Chem. 1A Midterm 1 Version A Practice Test 2

Chem. 1A Midterm 1

Version A

Practice Test 2

Name__________________________________________ Student Number _________________________________

All work must be shown on the exam for partial credit. Points will be taken off for incorrect or no units. Calculators are allowed. Cell phones may not be used for calculators. On short answer problems you must show your work in order to receive credit for the problem. If your cell phone goes off during the exam you will have your exam removed from you.

Fundamentals

(of 24 possible)

Problem 1

(of 8 possible)

Problem 2

(of 16 possible)

Problem 3

(of 16 possible)

Problem 4

(of 12 possible)

Multiple Choice

(of 24 possible)

Midterm Total

2

Fundamental Questions

Each of these fundamental chemistry questions is worth 3 points. You must show work to get credit, if applicable. Little to no partial credit will be rewarded. Make sure to report answers in the correct number of significant figures and indicate the proper units. 1) 3 pts _{How many kilograms are in 3251 mg?}

kg g

kg mg

g

mg 0.003251

1000 1 1000

1

3251 _

  

     

 

or 3.251103kg 2) 3 pts _{What ionic compound is formed between a) calcium and bromine and b)}

potassium and oxygen?

CaBr2

K2O

3) 3 pts _{Balance the following equations?}

a) __MnO2(s)+ __HCl(aq) __Cl2(aq)+ __MnCl2(aq)+__H2O(l) b) __BCl3(g)+__H2O(l)__B(OH)3(aq)+__HCl(aq)

MnO2(s) + 4HCl(aq)  Cl2(aq) + MnCl2(aq) + 2H2O(l) BCl3(g) + 3H2O(l)  B(OH)3(aq) + 3HCl(aq)

4) 3 pts _{What are the names of the following elements: Mo, Mg, and Sn?}

Mo = molybdenum, Mg = magnesium, and Sn = tin

5) 3 pts What are the names of the following compounds: a) SF6 and b) AlF3?

SF6 = sulfure hexaflouride AlF3 = aluminum flouride 6) 3 pts How many moles of Cu are in 3.20 g?

molCu gCu

molCu

gCu 0.0504

55 . 63

1 20

.

3 _

  

 

7) 3 pts _{How many protons, neutrons, and electrons does a) neon-22 and}

b) 209Bi3have?

neon-22 10 protons, 10 electrons, and 12 neutrons



3 209

Bi 83 protons, 80 electrons, and 126 neutrons 8) 3 pts _{How many widgets are in 2.3×10}-20 _moles?





widgets

mol mol

nN

N _a 2.3 10 20 6.02214 1023 1 1.4104

  



 _

 

Short Answer Questions

Each of the following short answer questions are worth the noted points. Partial credit will be give. You must show your work to get credit. Make sure to report answers in the correct number of significant figures and indicate the proper units.

1) 8 pts _{An iron bar weighed 664 g. After the bar had been standing in moist air for}

a month, exactly one-eighth of the iron turned to rust (Fe2O3). Calculate the final mass of the iron bar and rust.

Calculate the mass of Fe that when into forming Fe2O3 g

g 83.0 8

1

664 

    

Calculate the mass of rust formed g Fe  g Fe2O3

g Fe  mol Fe  mol Fe2O3 g Fe2O3

mol g O Fe mol g Fe M M 70 . 159 85 . 55 3 2   119 1 70 . 159 2 1 85 . 55 1 0 . 83 3 2 3 2 3 2                    O molFe O gFe molFe O molFe gFe molFe gFe

Calculate the mass of the bar remaining g

g g 83.0 581

664  

Calculate the mass of the bar and the rust g

g

g 119 700.

581  

2) A solid compound x contains 63.3% Mn and 36.7% O by mass. When X is heated, oxygen gas is evolved and a new solid compound Y containing 72.0% Mn and and 28.0% O is formed

2a) 6 pts Determine the empirical formula of X. Assume X weighs 100

Moles of Mn

mol g Mn

M 54.94

molMn gHg

molHg

gMn 1.15

94 . 54 1 3 .

63 _

  

  Moles of O

mol g O

M 16.00

molO gO

molO

gO 2.29

00 . 16 1 7 . 36 2 2       

Divide through by smallest number of moles (1.15 mol)

Mn O

1 15 . 1 15 . 1 _ mol mol

1.99

15 . 1 29 . 2 _ mol mol

Empirical Formula

4 2b) 6 pts Determine the empirical formula of Y.

Assume Y weighs 100 Moles of Mn

molMn gHg

molHg

gMn 1.31

94 . 54 1 0 .

72 _

  

  Moles of O

molO gO

molO

gO 1.75

00 . 16 1 0 . 28 2 2       

Divide through by smallest number of moles (1.31 mol)

Mn O

1 31 . 1 31 . 1  mol mol

1.34

31 . 1 75 . 1  mol mol

Multiple through by 3 to get whole numbers Empirical Formula

Mn3O4

2c) 4 pts _{Write a balanced equation for the conversion of X to Y.}

3MnO2(s)  Mn3O4(s) + O2(g)

3) DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction:

2C6H5Cl(l) + C2HOCl3(l)  C14H9Cl5(l) + H2O(l)

C6H5Cl chlorobenzene, C2HOCl3 chloral, and C14H9Cl5 DDT In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral.

3a) 7 pts _{What is the mass of DDT formed?} mol

g Cl

H C

M 112.56

5

6  , mol

g chloral

M 147.38 , andM_DDT 354.48_molg Amount of C6H5Cl and C2HOCl present at the beginning of r eaction. Cl H molC Cl H gC Cl H molC Cl H

gC _{6 5}

5 6

5 6 5

6 10.15

56 . 112

1

1142 _

     3 2 3 2 3 2 3

2 3.29

38 . 147

1

485 molC HOCl

HOCl gC

HOCl molC HOCl

gC _

  

 

Moles of C2HOCl3 needed to fully react 10.15 moles of C6H5Cl

3 2 5 6 3 2 5

6 5.075

2 1 15

.

10 molC HOCl

Cl H molC HOCl molC Cl H

molC _

  

 

Since we only have 3.29 mol C2HOCl3, it is our limiting reagent Calculate the mass of DDT formed from 3.29 mol of HOCl3

gDDT molDDT gDDT HOCl molC molDDT HOCl molC 3 3 2 3

2 1.17 10

1 48 . 354 1 1 29 .

3  

          

3b) 1 pts Which reactant is limiting? Which is in excess?

The limiting reagent is C2HOCl3 and C6H5Cl is present in excess.

3c) 4 pts _{What mass of the excess reactant is left over?}

Amount of C6H5Cl used

g g

g

Cl H gC Cl

H molC

Cl H gC HOCl

molC

Cl H molC HOCl

molC

401 741

1142

741 1

56 . 112 1

2 29

.

3 6 6

6 6

6 6 3

2 6 6 3

2

 

    

     

 

3d) 4 pts _{If the actual yield of DDT is 200.0 g, what is the percent yield?}

% 1 . 17 % 100 1170

0 . 200 %

100

% _ 

  

   

   

   

g g yield

l theoretica

yield actual yield

4) Convert the following into balanced equations

4a) 4 pts _{When gallium metal is heated in oxygen gas, it melts and forms solid}

gallium(III) oxide.

4Ga(l) + 3O2(g)  2Ga2O3(s)

4b) 4 pts _{When solutions of calcium chloride and sodium phosphate are mixed, solid}

calcium phosphate forms and sodium chloride remains in solution. 3CaCl2(aq) + 2Na3PO4(aq)  Ca3(PO4)2(s) + 6NaCl(aq)

4c) 4 pts _{When crystalline potassium chlorate is heated to just above its melting}

point, it reacts to form two different crystalline compounds, potassium chloride and potassium perchlorate.

4

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X2Y2. A) 50% B) 20% C) 30% D) 40% E) None 2. Which o

named? A) All o B) Fe2O C) PBr5 D) CoO E) CaSO

3. For a new with mas average a A) 280. B) 282. C) 283. D) 313.4 E) None 4. What is t

A) CrO3 B) Cr2O C) Cr6O D) CrO6 E) None

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5. Consider the following balanced equation: A(g) + 5B(g)  3C(g) + 4D(s)

When equal masses of A and B are reacted, which is limiting?

A) If the molar mass of A is less than the molar mass of B, then B must be limiting. B) More information is needed.

C) If the molar mass of A is greater than the molar mass of B, then A must be limiting.

D) If the molar mass of A is less than the molar mass of B, then A must be limiting. E) If the molar mass of A is greater than the molar mass of B, then B must be

limiting.

6. How many of the following chemical formulas are written incorrectly? ClNa BaBr2 LaF2 ClRb

A) All are written incorrectly. B) One is written incorrectly. C) Two are written incorrectly. D) All are written correctly. E) Three are written incorrectly.