City College, Chemistry Department
Chemistry 10301, sections E*, Prof. T. Lazaridis
First Midterm exam, Sep 29, 2010
Last Name: _____________________________________________
First Name: _____________________________________________
SSN last 4: ______________________________________________
Note: There are 7 questions in this exam. Fill in your answer in the blank space provided
immediately following each question. Half a point will be subtracted every time you report
a numerical result with an incorrect number of significant figures. A copy of the periodic
table is attached. Good luck!
1. (20)
a. (4) What is the name of the compound K2SO4 ?
Potassium Sulfate
b. (4) Is the above compound ionic or molecular?
Ionic
c. (4) What is the molar mass of the above compound ?
2*39.10+32.07+4*16.00= 174.27 g/mol
d. (4) What is the chemical formula of calcium carbonate?
CaCO3
e. (4) Give the names of the elements with the following atomic
symbols:
B:
Boron
S:
Sulfur
Fe:
Iron
Mg: Magnesium
2. (a) (4) Indicate the numbers of protons, neutrons, and electrons in
47 protons, 47 electros, 109-47=62 neutrons
€
109
47
Ag
(b) (6) Write a balanced equation for the reaction of burning methanol
(CH3OH) in oxygen:
2 CH3OH + 3 O2  2 CO2 + 4 H2O
3. (20) Balance the following chemical equations:
a. (5)
Cl2O5 +
b. (5)
V 2O5 +
c. (5)
H2O
2 H2 --->
4 Al + 3 O2 ----->
d. (5)
-----> 2 HClO3
V 2O3
+ 2 H2O
2 Al2O3
MnO2 + 4 HCl ----->
MnCl2 +
Cl2
+ 2 H2O
4. (15) Lead(II) oxide reacts with ammonia as follows:
3 PbO (s) + 2 NH3 (g) -----> 3 Pb (s) + N2 (g) + 3 H2O (l)
a) (5) Balance the above equation
b) (5) How many grams of NH3 are consumed in the reaction of 8.16 g PbO?
MM: NH3: 14.01+3*1.008= 17.03 Pb: 207.2 PbO: 207.2 + 16.00 = 223.2
N2: 2X14.01= 28.02
8.16 g PbO X
1molPbO
2molNH 3
X
X17.03 g NH3/mol NH3 = 0.415 g NH3
223.2gPbO 3molPbO
c) (5) If 928 g Pb are produced in this reaction, how many grams of nitrogen
are also formed? €
€
928 g Pb X
1molPb
1molN2
X
X28.02 g N2/mol N2 = 41.8 g N2
207.2gPb 3molPb
€
5. €
(15) A fertilizer has mass percent composition 20.00 % C, 6.71 % H,
46.65 % N, and 26.64 % O. What is its empirical formula?
In 100 g of this fertilizer we have:
C: 20.00 g/12.01 (g/mol) = 1.665 mol
H: 6.71 g/1.008 (g/mol) = 6.657 mol
N: 46.65 g/14.01 (g/mol) = 3.330 mol
O: 26.64 g/16.00 (g/mol) = 1.665 mol
Divide by smallest :
C: 1
H: 4
N: 2
O: 1
Empirical formula: CH4N2O
6. (10) Consider this reaction:
4BF3 + 3 H2O 
H3BO3 + 3 HBF4
The reacting mixture contains 0.496 mol BF3 and 0.313 mol H2O
a. (5) which compound is the limiting reactant?
0.496 mol BF3 X (3/4) = 0.372 mol HBF4
0.313 mol H2O X (3/3) = 0.313 mol HBF4
 H2O is limiting
b. (5) How many moles of HBF4 can be produced?
0.313 mol
7. (10) A student needs 625.0 g of zinc sulfide, a white pigment, which she
can synthesize using the following reaction:
Na2S (aq) + Zn(NO3)2 (aq)  ZnS (s) + 2 NaNO3 (aq)
Assuming that she has plenty of sodium sulfide, how many grams of zinc
nitrate will she need if she can make the zinc sulfide in 85.0 % yield?
85.0% = Actual/Theoretical => Theoretical = 625.0/0.8500 = 735 g
MM: ZnS: 65.39+32.07= 97.46 Zn(NO3) 2= 65.39+2X(14.01+3X16.00)=
189.41 g/mol
735 g ZnS X
1molZnS
1molZn(NO3)2
X
X189.41 g Zn(NO3)2/mol Zn(NO3)2
97.46gZnS
1molZnS
= 1430 g Zn(NO3)2
€
€