Download Thinking It Through

Chapter Six
Thinking It Through
T6.1
The oxidation states of the elements in the reactants need to be compared to the oxidation states of the
elements in the products. If there is a change, then the reaction is an oxidation reduction reaction.
For HIO4 + 2H2O  H5IO6:
Element
Reactant
Product
H
+1
+1
I
–2
–2
O
+7
+7
T6.2
After finding both nickel and cadmium in the activity series, we decide which will be more readily
oxidized, that is, which metal is higher in the activity series. If Cd is more easily oxidized than Ni, then Cd
metal will react with Ni2+, giving Ni metal.
T6.3
Find Au and Fe in the activity series and determine which one is more easily oxidized. The more active
metal will react with the ions of the other metal and the metallic strip will corrode.
T6.4
The combustion of hydrocarbon fuels forms water as one of its products. It is formed at a high temperature
and then condenses in the cold exhaust pipe. After the exhaust system becomes hot enough, the water is
expelled as vapor (gas).
T6.5
First, the overall equation should be balanced using the ion-electron method.
IO3–(aq) + 6NO2(g) + 3H2O(l)  I–(aq) + 6NO3–(aq) + 6H+(aq)
The molecular mass of NO2 must be used to convert grams of NO2 to moles of NO2. The coefficients in the
balanced equation will be needed to convert from moles of NO2 to moles of IO3–. Finally, moles of IO3–
are converted to volume of IO 3– solution by dividing moles by molarity.
 1 mol NO 2  1 mol IO 3   1000 mL IO 3  


# mL IO 3 = 0.230 g NO 2 

 46.01 g NO 2  6 mol NO 2  0.0200 mol IO 3 
T6.6
It is first necessary to balance the equation for the reaction.
4Cl2(g) + S2O32–(aq) + 5H2O(l)  8Cl–(aq) + 2SO42–(aq) + 10H+(aq)
The number of moles of Na2S2O3 can be calculated using the coefficients in the balanced equation. Next,
these numbers of moles can be converted to a volume by dividing by molarity.
 

1 mol S 2O3 
# mL Na 2S2O3 = 0.020 mol Cl 2 
 4 mol Cl 2 




1 mol Na 2S2O3 1000 mL Na 2S2O3 
 


 

 1 mol S 2O3 0.500 mol Na 2S2O3 

T6.7

It is first necessary to balance the equation for the reaction.

I3–(aq) + 2S2O32–(aq)  3I–(aq) + S4O62–(aq)
The number of moles of Na2S2O3 can be calculated using the coefficients in the balanced equation. Next,
these numbers of moles can be converted to a volume by dividing by molarity.
105
Chapter Six
 

  2 mol S 2O3


# mL Na 2S2O3 = 0.020 mol I 3 

 1 mol I  


3


1 mol Na 2S2O3 1000 mL Na 2S2O3 




 

 1 mol S 2O3 0.200 mol Na 2S2O3 
T6.8
The balanced equation is the place to start.

6Fe2+(aq) + Cr2O72–(aq) + 14H+(aq)  6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
Because amounts of both reagents are specified, we must work a limiting reactant problem to find out
which of the two reactants is completely consumed. From the number of moles of this reactant that
disappear, we can calculate the number of moles of H+ that react. This amount is subtracted from the initial
number of moles of hydrogen ion, and the amount of titrant is calculated by dividing moles by molarity of
NaOH solution.
If Fe2+ is the limiting reactant:

0.060 mol Fe 2 14 mol H + 1 mol NaOH 


# mL NaOH = 0.280 mol H   400 mL Fe 2 


2 
2 



 1000 mL Fe
6 mol Fe  1 mol H

 1000 mL NaOH 
 

0.0100 mol NaOH 
If Cr2O72– is the limiting reactant:


T6.9

# mL NaOH =


2 

2  0.0200 mol Cr 2O 7
14 mol H + 1 mol NaOH 
0.280 mol H   



300 mL Cr 2O7 



 1000 mL Cr O 2 1 mol Cr O 2  1 mol H  




2 7
2 7 
 1000 mL NaOH 
 

0.0100 mol NaOH 
To begin, determine balanced equations for the reaction of SO 32– with CrO42– and for S2O32– with CrO42–.

3SO32– + 2CrO42– + H2O  3SO42– + 2CrO2– + 2OH–
3S2O32– + 8CrO42– + H2O  6SO42– + 8CrO2– + 2OH–
We know that the amount of CrO42– that reacted is
# moles CrO
2
4
 (80.0 mL CrO
2
4

2
0.0500 moles CrO 4

)
2

 1000 mL CrO 4




2
 4.00  10 3 moles CrO 4
We also know the amount of SO42– produced

# moles SO 4
2
 1 mol BaSO 4
 (0.9336 g BaSO 4 )
 233.39 g BaSO 4
 4.00  10
3
moles SO 4
106
2
 1 mol SO 4 2

 1 mol BaSO 4




Chapter Six
Let x = # moles SO32– and y = # moles S2O32– , from the balanced equations and the known quantities we
can write:
2 
2 


2
1 mol SO 4 
2 mol SO 4 
4.00  10 3 moles SO 4  x
 y
2 
2 


1 mol SO 3 
1 mol S 2O3 
2 
2 


2
2 mol CrO 4 
8 mol CrO 4 
4.00  10 3 moles CrO 4  x
 y
2 
2 


 3 mol SO 3 
3 mol S 2O3 
We can solve for x and y from these and calculate the initial concentration of SO 32– and S2O32–.
T6.10

T6.11
It is first necessary to write a balanced equation for the reaction of MnO 4– with Sn2+. Then, we calculate
the original moles of Sn2+ in the 50.0 mL of 0.0300 M SnCl2 solution. The moles of Sn2+ that were titrated
are calculated by multiplying (remembering to include stoichiometry) molarity (0.0100 M) by volume of
titrant (0.02728 L). This number of moles of tin ion remaining is subtracted from the total that was
available in the 50.0 mL portion that was titrated, and the answer is converted to the number of moles of
MnO4– that had reacted with this number of moles of Sn2+. Multiply this number by 10 to get the moles of
MnO4– that had not reacted with the SO2 in the original 500 mL of 0.0200 M KMnO4. By difference,
calculate the moles of SO2 that had reacted, which is equal to the number of moles of S in the original
sample. The mass of S is calculated by dividing moles by atomic mass, and the percentage of S in the
original sample is the mass of S divided by the total sample mass, times 100.
The reaction that occurs is 2Ag+(aq) + Cu(s)  2Ag(s) + Cu2+(aq). If we assume that there is excess
copper available, we need to determine the number of moles of Ag that will be produced.
The number of moles of Ag+ available for the reaction is
0.250 M  0.0500 L = 0.0125 mol Ag+
We can determine the amount of copper consumed from the balanced equation. Since the stoichiometry is
2/1, the number of moles of Cu2+ ion that are consumed is 0.0125  2 = 0.00625 mol. Convert this number
of moles to a number of grams: 0.00625 mol  63.546 g/mol = 0.397 g. The amount of unreacted copper is
thus: 32.00 g – 0.397 g = 31.60 g Cu.
The mass of Ag that is formed is: 0.0125 mol  107.9 g/mol = 1.35 g Ag.
The final mass of the bar will include the unreacted copper and the silver that is formed: 31.60 g + 1.35 g =
32.95 g.
T6.12
Place a clean metal bar of each metal in each of the other metal nitrate solutions. The more active metals
will react (dissolve) in the nitrate solution. The most active metal will react with each of the other three
solutions, the least active metal will react with none of the solutions, and the other two metals will react
with one or two of the solutions. Arrange the metals in the order that corresponds to the number of
solutions in which it reacted.
107