CHEM 105 name (print)

HOUR EXAM II

7-OCT-99

(last)

(first)

1.

General Stuff a. Word Associations mole Avogadro mole atomic weight mole b. What is the pertinent relationship between 160 grams of liquid bromine and 28 grams of carbon monoxide?

2.

a. b.

What mass of acetone (MW = 58) will contain 2 / 7 as many molecules as are present in 0.50 grams of raffinose pentahydrate (MW = 595)? Suppose the density of a diatomic gas is 1.873 g/L, and that the density of an organic compound in the gas state (measured under same conditions of temperature and pressure) is 3.459 g/L. (i) What are the relative masses of the two compounds? Label your response to clearly indicate what gas has what specific value. (ii) If the atomic weight of the diatomic gas is 25, then what would be the molecular weight of the organic compound?

3.

For the combustion of the compound C 11 H 25 O 3 S how many moles of oxygen would be required to react with 2.615 g of the organic compound?

4.

for the reaction shown with an unbalanced chemical equation OPCl 3

+

→

NH 3

OP(NH 2 ) 3

+

NH 4 1+

+

Cl 1 –

what mass of OP(NH 2 ) 3 would be formed from 55.0 grams of ammonia, if the yield were 67 %? 5.

For the following reaction shown with an unbalanced chemical equation: iron(III) chloride

+

MnO 2

=

FeO 3

+ manganese(II) chloride

If 2.57 g of iron(III) chloride were reacted with 2.19 g of manganese(IV) oxide, how many grams of FeO 3 would be formed? 6.

with reference to a volume of 350 mL of a 0.0926 M solution of Scandium(III) chloride: ( scandium has the symbol Sc ) a. b. c. d. e.

write a balanced chemical equation showing substances present in this solution. How many grams of solute needed to be weighed out in order to prepare this solution? If enough water to added to 80.00 mL of this solution to achieve a final volume of 350 mL, what would be the concentration of the resulting solution? If 100 mL of the solution prepared in part c were to be further diluted to achieve a concentration of 0.010M , what would be the final total volume of the solution? What is the concentration (molarity) of chloride ion in: (i) the original starting solution? (II) the resulting solution prepared in part c? (III) the final solution prepared in part d?

7.

What is the molarity of a hydrochloric acid solution if 25.00 mL is completely neutralized by titration with 31.72 mL of a 0.0745 M barium hydroxide solution? Ba(OH) 2 is a strong base.

chemistry (i). (ii). (iii). mathematics

write a balanced chemical equation showing substances present in the hydrochloric acid solution. write a balanced chemical equation showing substances present in the barium hydroxide solution write a balanced chemical equation showing the neutralization reaction. Show the complete expression used to solve this problem with its full compliment of conversion factors.

8.

Mixing solutions of sodium oxalate, Na2C2O4, and lanthanum(III) chloride, LaCL3, precipitates lanthanum(III) oxalate, La2(C2O4)3. What is the molarity of a solution of sodium oxalate if 25.0 mL is required to react with 13.85 mL of 0.0225 M lanthanum(III) chloride ? How many grams of precipitate are formed?

chemistry (i). write a balanced chemical equation showing substances present in the lanthanum(III) chloride solution. (ii0. write a balanced chemical equation showing substances present in the sodium oxalate solution. (iii). write a balanced chemical equation for the precipitation reaction. mathematics Show the complete expression used to solve this problem with its full compliment of conversion factors.

9.

25.00 mL of a 0.0991 M solution of hydroiodic acid ( HI , a strong acid) is added to 35.00 mL of a 0.0843 M aqueous solution of ammonia ( NH 3 ). The neutralization reaction produces ammonium cation, NH 4 1+ . a. what is the final concentration of ammonia in the resulting solution? b. what is the final concentration of ammonium cation in the resulting solution?

chemistry (ii0. (iii). mathematics

10.

When 0.3198 g of ammonium carbonate, (NH 4 ) 2 CO 3 , are added to 50.00 mL of a 0.0681 M solution of uranium(III) nitrate, a precipitate of uranium(III) carbonate forms. What mass of precipitate is formed?

chemistry (i) (ii) mathematics

11.

(i). write a balanced chemical equation showing substances present in the ammonia solution. write a balanced chemical equation showing substances present in the hydroiodic acid solution write a balanced chemical equation for this neutralization reaction.

write a balanced chemical equations for substances present in the uranium(III) nitrate solution. write a balanced chemistry equation for the precipitation reaction.

For the reaction shown with unbalanced chemical equation Cr

+

C6 H6

=

Cr ( C 6 H 6 ) 2

Suppose the yield of the reaction is 37%. What mass of benzene ( C 6 H 6 ) is necessary to obtain 4.50 grams of the product?

Answers

1.

a.

mole

a mass amount expressed as some part of the standard mass.

Avogadro

Equal volume of gases, measured under same conditions of temperature and pressure, contain equal numbers of molecules.

mole

a ratio of

atomic weight

mole b.

mass std.mass

relative weight of an atom as approximately compared to hydrogen atom (relative mass of H is about 1.0), and compared to carbon–12 isotope (relative mass is defined as 12 ) for solutions, moles = ( Molarity )( Volume in Liters)

Bromine is an element that exists in the form of a diatomic molecule, Br 2 . The molecular mass of Br 2 is 2x80 = 160 g/mole. Carbon monoxide is a molecular compound. The molecular mass of CO is 28 g/mole. These two quantities, 160 g Br 2 and 28 g CO , contain the same number of molecules.

2.

a.

Standard masses of compound contain the same number of molecules. So 58 g acetone and 595 g of raffinose pentahydrate contain the same number of molecules. These masses are part of the conversion factor that relates same numbers of molecules.

step 1.

First find what mass of acetone has the same number of molecules as 0.50 g of raffinose pentahydrate…

58 g acetone  (have same numer of moecules)   595 g raff. 

? g acetone = 0.50 g raff. 

= 0.0487 g acetone step 2.

Secondly, reduce this amount by 2 / 7 g acetone = ( 2 / 7 ) x 0.0487 = 0.0139 g acetone

b.

use Avogadro’s findings: equal gas volumes, measured under same conditions, contain equal numbers of molecules. Both density values are expressed in grams per liter i.e., one liter. So the ratio of density masses yields the ratio of relative masses…

org. gas 3.459 1.847 = = diatomic gas 1,873 1.000

(i)

relative masses of

(ii)

molecular weight of the diatomic gas is 2 x 25 = 50 g/mole. Organic compound standard mass is 1.847 times greater, or MW (org) = 1.847 x 50 = 92.3 g/mole.

2 C 11 H 25 O 3 S + 63 / 2 O 2 →

3.

22 CO 2

+ 25 H 2 O +

S

 63/2 moles O 2   2 moles org.cmpd 

moles O 2 = moles org.cmpd 

moles O 2 =

4.

2.615 g org.cmpd  63/2 moles O 2  2 moles org.cmpd  = 0.174 mole O 2 237 g/mole  

OPCl 3

+

→

6 NH 3

+ 3 NH 4 1+ + 3 Cl 1 -

OP(NH 2 ) 3

1 mole OP(NH 2 ) 3    6 mole NH 3 

moles OP(NH 2 ) 3 = moles NH 3 

mass OP(NH 2 ) 3 55 g NH 3 1  = 95 g/mle 17 g/mole  6  

mass OP(NH 2 ) 3 = 51.2 g @ 100 %

mass OP(NH 2 ) 3 = 51.2 g x ( 0.67 ) @ 67 % = 34.3 g

5.

2 FeCl 3

+

3 MnO 2

→

2 FeO 3

+ 3 MnCl 2

This is a limiting reagent type of a problem so first calculate moles of given amounts for both reactants…

 1 mole   = 15.8 millimoles 162.3 g/mole  

mole FeCl 3 = 2.57 g 

 1 mole   = 25.2 millimoles 87 g/mole 

mole MnO 2 = 2.19 g  2 FeCl 3

+

→

3 MnO 2

2 FeO 3

+ 3 MnCl 2

B4

15.8 (lim)

25.2

none

none

REACT

-15.8

-23.7

+15.8

+23.7

1.5

15.8

23.7

AFTR

none

103.8 g/mole  = 1.64 g FeO 3  1 mole  

? mass FeO 3 = 0.0158 moles 

6.

a.

ScCl 3 → Sc 3 + (aq)

b.

moles solute = M x L = 0.0926 x 0.350 = 0.03241

+

3 Cl 1 - (aq)

151.4 g/mole  = 4.91 g ScCl 3  1 mole  

? mass solute = 0.03241 mole  c.

before dilution:

V = 80.00 mL

M = 0.0926

after dilution:

V = 350 mL

M=?

( ? M )( 350 ) = ( 0.0926 M )( 80 ) d.

M = 0.0212

before dilution:

V = 100 mL

M = 0.0212

after dilution:

V=?

M = 0.010

( ? V )( 0.010 M ) = ( 100 mL )( 0.0212 ) e.

7.

(i)

[ Cl 1 - ] = 3 x 0.0926 = 0.2778 mole/L

(ii)

[ Cl 1 - ] = 3 x 0.0212 = 0.0636 mole/L

(iii)

[ Cl 1 - ] = 3 x 0.010 = 0.030 mole/L

(i)

HCL → H 1+(aq) +

(ii)

Ba(OH) 2 →

(iii)

H 1+(aq) + OH 1-(aq) →

V = 211.7 mL

Cl 1 - (aq)

Ba 2+(aq) + 2 HOH

 2 mole OH−  1 mole H+ mole HCl = mole Ba(OH) 2   − 1 mole Ba(OH)2 1 mole OH

 2 x1x1 ( ? M )( 25.00 ) = ( 0.0745 M )( 31.72 )   1x1x1  M (HCl) = 0.189 mole/Liter 8.

(i)

LaCl 3 → La 3 + (aq)

(ii)

Na 2 C 2 O 4 →

(iii)

2 La 3 + (aq) +

+

3 Cl 1 - (aq)

2 Na 1+(aq) + C 2 O 4 2 –(aq) 3 C 2 O 4 2 –(aq) →

La 2 ( C 2 O 4 ) 3

1 mole HCl   +   1 mole H 

a.

 1 mole La3 +  3 mole C 2O 4 2− 1 mole Na2C 2 O 4  mole Na 2 C 2 O 4 = mole LaCl 3     2− 3+  1 mole C 2O 4  1 mole LaCl3  2 mole La

1x3x1 ( ? M )( 25.0 ) = ( 0.0225 M )( 13.85 )  1x2x1 ? M = 0.0187 mole/Liter b.

 1 mole ppt  mole ppt = mole Na 2 C 2 O 4    3 mole Na2 C 2 O 4 

 1 mole ppt  ? mass ppt = (0.0187)(0.025)   542 g / mole  3 mole Na2 C 2O 4 

mass ppt. = 0.0845 g ppt

9.

(i)

NH 3 (aq) + HOH

(ii)

HI

(iii)

NH 3 (aq) + H 1+(aq)

→

↔

NH 4 1+ (aq) + OH 1- (aq)

H 1+(aq) + I 1-(aq) →

(vast majority is present as ammonia, NH 3 )

NH 4 1+(aq)

This is a limiting reagent type of a problem, suggest first calculating moles of each reagent present… moles HI = (0.0991)(25.00) = 2.48 millimoles moles NH 3 = (0.0843)(35.00) = 2.95 millimoles NH 3 (aq) + H 1+(aq)

a.

→

NH 4 1+(aq)

B4

2.95

2.48 (lim)

none

REACT

-2.48

-2.48

+2.48

AFTR

0.47

none

+2.48

Total volume of solution after reaction is 25+35= 60 mL. Molarity NH 3 = 0.47 millimoles / 60 mL = 0.00783 molar

b.

Molarity NH 4 1+ = 2.48 millimoles / 60 mL = 0.0413 molar

10.

(i)

U(NO 3 ) 3 →

(ii)

3 (NH 4 ) 2 CO 3 or

U 3+(aq)

+ 3 NO 3 1 – (aq) → U 2 (CO 3 ) 3 (s)

+ 2 U 3+(aq)

3 CO 3 2 –(aq) + 2 U 3+(aq)

→

U 2 (CO 3 ) 3 (s)

is also just fine

This is a limiting reagent type of a problem. Suggest converting to moles of both reactants… 1 mole  ? mole (NH 4 ) 2 CO 3 = 0.3198 g   = 3.33 millimoles  98 g  ? mole U(NO 3 ) 3 = ( 0.0681 )( 50.00 ) = 3.41 millimoles 3 CO 3 2 –(aq) + 2 U 3+(aq)

→

U 2 (CO 3 ) 3 (s)

B4

3.33 (lim)

3.41

none

REACT

-3.33

-2.22

+1.11

AFTR

none

1.19

1.11

 656 g ppt  ? mass ppt = 0.00111 moles ppt   = 0.728 g ppt 1 mole ppt  11.

Cr + 2 C 6 H 6

→

Cr ( C6 H 6 ) 2

The 4.50 g product is associated with 37 %, because that is the final sought amount. The 37 % is associated with 100 % and with 63 %. these percentages in this problem...

Identify the masses associated with

%

mass

how was this determined?

37

4.50

given by problem

100

12.16

63

7.66

100  conversion: mass at 100 % = mass at 37 %   37  mass at 100 % = 4.5 / 0.37 = 12.16 by difference 12.16 – 4.50

The 100 % amount is used to determine the amount of reactant necessary to furnish 4.50 g product … 12.162 g product  2 mole benzene   2 mole benzene  mole benzene = mole product  =  208 g / mole  1 mole product   1 mole product  mass benzene = 78 x 0.1169 = 9.12 g benzene