AP Exam Prep: Stoich Answers
1991 B:
1 mol H 2 O
2 mol H
(a) 7 .2 g H 2 O
0 . 8 0 mol H
18 . 0 g H 2 O
1 mol H 2 O
1 mol CO 2
1 mol C
7 .2 L CO 2
0 . 32 mol C
2 2.4 L CO 2
1 mol CO 2
(or PV=nRT could be used to solve for n)
0. 80 m ol H
2. 5 H
5 H
0. 32 m ol C
1 C
2 C
C
H
2
5
(b) 0.40 mol oxygen in water + 0.64 mol oxygen in CO
= 1.04 mol O = 0.52 mol O
= 16.64 g = 17 g of
2
2
oxygen gas (alternative approach for mol O
from balanced equation)
2
(c) 63.5C - 64.0C = 0.5C
1 mol a l
0.107 mol s olute
0 . 5 C
0 . 107 molal
4.68 C
1 . 0 kg solvent
0.107 mol H C
0 . 100 kg CHCl 3 0 . 0107 mol
1 . 0 kg CHCl 3
0 . 6 0 g H C
g
m ol . wt .
56 . 2
m o l
0 . 01 07 m o l
OR
solve for mol. wt. using
g
K
mo l.w t.
T
k g solve nt
-1
(d) C
H
= 29 g mol
2
5
56.2/29 = 1.93 = 2, (C
H
)
= C
H
2
5
2
4
10
2001 B:
0.325 g
2.00 g 100% = 16.3%
(a)
(1.0079)(2) g H
O
(b) 1.200 g H
+ 16 g H
O) = 0.134 g H
(1.0079)(2)
2
2
750
atm (3.72 L)
P•V
760
)(298 K) = 0.150 mol CO
n =
R•T =
-1
-1
2
(0.0821L
atm·mol
K
•
•
12.0 g C
0.150 mol CO
= 1.801 g C
1 mol CO
2
2
3.000 g ASA – (1.801 g C + 0.134 g H) = 1.065 g O
0.102 mol
(c) 0.08843 L
= 0.00902 mol base
1L
1 mol base = 1 mol acid
1.625 g ASA
= 180 g/mol
0.00902 mol
–
(d) (i) HAsa Asa
+
+ H
2.00 10
-3
mole
= 0.133 M
0.015 L