ENVI 1106
Sample Midterm 2
ANSWERS
A periodic table and solubility table/activity series
(http://jenniferwolfchemistry.files.wordpress.com/2013/02/solubility-and-activity.pdf ), will also
be provided and a calculator will be allowed. All numerical answers should be given with the
correct number of significant figures and units. All work should be shown unless otherwise
indicated.
This sample midterm would take the full class period, or 1 hour 50 minutes
1. Give the formula or name for the following compounds:
(a) potassium hydrogen phosphate
K2HPO4
(b) hypoiodous acid HIO (aq)
(c) mercury (II) nitrite Hg(NO3)2
(d) calcium permanganate Ca(MnO4)2
(e) lithium perchlorate pentahydrate LiClO4 . 5H2O
(f) NH4BrO2 ammonium bromite
(g) Co(C2H3O2)3 cobalt (III) acetate
(h) SiF4
silicon tetrafluoride
(i) MgCrO4 magnesium chromate
2.
Balance the following equations:
a)
b)
2 Fe2O3(s) +
3 N2 (g) 
Ba3N2 (s) + 6 H2O (l) 
4 Fe(s) +
6 NO (g)
3 Ba(OH)2 (s) +
2 NH3(g)
3. Complete and balance the following chemical reactions (if there is no reaction write ‘NR’):
a)
Mg (s) +
2 HCl (aq) →
b)
Cu (s) +
LiCl (aq) →
c)
K2CO3 (aq) + 2 HBr (aq) →
d)
PbCl2
e)
2 Na (s) +
f)
3 NaOH (aq) +
+
2 C4H10 (l)
h)
NaCl (aq) +
2 H2O (l)
→
2 KBr2 (aq) + H2O (l) + CO2 (g)
2 NaCl (aq) + PbSO4 (s)
2 NaOH (aq) +
H3PO4 (aq) →
+
+ H2 (g)
NR
Na2SO4 (aq) →
g)
4.
MgCl2 (aq)
13 O2 (g)
KBr (aq)
→
H2 (g)
Na3PO4 (aq) +
8 CO2
H2O (l)
+ 10 H2O
→ NR
Pick one reaction from question 3 above which fits the following:
more than one correct answer.)
Acid-base:
c or f
Oxidation-reduction (redox):
a, e, or g
(Note: there may be
Precipitation: d
5. For the following reactions, write the net ionic equations:
(a)
CaCl2 (aq) + 2 AgNO3 →
2 Ag+ (aq) + 2 Cl- (aq) →
(b)
Ca(NO3)2 (aq) + 2 AgCl (s)
+ 2 AgCl (s) (note: you can leave out the 2s)
3 KOH (aq) + AlCl3 (aq) → Al(OH)3 (s)
+
3 KCl (aq)
3 OH- + Al3+ → Al(OH)3 (s)
(c)
In reaction (b) above, which ions are spectator ions?
K+ and Cl-
6. Consider the following reaction:
2 KOH (aq) + H2SO4 (aq) → 2 H2O (l)
(a)
+
K2SO4 (aq)
What volume of 0.540 M KOH is needed to completely react with
100.0 mL 0.150 M H2SO4?
0.100 L H2SO4 x 0.150 mol/L x 2 mol KOH/1 mol H2SO4 = 0.0300 mol KOH
0.0300 mol KOH/ 0.540 M = 0.0556L or 0.55.6 mL
(b)
In the above reaction, what would be the molarity of sulfate after the reaction
finishes?
0.100 L H2SO4 x 0.150 mol/L x 1 mol SO42-/1 mol H2SO4 = 0.0150 mol SO4-2
Total volume = 155.6 mL = 0.1556 L
Sulfate molarity = 0.0150 mol SO4-2 / 0.1556 L = 0.0964 M
7.
Drain cleaners are mostly sodium hydroxide, but also sometimes include aluminum
crystals. The sodium hydroxide breaks up grease clogs and also allows the aluminum to
react with water. This reaction produces hydrogen gas which can aid in the breaking of
the clogs:
2 Al (s) + 6 H2O (l) 
3 H2 (g) + 2 Al(OH)3 (aq)
(a) How many moles H2 can be produced from 7.00 mol Al?
7.00 mol Al x (3 mol H2/ 2 mol Al) = 10.5 mol H2
(b) How many moles H2 can be produced from 32.6 g Al?
32.6 g Al x (1 mol Al/ 26.89 g Al ) x (3 mol H2/2 mol Al ) = 1.81 mol H2
(c) How many grams of water react with 2.39 g Al?
2.39 g Al x (1 mol Al/ 26.89 g Al ) x (6 mol H2O/2 mol Al ) x (18.02 g H2O/ 1 mol H2O )
= 4.80 g H2O
(d) If 8.68 g H2 are produced, how many grams Al(OH)3 are also produced?
8.68 g H2 x (1 mol H2/ 2.02 g H2 ) x (2 mol Al(OH)3/3 mol H2 )
x (18.02 g Al(OH)3/ 1 mol Al(OH)3 ) = 51.6 g Al(OH)3
8.
Coniine, a toxic substance isolated from poison hemlock, contains only carbon,
hydrogen, and nitrogen. Combustion analysis of a 5.024 mg sample yields 13.90 mg CO2 and
6.048 mg H2O. What is the empirical formula of coniine?
C8H17N
9.
A 2.00 g sample of XCl3(s) (where X is a metal) was dissolved in water and excess
AgNO3(aq) was added. The reaction is:
XCl3(aq) + 3 AgNO3(aq) → 3 AgCl(s) + X(NO3)3(aq)
A total of 5.20 g of AgCl (molar mass = 143.4 g/mol) was isolated. Determine the molar
mass and the identity of the metal X.
Molar mass of XCl3 = 165.5 g/mol, molar mass of X = 165.5 – 3(35.45) = 59.15
X = Co