule 1 = soluble

Rule 1 beats Rule 5

phosphate PO4 3− Rule 5 = insoluble

CuCl2

copper(II) chloride

copper(II) Cu2+ Rule 7 = insoluble Rule 4 beats Rule 7

chloride Cl − Rule 4 = soluble

chromium(III) hydroxide

Cr(OH)3

chromium(III) Cu2+ Rule 7 = insoluble Rule 5 beats Rule 7

hydroxide OH− Rule 5 = insoluble

Au2(Cr2O7)3 Gold(III) dichromate

gold(III) Au3+ Rule 7 = insoluble Rule 7 (tie)

dichromate Cr2O7 2− Rule 7 = insoluble


Page 10 of 13

24) Balance the following two equations:

24a) 2 C4H10(l) + 9 O2(g) ∆→ 8 CO(g) + 10 H2O(g)

Note: For Question 24a, I purposely did not tell you how much oxygen was present. It does not matter, since the products are

already written. Sorry about the wrong answer.

24b) 3 Ca(OH)2(s) + 2 H3PO4(aq) 1 Ca3(PO4)2(s) + 6 HOH(g)

Hint: Rewrite water as _____ to help you balance.

Time Limit: 4 minutes maximum – all levels.

25) Write the skeleton equation: (Don’t forget states: s, l, g, aq)

phosphorus(s) + chlorine(g) phosphorus pentachloride(s)

P(s) + Cl2(g) PCl5(s)

Time Limit: 2 minutes to 4 minutes

26) Write the word equation: (Don’t forget states: s, l, g, aq)

Cl2O7(s) + H2O(l) HClO4(aq)

dichlorine heptoxide(s) + water(l) perchloric acid(aq)

Time Limit: 2 minutes to 4 minutes


Page 11 of 13

27) Predict the products (chemical formulas):

Do not balance equations.

Include states: (s), (l), (g), or (aq)

Hint: Name the products first. Write the names in the margins. Then write the chemical formulas from the names.

27a) FeBr3(aq) + Cl2(g) FeCl3(aq) + Br2(l)

Time Limit: 2 minutes maximum – all levels

Solution

The reactants are: ionic compound + an element. Therefore, this is a Single Replacement (SR)

Before anything else, name the reactants: iron(III) bromide + chlorine(g)

The element (chlorine gas) is a nonmetal. Therefore, it will try to replace bromide.

Check the Activity Series Table (Appendix 8): chlorine beats bromine. Therefore, chlorine and bromide switch places.

Before anything else: name the products:

iron(III) bromide + chlorine iron(III) chloride + bromine Note that the suffixes –ine and –ide stay put.

FeBr3(aq) Cl2(g) FeCl3(aq) + Br2(l)

Write the chemical formulas from the names.

iron(III) chloride is a salt. Therefore, you must determine if water is hanging around, and if iron(III) chloride is soluble.

The “aq” in the reactants tells you that water is present. Therefore, you must determine the solubility of iron(III) chloride.

Solubility Table, Rule 4, chlorides are soluble. Therefore, iron(III) chloride is aqueous.

Don’t forget states

iron(III) chloride

The “aq” in the reactants tells you that water is present. Therefore, you must determine the solubility of iron(III) chloride.

Solubility Table, Rule 4, chlorides are soluble. Therefore, iron(III) chloride is aqueous.

bromine

The states of all elements should be memorized. Elemental bromine, Br2, is a liquid.


Page 12 of 13

27b) XS

C5H8(l) + O2(g) ∆→ CO2(g) + H2O(g)

Time Limit: 1 minute maximum – all levels

Solution

C5H8 is a hydrocarbon (contains carbon and hydrogen).

The other reactant is oxygen, and the reaction is being heated.

Therefore, this is a HYDROCARBON COMBUSTION reaction.

Since there is excess oxygen, the products are carbon dioxide gas and water vapor (memorized). There is no way to “figure this out.” You

either have learned/memorized this, or not.

27c) K(s) + Br2(l) KBr(s)

Time Limit: 1 minute maximum – all levels

Solution

This is a simple COMBINATION reaction. Metal + nonmetal salt

Name the salt: potassium bromide (common sense)

Write the chemical formula from the name.

States

If a product is a salt (ionic compound), the salt is a solid unless water is hanging around.

There are no “aq” in the reactants, water is NOT a reactant, and water is NOT present over the reaction arrow H2O →

Water is not present, so you do not need to check for solubility. This product is a solid.

Do not worry that the equation is not balanced. I asked for the unbalanced equation.


Page 13 of 13

28) The following reaction is NR. In three or less sentences, explain why no reaction occurs.

Time Limit: 3 minutes maximum – all levels

(NH4)3PO4(aq) + K2SO4(aq) (NH4)2SO4(aq) + K3PO4(aq)

This is a DOUBLE REPLACEMENT (DR) reaction. In DR reactions, if all substances (reactants and products) are soluble

(aqueous), then there is no reaction. All reactants and products in the reaction above are aqueous, so NR.

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EXAM 4 REVIEW CHEM 110 Last Name Professor BEAMER 4 Review Practice Exam... · EXAM 4 REVIEW CHEM 110 Page 1 of 13 Last Name Professor BEAMER First Name Which lab section are you