5 Steps to a 5 AP Chemistry

 Answers and Explanations

There are multiple “correct” ways to do these calculations. Only one calculation is shown

for each answer.

1. D—The reaction is H 2 SO 4 +2 KOH →K 2 SO 4 +2 H 2 O


2. C—The reaction is H 2 C 2 O 4 +2 NaOH→Na 2 C 2 O 4 +2 H 2 O


3. C—Moles acid =(50.0 mL)(0.20 mol acid/1000 mL) =0.0100 mol
   Moles base =(50.0 mL)(0.20 mol base/1000 mL) =0.0100 mol

There is sufficient base to react completely with only one of the ionizable hydrogens

from the acid. This leaves H 2 AsO 4 −.

4. A—Ox=oxidizing agent =Cr 2 O 72 −;Red=reducing agent =Fe^2 +


5. C—The reaction is WO 3 +3 H 2 →W +3 H 2 O
   (0.0500 mol WO 3 )(l mol W/l mol WO 3 ) (183.8 g W/1 mol W) =9.19 g W


6. D— 63.2% Mn leaves 36.8% O
   63.2/54.94 =1.15 Mn 36.8/16.0 =2.30 O
   Thus, there is 1 Mn/2 O.


7. C—V: 2.39/50.94 =0.0469 O:1.00/16.0 =0.0625
   0.0469/0.0469 =1 0.0625/0.0469 =1.33
   Multiplying both by three gives: 3 V and 4 O.


8. A—(25.0 g(NH 4 ) 2 SO 4 )(l mol(NH 4 ) 2 SO 4 /132 g) ×
   (2 mol N/l mol(NH 4 ) 2 SO 4 )(14.0 g N/1 mol N) =5.30 g

9. E—

10. C—[(6 mol H 2 O)(18 g/mol H 2 O)]/(250 g Na 2 SO 4 ·6 H 2 O) ×100% =43%


11. B—Calculate the moles of acid to compare to the moles of Cu:
    (10.0 mL)(12 mol/1000 mL) =0.12 mol
    The acid is the limiting reactant, and will be used to calculate the moles of NO formed.
    (0.12 mol acid)(2 mol NO/8 mol acid) =0.030 mol

(

%%

2

100 64

N 14.0 g /N)

44.0 g N O 2

×

×=

45 20

1000

6

.mL

0.1000 mol

mL

mol

Ox

Ox

Ox

××

Red

1 1mol

1

50.00 mL

1000

Ox

×

×

mmL

L

=0 5424. M

45 20

1000

.mLbase

0.1200 mol base

mL base

× ××

1

2

mol acid

mol base

.

××=.

1

mL

1000 mL

L

mL

20 00

0 1356

50 0

1000

1

.mLbase

0.200 mol base

mL base

××

mol acid

2 mol base

×

11000 mL acid

mol acid

mL

0 100

50 0

.

=.

98  STEP 4. Review the Knowledge You Need to Score High