Chemistry 201/211
Worksheet 3
Fall 2004
October 14, 2004
Oregon State University
1.) Consider the mixture of iodine gas and chlorine gas represented in the box below.
(I atoms are represented as squares, chlorine atoms as circles.)
What will the contents of the box look like if the molecules undergo the reaction
I2(g) + 3Cl2(g)  2ICl3(g)
There are 4 Cl2 molecules and 4 I2 molecules. Cl2 is the limiting reagent and will
allow only 2 ICl3 molecules to be formed. 6 Cl atoms are accounted for with ICl3
leaving one Cl2 molecule. Only one I2 was used so 3 will be left.
For the next two problems, circle the true statement and make the false statements true.
2.) If 1.00 mol of ammonia reacts with 1.00 mol of oxygen according to the reaction
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
a.) All the oxygen is consumed. True
b.) 4.00 mol of NO is produced. False, 4/5 mol NO is produced
c.) 1.50 mol of water is produced False, 6/5 mol H2O is produced
d.) 0.20 mol of ammonia is left over. True
e.) The statement does not provide enough information to determine percent yield.
True, one would need to know the experimental yield
4 mol NO
1mol NH 3 
 1mol NO produced
4 mol NH 3
1mol O 2 
4 mol NO 4

mol NO produced : O 2 is the limiting reagent
5
5 mol O 2
1 mol O 2 
6 mol H 2 O 6

mol H 2 O produced
5
5 mol O 2
1 mol O 2 
4 mol NH 3 4

mol NH 3 used 1 - 4  1 mol NH 3 left
5
5
5
5 molO 2
3.) In the reaction of 2.0 mol CCl4 with an excess of HF, 1.7 mol CCl2F2 is obtained
CCl4(l) + 2HF(g)  CCl2F2(l) + 2HCl(g)
a.) The theoretical yield for CCl2F2 is 1.7 mol. True
b.) The actual yield for CCl2F2 is 1.0 mol False
c.) The percent yield for the reaction is 85% True
d.) Theoretical yield cannot be determined unless the exact amount of HF used is
known. False
1 mol CCl 2 F2
2 mol CCl 4 
 2 mol CCl 2 F2
1 mol CCl 4
1.7 mol CCl 2 F2
2 mol CCl 2 F2
 0.85
4.) Which of the following best describes the preparation of 0.500 L of aqueous solution
of 1.00 M NaCl?
a.) place 29.2 g NaCl(s) in a flask; dilute to a total volume of 5.00102 mL
b.) place 58.4 g NaCl(s) in a flask; add 5.00102 mL of water
c.) place 0.500 L of water in a flask and add 29.2 g NaCl
d.) place 11.5 g Na(s) and 17.7 g Cl2(g) in a flask; dilute to 5.00102 mL
e.) none of the above
The answer is a. 1.00 mol of NaCl weighs 58.4g.
mol NaCl
29.2gNaCl 
 0.500 mol NaCl
58.4g
0.500 mol NaCl
 1.00 M
0.500L
5.) Phosphoric acid can be neutralized by sodium hydroxide according to the equation
H3PO4 + 3NaOH  Na3PO4 + 3H2O
What volume of 0.176 M NaOH would be required to neutralize 5.00 mL of 14.6 M
concentrated phosphoric acid?
0.00500L H 3 PO 4 
0.219 mol NaOH 
14.6 mol H 3 PO 4 3 mol NaOH

 0.219 mol NaOH
1L H 3 PO 4
1 mol H 3 PO 4
1 L NaOH solution
 1.24 L NaOH solution
0.176 mol NaOH
6.) Use the table on page 80 of your book to answer the following question.
Suppose you have a solution that might contain any or all of the following cations:
Na+,Ca2+, Ag+, Ba2+. Addition of HCl solution causes a precipitate to form. After
filtering off the precipitate, H2SO4 solution is added to the resultant solution and another
precipitate forms. This is filtered off, and a solution of Na3PO4 is added to the resulting
solution. No precipitate is observed. Which ions are present in each of the precipitates?
Which of the four ions listed above must be absent from the original solution?
Following the table, only Ag+ will react with the Cl- from HCl to form AgCl. Only
Ba2+ will react with the H2SO4 to form BaSO4. Na+ won’t react with anything. Ca2+
would react with Na2PO4, but since no precipitate is formed when the NaPO4 is
added, Ca2+ can’t be present.
7.) Write a balanced net ionic equation for each of the following reactions in water
a.) nitric acid and barium hydroxide
b.) barium nitrate and lithium sulfate
c.) analine (C6H5NH2) and sulfuric acid
a.) Nitric acid is a strong acid and barium hydroxide is a strong base.
HNO 3 (aq)  H  (aq)  NO 3 (aq)
BaOH 2(aq)  Ba 2 (aq)  2OH - (aq)
2H  (aq)  2NO 3
 Ba 2 (aq)  2OH - (aq)  2 NO 3
(aq)
(aq)
 Ba 2  2H 2 O
Net : 2H  (aq)  2OH - (aq)  2H 2 O (l)
b.) Barium Nitrate and lithium sulfate are ionic species.
BaNO 3(s)  Ba 2 (aq)  NO3- (aq)
Li 2 SO 4 (s)  2Li 2 (aq)  SO 4 (aq)
The major concern here is whether any of these ionic species precipitate out of solution
2Ba 2 (aq)  NO3- ( aq)  2Li 2 (aq)  SO 4 (aq)  BaSO 4(s)  NO3- ( aq)  2Li 2 (aq)
2-
Net : Ba 2 (aq)  SO 4
2(aq)
 BaSO 4(s)
c.) C6H5NH2 is a weak base and sulfuric acid is a strong acid. C6H5NH2 could be
deduced as being a weak base because it is similar to NH3. There cannot be
precipitation because there are no metal cations present in the reactants.

C 6 H 5 NH 2  H 2 O  OH  (aq)  C 6 H 5 NH 4 (aq)
H 2SO 4  H  (aq)  HSO 4
2H  (aq)  SO 4
2(aq)
(aq)
 OH  (aq)  C 6 H 5 NH 4

(aq)
 C 6 H 5 NH 4

(aq)
 SO 4
(aq)
 H 2 O (l)
8.) Calculate the molarity of 12.26 g H2SO4 in 250.0 mL of solution. One method used
commercially to peel potatoes is to soak them in a solution of NaOH for a short time,
remove them from the NaOH, and spray off the peel. The concentration of NaOH is
normally in the range 3 to 6 M. The NaOH is analyzed periodically to ensure quality. In
one such analysis, 45.7 mL of the above solution is required to react completely with a
20.0 mL sample of NaOH:
H2SO4(aq) + 2NaOH(aq)  2H2O(l) + Na2SO4(aq)
What is the concentration of the NaOH in solution?
12.26 g H 2 SO 4 
1 mol H 2 SO 4
 0.1250 mol H 2 SO 4
98.08 g H 2 SO 4
0.1250 mol H 2 SO 4
 0.500 M H 2 SO 4
0.2500 L
0.500 mol H 2 SO 4 2 mol NaOH
0.0457 L H 2 SO 4 
 0.0457 mol NaOH
1 L H 2 SO 4
1 mol H 2 SO 4
0.0457 mol NaOH
0.0200L
 2.285 M NaOH
9.) To determine the SO2 content of a preservative, 0.3111 g is weighed into 50.00 mL of
0.05238 M I2 where it reacts
SO2 + 2H2O + I2  HSO4- + 3H+ + 2IThe excess I2 is then titrated with 17.22 mL of 0.1101 M Na2S2O3
I2 + 2S2O32-  2I- + S4O62What is the weight of the SO2 in the preservative? Not all of the information given will
be needed.
0.05238 mol I 2
The total amount of I 2 added is 0.05000L I 2 
 2.619  10 3 mol I 2
1 L I 2 solution
Some of this I 2 reacts with the SO 2 , but not all of it. The leftover I 2 is titrated with Na 2 S 2 O 3
0.01722 L Na 2 S 2 O 3 
0.1101 mol Na 2 S 2 O 3
1 mol S 2 O 3
1 mol I 2


 9.480  10 -4 mol I 2
1 L Na 2 S 2 O 3
1 mol Na 2 S 2 O 3 2 mol S 2 O 3
If 9.480  10 -4 mol of I 2 was left after the reaction w ith SO 2 that means that (2.619  10 3 - 9.480  10 -4
 1.671  10 -3 ) mol of I 2 was used.
1.671  10 -3 mol I 2 
1mol SO 2 64.07 g SO 2

 0.1071 g SO 2 in the sample
1 mol I 2
1 mol SO 2
10.) Balance the following unbalanced redox equation in acid.
Zn(s) + VO2+ (aq)  Zn2+ (aq) + V2+(aq)
a.) Write the oxidation number for each atom in the reactants and products
b.) Write the balanced oxidation and reduction half reactions
c.) Write the net balanced equation
a.) Zn(s) = 0
Zn2+(aq)=2+
V in VO2+(aq) = 5
V2+(aq)=2+
b.) Zn(s) Zn2+(aq)+2e3e- + VO2+(aq) V2+(aq)
4H+ + 3e- + VO2+(aq) V2+(aq)+ 2H2O
c.) 3[Zn(s) Zn2+(aq)+2e-]
2[4H+(aq) + 3e- + VO2+(aq) V2+(aq)+ 2H2O(l)]
=3Zn(s)+8H+(aq)+ 2VO2+(aq)3Zn2+(aq) + 2V2+(aq) + 4H2O(l)