H 4 4 d c 2 h 2 1 e c 2 h 4 2 f k 2 cr o 4 6 g k 2 cr

H
4
=-4
d. C
2
H
2
=-1
e. C
2
H
4
=-2
f. K
2
Cr O
4
=+6
g. K
2
Cr
2
O
7
=+6
h. KMn O
4
=+7
i. NaHC O
3
=+4
j. O
2
=0
k. NaI O
3
=+5
l. K
2
O =-1
m. P F
6
=+6
n. Au Cl
4
=+3
(Reference: Chang 4.47)
5. For the following reactions, identify the type (combination, combustion,
decomposition, displacement, methathesis [double-displacement)] neutralization),
balance the reaction, and for the redox reactions identify the oxidizing agent and
reducing agent. (12 points)
(a) CH
4
+ O
2
→ CO
2
+ H
2
O
Combustion, CH4 + 2 O2 = CO2 + 2H2O
(b) HCl + NaOH → H
2
O + NaCl
Neutralization HCl + NaOH = H2O + NaCl
(c) H
2
+ O
2
→ H
2
O
Redox reactions 2H2 + O2 = 2 H2O oxidizing agent=02 reducing agent=H2
(d) CaCO
3
+ HCl → CaCl
2
+ CO
2
+ H
2
O
Decomposition CaCO3 +2HCl = CaCl2 + CO2 + H2O
2
Copyright © 2014 by Thomas Edison State College. All rights reserved.

(e) Zn + HCl → ZnCl
2
+ H
2
Displacement Zn + 2 HCl = ZnCl2 + H2
(f) NaCl → Na + Cl
2
Decomposition 2NaCl = 2Na+ + Cl2
6. Calculate the molarity of each of the following solutions: (4 points)
a. 29.0 g of ethanol (C
2
H
5
OH) in 545 mL of solution
Number of mole of C2H5OH = 29.0 / (12*2+1.0*6+16.0) = 0.630434782mole
Molarity = number of mole/volume = 0.630434782 / (545/1000) = 1.26moldme-
3 = 1.26M
b. 9.00 g of sodium chloride (NaCl) in 86.4 mL of solution
Number of mole of NaCl = 9.00 / (58.44) = 0.154 mol
Molarity = number of mole/volume = 0.154 / (86.4/1000) = 1.78 M
7. Calculate the mass of KI in grams required to prepare 5.00 × 10 2 mL of a 2.80 M
solution. (3 points)
5x10^2 ml = 500 ml = .5L
2.8 = moles/.5L
moles=1.4
(1.4)(166) = 232.4
232.4 grams KI