Answers - Avon Chemistry

AP CHEMISTRY
TOPIC 2: STOICHIOMETRY TEST REVIEW Day 30:
Limiting Reactants
Solution concentration
Electrolytes
Precipitation Reactions
Acids and Bases
Oxidation Numbers
Oxidation / Reduction
Acid Redox
Base Redox
1) Calculate the molarity, M, of a 73.55 g sample of Na 2 CO 3 that is dissolved in enough water to make 550 mL of
solution.
Answers:
73.55 g Na CO 1 mol Na CO
105.991 g
2 3 2 3
0.69393
mol
; 550 mL 1 L
0.550
1000 mL
L
M
0.69393
0.550
mol
L
1.3
M
2) Describe how you would prepare (give instructions) 6.00 liters of a 3.50 M CuSO 4 from solid CuSO 4
Answers:
6.00 L 3.50 mol CuSO4
159.61_ g
L 1 mol CuSO
4
3350
g CuSO
4
Place 3350 grams of CuSO 4 in a flask, then add enough water so that the total volume is equal to 6.00 liters.
3) Describe how you would prepare (give instructions) 3.25 liters of a 2.23 M AgNO 3 from a 6.00 M AgNO 3 stock
solution.
Answers:
3.25
L
2.23
mol AgNO
L
3
7.2475
mol AgNO
3
Volume x M = moles ;
Volume
6.00
mol AgNO
L
3
7.2475
mol AgNO
3
Take 1.21 liters of the 6.00 M stock solution of AgNO 3 and dilute with water until the total volume is 3.25 liters.

4) Calculate the mass, in grams, of the precipitate formed when 135.22 mL of a 3.45 M copper(II) sulfate solution is
mixed with 123 mL of a 3.30 M sodium hydroxide solution.
Answers:
CuSO 4 + 2 NaOH Cu(OH) 22
+ Na 2 SO 4
123 mL 1 L 3.30 mol NaOH 1 mol CuSO4
L 1000 mL
1000 mL L 2 mol NaOH 3.45 mol CuSO 1 L
4
58.8
mL
NaOH is the Limiting Reactant
123 mL 1 L 3.30 mol NaOH 1 mol Cu( OH )
2
97.8658 g
1000 mL L 2 mol NaOH 1 mol Cu( OH )
2
19.8 g Cu( OH )
2
5) A stock solution containing Ni +3 ions was prepared by dissolving 425 grams of pure nickel metal in nitric acid and
diluting to a final volume of 4.50 L. Calculate the concentrations of the stock solution and the diluted solution if
250 mL of the stock solution was diluted to 733.0 mL.
Answers:
Stock Solution:
425 g Ni 1 mol Ni
58.69 g
3 3
7.24144
mol Ni
3
Dilute Solution:
M
7.24144 mol Ni
4.50 L
3
1.61
M
250 mL Ni 1 L 1.61 mol Ni
1000 mL L
3 3
0.402
mol Ni
3
M
0.402 mol Ni
0.733 L
3
0.55
M

6) Identify the oxidizing agent, the reducing agent, the substance being reduced, and the substance being oxidized.
Mg (s) + HCl (aq)
MgCl 2 (aq) + H 2 (g)
1(2)
0 1 1 2 0
Mg 2 H Cl Mg Cl H
2 2
Mg 0 is oxidized, H +1 is reduced, HCl is the oxidizing agent, Mg is the reducing agent
7) Write out the complete equation.
a) HNO 3 (aq) + Mg(OH) 2 (s)
2 HNO 3 (aq) + Mg(OH) 2 (s) 2 HOH (l) + Mg(NO 3 ) 2 (aq)
b) H 2 SO 4 (aq) + LiOH (aq)
H 2 SO 4 (aq) + 2 LiOH (aq)
Li 2 SO 4 (aq) + 2 HOH (l)
8) A 96.2 mL sample of hydrosulfuric acid solution requires 33.7 mL of a 4.80 M potassium hydroxide for complete
neutralization. What is the concentration of the original hydrosulfuric acid solution
H 2 S (aq) + 2 KOH (aq)
2 HOH (l) + K 2 S (aq)
33.7 mL KOH 1 L 4.80 mol KOH 1 mol H2S
1000 mL L 2 mol KOH
0.08088
mol H S
2
M
0.08088 mol H2S
0.0962 L
0.841
M
9) Determine the oxidation number for each ELEMENT in the chemical formula
a) Na 2 SO 3
b) Sn(ClO 3 ) 4
Na: (2) +1 = +2
S: (1) +4 = +4
O: (3) –2 = - 6
0
Sn: (1) +4 = +4
Cl: (4) +5 = +20
O: (12) –2 = - 24
0

10) Balance the following REDOX equation that occur in an Acidic solution:
Cr 2 O 2– 7 + I – Cr 3+ + IO – 3
Cr2O7 2– + I– Cr3+ + IO3 –
3 H 2 O + I– IO3 – + 6 e - + 6 H +1
6 e - + 14 H +1 + Cr 2 O 7
2-
2 Cr 3+ + 7 H 2 O
6 e - + 14 H +1 + Cr 2 O 7
2-
+ 3 H 2 O + I– IO3 – + 6 e - + 6 H +1 + 2 Cr 3+ + 7 H 2 O
8 H +1 + Cr 2 O 7
2-
+ I– IO3 – + 2 Cr 3+ + 4 H 2 O
11) Balance the following REDOX equation that occur in an Basic solution:
MnO 4
– + Br – MnO 2 + BrO 3
–
MnO4 – + Br– MnO2 + BrO3 –
3 H 2 O + Br– BrO3 – + 6 e - + 6 H +1
2 ( 3 e - + 4 H +1 + MnO4 – MnO2 + 2 H 2 O )
6 e - + 8 H +1 + 2 MnO4 – + 3 H 2 O + Br– 2 MnO2 + 6 e - + 6 H +1 + BrO3 – + 4 H 2 O
2 H +1 + 2 MnO4 – + Br– 2 MnO2 + BrO3 – + H 2 O
2 HOH + 2 MnO4 – + Br– 2 MnO2 + BrO3 – + H 2 O + 2 OH -1
HOH + 2 MnO4 – + Br– 2 MnO2 + BrO3 – + 2 OH -1