Chemistry 120
First Exam
Name_____________________
October 2, 2014
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1(10)
2(15)
3(18)
4(15)
5(10)
6(20)
7(6)
8(6)
TOTAL
1) Please fill in the blanks,
Atomic #
129
2Te _______
# of Neut.
# of Prot.
# of Elect.
Mass#
_______
_______
_______
_______
52
2) Name the following compounds and circle whether they are soluble or insoluble.
a) BaSO4
_______________________
Soluble
Insoluble
b) Fe2(CO3)3
_______________________
Soluble
Insoluble
c) P2O5
_______________________
Soluble
Insoluble
d) (NH4)2S
_______________________
Soluble
Insoluble
e) Zn(NO3)2
_______________________
Soluble
Insoluble
3a) Balance the following reactions
i)
Fe3S4 +
ii)
Al +
O2
→
Fe3O4 +
H2SO4
→
Al2(SO4)3 +
SO3
H2
3b) Complete and balance the following reactions;
iii)
CuCl2 + Al2(SO4)3 →
iv)
Mg + HCl →
3c) Predict the product and balance each of the following reactions;
v)
N2
+ S →
vi)
Li
+
O2
→
4) When a 9.474 g sample of an unknown organic acid is subjected to combustion analysis
13.895 grams of CO2 and 5.684 g of H2O are produced. What is the empirical formula of the
acid?
5) What is the concentration of H3PO4 and if it takes 36.00 mL of 0.75 M NaOH to completely
neutralize 25.00 mL of the H3PO4?
6) When 60.00 mL of 0.6 M Fe(NO3)3 is mixed with 40.00 mL of 0.80 M K2S a black solid
forms. How many grams of precipitate (solid) will form and what is the concentration of all the
ions left in solution?
What is the net ionic reaction?
What is the concentration of all ions left in solution?
How many grams of the black solid will be made?
7) Please propagate the error of the following mathematical operation and give the answer to the
correct number of significant figures.
(26.24 +/- 0.02) - ( 2.564 +/- 0.004)
35.00 +/- 0.02
8) Starting with solid CaCl2 please describe how you would make 250 mL of 1.2 M CaCl2.
Chemistry 120
First Exam
Name___Answer Key__________
October 2, 2014
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1(10)
2(15)
3(18)
4(15)
5(10)
6(20)
7(6)
8(6)
TOTAL
1) Please fill in the blanks,
Atomic #
129
2-
Te
52
# of Neut.
77
# of Prot.
# of Elect.
52
54
Mass#
129
52
2) Name the following compounds and circle whether they are soluble or insoluble.
a) BaSO4
barium sulfate
Soluble
Insoluble
b) Fe2(CO3)3
iron (III) carbonate
Soluble
Insoluble
c) P2O5
diphosphorus pentoxide
Soluble
Insoluble
d) (NH4)2S
ammonium sulfide
Soluble
Insoluble
e) Zn(NO3)2
zinc nitrate
Soluble
Insoluble
3a) Balance the following reactions
i)
Fe3S4 + 8 O2
ii)
2 Al +
→
Fe3O4 +
→
3 H2SO4
4 SO3
Al2(SO4)3 +
3 H2
3b) Complete and balance the following reactions;
iii)
3 CuCl2 + Al2(SO4)3 → 3 CuSO4 + 2 AlCl3
iv)
Mg + 2 HCl → MgCl2 + H2
3c) Predict the product and balance each of the following reactions;
v)
N2
vi)
2 Li
+ 5 S → N2S5
+
O2
→ 2 Li2O
4) When a 9.474 g sample of an unknown organic acid is subjected to combustion analysis
13.895 grams of CO2 and 5.684 g of H2O are produced. What is the empirical formula of the
acid?
3
5
5 g CO2
g
l
3 5
g H2 O
g
l
3 5
l CO2
l H2 O
3 5
lC
2H
H2 O
3
2 g⁄
l 3
g⁄
lH
gC
l
3
gH
9.474 g sample – 3.7896 g C – 0.6316 g H = 5.053 g oxygen (O)
5 53 g O
g
l
3 5
3 5
3 5
lC
lO
lO
C
O
3
3 5
lH
lO
2H
O
ee
e C H2 O
5) What is the concentration of H3PO4 and if it takes 36.00 mL of 0.75 M NaOH to completely
neutralize 25.00 mL of the H3PO4?
MH+
MH+
H+
MOH
OH
M H+ u
(
H3 O
3H+
5 M N OH)(
H3 O
M H+
3
L)
(MH+ )
3
25 L)
M H3 O
6) When 60.00 mL of 0.6 M Fe(NO3)3 is mixed with 40.00 mL of 0.80 M K2S a black solid
forms. How many grams of precipitate (solid) will form and what is the concentration of all the
ions left in solution?
What is the net ionic reaction?
Overall:
2 Fe(NO3)3 + 3 K2S → Fe2S3(s) + 6 KNO3
Net Ionic:
2 Fe3+ + 3 S2- → Fe2S3
What is the concentration of all ions left in solution?
Note: K+ and NO3- are spectator ions
Moles of Fe(NO3)3
Moles of K2S
Ion
Fe3+
NO3K+
S2-
= (0.60 M)(0.060 L) = 0.036 mol Fe(NO3)3
= (0.80 M)(0.040 L) = 0.032 mol K2S
Moles
Before
Rxn
Moles After
Rxn
Conc.
0.036
0.108
0.064
0.032
0.01467
0.108
0.064
0
0.01467 mol/0.100 L = 0.1467 M Fe3+
Comments
3  0.036 M = 0.108 M
0.108/0.100 L = 1.08 M NO30.064/0.100 L = 0.64 M K
There are more negative ions than positives. You need positives
therefore Fe3+ cannot be limiting. S2- is limiting
0/0.100 L = 0 M S2-
Limiting reactant
+
Determine the limiting reactant –
Using charges: There are 0.108 mole of negative charges and 0.064 mol of positive charges so
we have 0.44 mole too many negative charges. We need another 0.44 mole of positive charges
and our source is Fe3+ so there must be,
+
3+
ge
ge
e Fe
l Fe3+ le
3+
ve
S2 i li i i g
Standard method: compare the two reacting ions
2 Fe3+
2
3S
l
32
2 33
l
le Fe3+ ee e
l Fe3+ le
ve
How many grams of the black solid will be made?
Use the limiting reactant –
Fe2 S3
3S
2
l
32
l
leFe2 S3
2
le Fe2 S3
g⁄
l
22
g Fe2 S3
S2 i li i i g
7) Please propagate the error of the following mathematical operation and give the answer to the
correct number of significant figures.
(26.24 +/- 0.02) - ( 2.564 +/- 0.004)
35.00 +/- 0.02
Propagate the subtraction:
√(
)
(
)
Therefore, 26.24 – 2.564 = 23.676 = 23.68 +/- 0.02
Propagate the division:
√(
)
(
)
So,
e
2
e
2
e ig ig
Answer = 0.6766 +/- 0.0007
8) Starting with solid CaCl2 please describe how you would make 250 mL of 1.2 M CaCl2.
Note: CaCl2 = 110.98 g/mol
MV = moles = (1.2 M CaCl2) (0.250 L) = 0.300 mole CaCl2
0.300 mole CaCl2  110.98 g/mol = 33.294 g CaCl2
So, put some water into a 250 mL volumetric flask and then weigh out 33.294 g CaCl2
and add it to the flask. Stir the mixture until all of the CaCl2 dissolves and then add
enough water to fill the flask to the 250 mL mark.