Final Exam Questions with Solutions - General Chemistry I | CHEM 151, Exams of Chemistry

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Solutions 4a (Chapter 4 problems) Chem151 [Kua]

4.10 A balanced chemical equation must have equal numbers of atoms of each element on each

side of the arrow. Balance each element in turn, beginning with those that appear in only

one reactant and product, by adjusting stoichiometric coefficients. Generally, H and O are

balanced last. When balancing the equation, start by determining the number of atoms on

each side of the chemical equation.

(a) P

4

• 12 Na → 4 Na

3

P

(b) Na

3

P + 3 H

2

O → PH

3

• 3 NaOH

(c) 4 PH

3

+ 8 O

2

→ P

4

O

10

+ 6 H

2

O

(d) P

4

O

10

+ 6 H

2

O → 4 H

3

PO

4

4.16 This problem tells you how much of the reactant you have (in kg) and asks you to

calculate the amount of product formed. Remember that calculations of amounts in

chemistry always center on the mole:

375 kg C

6

H

12

3

g

1 kg

!

"

$

%

&

1 mol

84.16 g

!

"

$

%

&

2 mol C

6

H

10

O

4

2 mol C

6

H

12

!

"

$

%

&

146.14 g

1 mol

!

"

$

%

&

1 kg

3

g

!

"

$

%

&

= 651 kg C

6

H

10

O

4

4.24 To calculate amounts required for a synthesis that is less than 100% efficient, first

convert the actual yield into a theoretical yield if 100% efficient and then do the usual

stoichiometric calculations. (Adipic acid = AA)

Theoretical yield (if 100% efficient) = 4.575kg

76.5%

100%

3.50 kg = !

"

$

%

&

;

Amt required: !

"

$

%

&

!

"

$

%

&

!

!

"

$

$

%

&

1 mol

84.16 g

2 mol AA

2 molC H

146.14 g

1 mol

4.575 kg AA

6 12

=2.63 kg C

6

H

12

4.78 The problem asks about amount of a product that forms from a given amount of reactant.

First balance the chemical equation. The reaction is:

C

8

H

18

+ O

2

→ CO

2

+ H

2

O

8C + 18H + 2O → 1C + 2H + 3O

Give CO

2

a coefficient of 8, and H

2

O a coefficient of 9 to balance C and H:

C

8

H

18

+ O

2

→ 8 CO

2

+ 9 H

2

O

8C + 18H + 2O → 8C + 18H + 25O

There are 25 O on the product side, so give O

2

a coefficient of 25/2 to balance O and

multiply through by 2 to eliminate the fraction:

2 C

8

H

18

+ 25 O

2

→ 16 CO

2

+ 18 H

2

O

Use the balanced equation to do the appropriate mass-mole-number calculations:

(a) n

CO

2

= 3.5 g

1 mol

114.2 g

!

"

$

%

&

16 mol CO

2

2 mol C

8

H

18

!

"

$

%

&

= 0.25 mol CO

2

(b) # = n N

A

= 0.25 mol

6.022 x 10

23

molecules

1 mol

!

"

$

%

&

= 1.5 x 10

23

molecules

(c) m = n MM = 0.25 mol

44.01 g

1 mol

!

"

$

%

&

= 11 g

4.28 The problem gives information about the amounts of both starting materials, so this is a

limiting reactant situation. We must calculate the number of moles of each species,

construct a table of amounts, and use the results to determine the mass of the product

formed.

Starting amounts are in kilograms, so it will be convenient to work with 10

3

mol amounts.

The balanced equation is given in the problem.

Calculate the initial amounts:

1.50 x 10

3

kg propene

!

!

"

$

$

%

&

!

!

"

$

$

%

&

4 2.08 g

1 mol

1 kg

10 g

3

= 35.6 x 10

3

mol;

6.80 x 10

2

kg NH

3

!

!

"

$

$

%

&

!

!

"

$

$

%

&

1 7.04 g

1 mol

1 kg

10 g

3

= 39.9 x 10

3

mol;

1.92 x 10

3

kg O

2

!

!

"

$

$

%

&

!

!

"

$

$

%

&

3 2.00 g

1 mol

1 kg

10 g

3

= 60.0 x 10

3

mol;

Next construct an amounts table:

Reaction 2 C

3

H

6

+ 2 NH

3

+ 3 O

2

→ 2 C

3

H

3

N + 6 H

2

O

Amt (

3

mol) 35.6 39.9 60.0 0 0

kmol/coeff 17.8 (LR) 19.95 20.

Change (

3

mol) - 35.6 - 35.6 - (

2

3

)(35.6) +35.6 +3(35.6)

Final (

3

mol) 0.0 4.3 6.6 35.6 106.

The mass of acrylonitrile that could be produced is:

35.6 x 10

3

mol

!

!

"

$

$

%

&

!

"

$

%

&

'

1 g

10 kg

1 mol

5 3.06 g

3

= 1.89 x 10

3

kg.

4.30 The problem gives information about the amounts of both starting materials, so this is a

limiting reactant situation. We must calculate the number of moles of each species,

construct a table of amounts, and use the results to determine the final masses of the

products.

Starting amounts are in 10

3

kilograms (

3

kg = 10

6

g), so it will be convenient to work

with 10

6

mol amounts. See the answers to problem 4.6 for the balanced equations.

(a) Begin by calculating the initial amounts:

7.50 x 10

3

kg NH

3

!

!

"

$

$

%

&

!

!

"

$

$

%

&

1 7.03 g

1 mol

1 kg

10 g

3

= 0.440 x 10

6

mol;

7.50 x 10

3

kg O

2

!

!

"

$

$

%

&

!

!

"

$

$

%

&

32 .00 g

1 mol

1 kg

10 g

3

= 0.234 x 10

6

mol;

Next construct the table of amounts using the balanced equation from 4.6:

which is 4 NH

3

+ 5 O

2

→ 4 NO + 6 H

2

O

Reaction

4 NH

3

+ 5 O

2

→ 4 NO +

6 H

2

O

Amt (

6

mol)

0.440 0.234 0 0

6

mol/coeff

0.110 0.0468 (LR)

Change (

6

mol)

- (

5

4

)(0.234) - 0.234 +(

5

4

)(0.234) +(

5

6

)(0.234)

Final (

6

mol)

0.253 0 0.187 0.

Now using the results from the amounts table, determine the mass of products produced:

Mass NO:

!

!

"

$

$

%

&

!

"

$

%

&

10 g

1 ton

1 mol

3 0.01 g

0 .187 x 10 mol

6

6

= 5.61 metric ton or 5.61 x 10

3

kg

Mass H

2

O:

!

!

"

$

$

%

&

!

"

$

%

&

10 g

1 ton

1 mol

1 8.02 g

0 .281 x 10 mol

6

6

= 5.06 metric ton or 5.06 x 10

3

kg

4.76 Molecular pictures and limiting reactant calculations require a balanced chemical

equation for the reaction under consideration. For this process, the balanced equation is

CH

4

+ 2 H

2

O → CO

2

+ 4 H

2

.

(a) There are 6 H

2

O molecules and 5 CH

4

molecules in the picture. Divide number of

molecules by coefficient to see which reactant is limiting:

= 5 for CH

4

,

= 3 for H

2

O.

The smaller number identifies the limiting reactant, H

2

O.

(b) The new molecular picture must show that all the H

2

O has been consumed and CO

2

and H

2

have been produced, but the number of atoms of each element must be the same

as before: 5 atoms of C, 32 of H, and 6 of O. Six H

2

O molecules react with three CH

4

molecules to form three molecules of CO

2

and 12 of H

2

:

(c) Construct an amounts table for the stoichiometric calculations:

Reaction CH

4

+ 2 H

2

O → CO

2

4 H

2

Amt (mol) 5 6 0 0

Change (mol) - 6/2 - 6 +6/2 +(4/2)(6)

Final (mol) 2 0 3 12

Use the results from the amounts table to calculate the desired masses:

2.0 mol CH

4

!

"

$

%

&

1 mol

16.04 g

= 32.08 g CH

4

remaining

3 mol CO

2

!

"

$

%

&

1 mol

44.01 g

=132.0 g CO

2

produced

12 mol H

2

!

"

$

%

&

1 mol

2.016 g

=24.19 g H

2

produced

4.98 Because amounts of each reactant are provided, this is a limiting reactant problem, for

which it is convenient to use an amounts table. We are not asked about the second

product, CO, so its amounts do not need to be tabulated.

Calculate the initial amounts:

SiO

2

: 75.0 g

!

!

"

$

$

%

&

60.09 g

1 mol

=1.25 mol

C: 75.0 g

!

!

"

$

$

%

&

12.01 g

1 mol

= 6.24 mol

Cl

2

: 75.0 g

!

!

"

$

$

%

&

70.90 g

1 mol

= 1.06 mol

Reaction SiO

2

• 2 C + 2 Cl

2

→ SiCl

4

MM (g/mol) 60.09 12.01 70.90 169.

Amt (mol) 1.25 6.24 1.06 0

mol/coeff 1.25 3.12 0.530 (LR)

Change (mol) - (1/2)(1.06) - 1.06 - 1.06 +(1/2)(1.06)

Final (mol) 0.72 5.18 0 0.

The theoretical yield is 0.530 mol !

"

$

%

&

1 mol

169.89 g

= 90.0 g

The actual production is 90.0 g !

"

$

%

&

100%

95.7%

= 86.1 g.

To determine the amount of each reactant that remains unreacted, redo the amounts table

using the correct change:

86.1 g

!

!

"

$

$

%

&

169.9 g

1 mol

= 0.507 mol

Reaction SiO

2

• 2 C + 2 Cl

2

→ SiCl

4

MM (g/mol) 60.09 12.01 70.90 169.

Amt (mol) 1.25 6.24 1.058 0

Change (mol) - 0.507 - 2(0.507) - 2(0.507) 0.

Final (mol) 0.74 5.23 0.046 0.

Calculate the final amounts:

SiO

2

: 0.74 mol !

"

$

%

&

1 mol

60.09 g

=44.5 g

C: 5.23 mol !

"

$

%

&

1 mol

12.01 g

= 62.8 g

Cl

2

: 0.046 mol !

"

$

%

&

1 mol

70.90 g

= 3.26 g

4.40 A net ionic equation shows which ions combine to give new products. Spectator ions do

not appear in the net equation, and charges must be balanced. Start by determining what

ions are present in solution. Then use the Solubility Guidelines from your text to

determine what precipitates can form.

(a) The ions present are: Ag

, NO

3

, Li

, and Cl

• . So the possible combinations are:

AgNO

3

, AgCl, LiCl, and LiNO

3

. Guidelines 1 and 2 predict that salts of lithium and

nitrate are soluble. Since AgCl is not soluble by the guidelines, it must be an insoluble

salt:

Ag

(aq) + Cl

(aq)→AgCl(s); spectator ions: Li

,

NO

3

(b) The ions present are: Mg

2+

, SO

4

2 -

, Na

, and PO

4

3 -

. Guidelines 1 and 3 state that all

salts of sodium and most of sulfate are soluble. Mg

3

(PO

4

)

2

is not covered by guidelines

1 or 2, and is not in the exceptions of 4 and 5, thus by guideline 3 it is insoluble:

3 Mg

2+

(aq)+ 2 PO

4

3 -

(aq) →Mg

3

(PO

4

)

2

(s); spectator ions: Na

,

SO

4

2!

(c) The ions present are: Ba

2+

, SO

4

2 -

, Na

, and OH

• . Guidelines 1 and 3 state that all salts

of sodium and most of sulfate are soluble. Ba(OH)

2

is soluble by guideline 5. BaSO

4

is

an exception in guideline 4 and is thus insoluble:

Ba

2+

(aq)+ SO

4

2 -

(aq)→ BaSO

4

(s); spectator ions: Na

, OH

(d) The ions present are: Al

3+

, Cl

, K

, and OH

• . Guidelines 1 and 2 state that all salts of

potassium and most of chloride are soluble. Al(OH)

3

is not soluble by the guidelines and

is thus insoluble by guideline 3:

Al

3+

(aq) + 3 OH

(aq)→ Al(OH)

3

(s); spectator ions: K

, Cl

.

4.42 The problem gives information about the amounts of both starting materials, so this is a

limiting reactant situation. We must calculate the number of moles of each species,

construct a table of amounts, and use the results to determine the final product mass.

Start by determining the balanced net ionic reaction using the solubility guidelines.

The ions present are Pb

2+

, NO

3

, NH

4

, and Cl

• . Guideline 1 and 2 state that all salts

containing ammonium and nitrate are soluble. PbCl

2

is an exception of guideline 4 and

thus insoluble:

2 Pb

2+

• Cl

→ PbCl

2

.

Calculations of initial amounts:

For Pb

2 +

, 75.0 ml = 0.0750 L and 0.750 M = 0.750 mol Pb

2+

/L

=

!

"

$

%

&

• 1 2

1. 5625 x 10 molPb

1 L

1. 750 mol

0. 0750 L

For Cl

, 125 mL = 0.125 L and 0.855 M = 0.855 mol Cl

/L

• 1 -

1. 069 x 10 molCl

1 L

1. 855 mol

0. 125 L =

!

"

$

%

&

Next set up the amounts table:

Reaction Pb

2+

• 2 Cl

→

PbCl

2

(s)

Start (

• 1

mol)

0.5625 1.069 0

Mol/coeff 0.5625 0.5345 (LR)

Change (

• 1

mol)

- 1.069/2 - 1.069 +1.069/

Final (

• 1

mol)

0.0280 0 0.

The mass of solid that forms is: 5.344 x 10

• 2

mol !

"

$

%

&

1 mol

278.1 g

= 14.9 g.

The ions remaining in solution are the excess Pb

2+

and the spectator ions, NO

3

and

NH

4

.

4.60 The first two parts of this problem ask about an ionic reaction. First identify the species

present in the solution, then use the identified products to find the net ionic reaction and

the spectator ions. The species present are H

2

O, NH

4

, SO

4

2 -

, Ca

2+

, and NO

3

• . All

ammonium and nitrate salts are soluble, so these ions do not precipitate.

(a) The solid product is CaSO

4,

and the net ionic reaction is Ca

2+

+ SO

4

2 -

→ CaSO

4

(s);

(b) The spectator ions are NH

4

and NO

3

(c) The second two parts of this problem involve stoichiometric calculations. The

problem gives information about the amounts of both starting materials, so this is a

limiting reactant situation. We must calculate the number of moles of each species,

construct a table of amounts, and use the results to determine the mass of product formed

and the final concentrations of all species.

Calculations of initial amounts:

Ca

2+

: n = 0.1000 L

!

"

$

%

&

1 L

1 .50 mol

= 0.150 mol

SO

4

2 -

: n = 0.0750 L

!

"

$

%

&

1 L

3 .00 mol

= 0.225 mol

The balanced equation shows that the starting materials react in a 1:1 mole ratio, so we

can identify the limiting reactant by inspection; the limiting reactant is Ca

2+

. Here is the

complete table of amounts:

Reaction Ca

2+

+ SO

4

2 -

→

CaSO

4

Start (mol) 0.150 0.225 0

Change (mol) - 0.150 - 0.150 +0.

Final (mol) 0 0.075 0.

Obtain the mass of precipitate from the final amount of CaSO 4

from the table:

MM = 40.08 g/mol + 32.07 g/mol + 4(16.00 g/mol) = 136.15 g/mol

The mass of precipitate is 0.150 mol

!

"

$

%

&

1 mol

1 36.15 g

= 20.4 g;

(d) To find ion concentrations in the final solution, first determine how many moles of

each ion remain in solution, then divide by the final volume of the solution, which is:

V

final

= 100.0 mL + 75.0 mL = 175.0 mL or 0.1750 L

The amounts of the two spectator ions are unaffected by the reaction:

[NH

4

] =

!

!

"

$

$

%

&

!

"

$

%

&

4 2 4

4 2 4 4

1 mol(NH ) SO

2 mol NH

0.1750 L

0.225mol(NH ) SO

= 2.57 M;

[NO

3

] =

!

!

"

$

$

%

&

!

"

$

%

&

'

3 2

3 2 3

1 molCa(NO )

2 mol NO

0.1750 L

0.150molCa(NO )

= 1.71 M;

The final moles of each reactant can be obtained from the amounts table in part c:

[Ca

2+

] =0 M; [SO

4

2 -

] =

0.175L

1. 075 mol = 0.43 M.

4.46 This problem describes an acid-base reaction. We are asked to determine the mass of

base required to completely neutralize the acid. The starting materials are HCl, a strong

acid, and Mg(OH)

2

, a weak base. In addition to water, the major species are:

H

3

O

, Cl

, Mg(OH)

2

The presence of hydronium ions and a weak base together as major species will result in

an acid-base reaction:

The balanced reaction is:

Mg(OH)

2

(s) + 2 H

3

O

(aq) → Mg

2+

(aq) + 4 H

2

O(l).

The problem gives information about the amount of acid present. This is a mol-mass

conversion problem:

Calculation of initial amount of acid:

For H

3

O

: 0.125 L

!

"

$

%

&

1 L

1. 115 mol

= 1.44 x 10

• 2

mol;

Now determine the mass of Mg(OH)

2

required:

MM [Mg(OH)

2

] = 24.31 g/mol + 2(16.00 g/mol) + 2(1.008 g/mol) = 58.33 g/mol

m

Mg(OH) 2

= 1.44 x 10

• 2

mol H

3

O

!

"

$

%

&

!

!

"

$

$

%

&

1 mol

5 8.33 g

2 molH O

1 mol Mg(OH)

3

2

= 0.420 g Mg(OH)

2

4.50 This problem describes an acid-base titration. We are asked to determine the

concentration of acid used to neutralize the base. The starting materials are HCl, a strong

acid, and NaOH, a strong base. In addition to water, the major species are:

H

3

O

, Cl

, Na

, OH

The presence of hydronium ions and hydroxide ions together as major species will result

in an acid-base reaction:

The balanced reaction is:

OH

(aq) + H

3

O

(aq) → 2 H

2

O(l).

Calculation of the amount of base added:

OH

: 32.45 mL

!

"

$

%

&

!

!

"

$

$

%

&

1 L

1. 0965 mol

1 mL

10 L

= 3.13 x 10

• 3

mol

Since this is a completed titration, the moles of base added to the solution will equal the

moles of acid in the solution.

n (OH

) = 3.13 x 10

• 3

mol = n (H

3

O

)

The concentration of the acid solution is obtained by dividing the number of moles of

H

3

O

by the volume of the HCl solution.

V

soln

= 0.01000 L; [H

3

O

] =

0. 01000 L

3. 13 x 10 mol

= 0.313 M

4.88 (a) The reaction is:

Mg + H

3

O

→ Mg

2+

+ H

2

+ H

2

O.

1Mg + 3H + 1O → 1Mg +4H + 1O

Magnesium and oxygen are already balanced. Give H

3

O

a coefficient of 2 to balance

the charges and H

2

O a coefficient of 2 to balance O. Then H is also balanced (6 on each

side):

Mg + 2 H

3

O

→ Mg

2+

+ H

2

+ 2 H

2

O.

(b and c)

The problem gives information about the amounts of both starting materials, so this is a

limiting reactant situation. We must calculate the number of moles of each species,

construct a table of amounts, and use the results to determine the final amounts.

Calculations of initial amounts:

Mg: 1.215 g Mg

!

!

"

$

$

%

&

2 4.305 g

1 mol

= 4.999 x 10

• 2

mol Mg

H

3

O

= 0.0500 L (4.0 M) = 20 x 10

• 2

mol H

3

O

Use the balanced equation and the initial amounts to construct the amounts table:

Reaction Mg + 2 H

3

O

→ Mg

2+

+

H

2

Amt (

• 2

mol)

4.999 20.0 0 0

• 2

mol/coeff

4.999 (LR) 10.

Change (

• 2

mol)

- 4.999 - 9.998 +4.999 +4.

Final (

• 2

mol)

0 10 4.999 4.

(b) m = n MM = 4.999 x 10

• 2

mol !

"

$

%

&

1 mol

2.0158 g

= 0.1008 g H

2

formed;

(c) The spectator ion concentration remains unchanged: [Cl

] = 4.0 M;

[Mg

2+

] =

50.0x 10 L

4.999 x 10 mol

• 3

= 1.00 M;

[H

3

O

] =

50.0x 10 L

1 x 10 mol

• 3

= 2.0 M.

4.66 When determining reaction products, first identify the type of substances present to

determine what kind of reaction can occur.

(a) Ca(OH)

2

is a strong base, HCl is a strong acid:

OH

(aq) + H

3

O

(aq) → 2 H

2

O(l) (acid-base reaction);

(b) Ni displaces hydrogen gas from strong acids such as HCl:

Ni(s) + 2 H

3

O

(aq) → Ni

2+

(aq) + H

2

(g) + 2 H

2

O(l) (redox reaction, but Ni also

acts as a base, so its also an acid-base reaction)

(c) AgOH is an insoluble salt so its a precipitation reaction. However Ag

also acts as a

weak acid (so this could also be an acid-base reaction).

Ag

(aq) + OH

(aq) → AgOH(s) (precipitation reaction);

(d) A strong acid and a strong base react to form water:

H

3

O

(aq) + OH

(aq) → 2 H

2

O(l) (acid-base reaction);

(e) Ca is a reactive metal, which reacts with water to generate H

2

gas:

Ca(s) + 2 H

2

O(l) → Ca

2+

(aq) + H

2

(g) + 2 OH

(aq) (redox reaction, but Ca also acts

as a base so this could be an acid-base reaction);

(f) NH

3

is a weak base, which reacts with strong acid:

NH

3

(aq) + H

3

O

(aq) → H

2

O(l) + NH

4

(aq) (acid-base reaction).

4.102 The statement of the problem indicates what reactions take place:

H

2

S + O

2

→ SO

2

+ H

2

O. H and S are balanced; give O a coefficient of 3/2 to balance O,

then multiply through by 2 to clear fractions:

2 H

2

S + 3 O

2

→ 2 SO

2

+ 2 H

2

O

SO

2

+ H

2

S → S + H

2

O. H is already balanced; give H

2

S and H

2

O coefficients of 2 to

balance O and S a coefficient of 3 to balance S:

SO

2

+ 2 H

2

S → 3 S + 2 H

2

O.

The second part of the problem asks for the amount of H

2

S required to form a given

amount of S. From the first reaction, each mole of SO

2

requires 1 mol of H

2

S, so the

amount of H

2

S required is the sum of the moles of H

2

S and SO

2

required in the second

reaction. From the second reaction, 3 mol S requires 2 + 1 = 3 mol (H

2

S + SO

2

), so n

H

2

S

= n

S

:

m

H 2

S

= 1.25 kg !

"

$

%

&

!

"

$

%

&

!

!

"

$

$

%

&

1 mol

34.09 g

1 mol S

1 molH S

32.07 g

1 mol

2

= 1.33 kg.