AP CHEM WKST KEY: STOICHIOMETRY
1)
0.423 g C (
2.50 g Cl (
1 mol C
) = 0.0352 mol C
12.01 g C
C0.0352 Cl0.0705 F0.0705
0.0352
1 mol Cl
) = 0.0705 mol Cl
35.45 g Cl
0.0352 0.0352
C1.00Cl2.00F2.00
CCl2F2
1 mol F
1.34 g F (
) = 0.0705 mol F
19.00 g F
2)
75.7% As  75.7 g As = 1.01 mol As
As1.01 O1.52
24.3% O  24.3 g O = 1.52 mol O
As1.00O1.52
1.01
1.01
As2O3
3)
Given:
4.22 g product
1.31 g S
S0.0408 Cl0.0821
1.31 g S = 0.0408 mol S
0.0408
4.22 g product – 1.31 g S = 2.91 g Cl = 0.0821 mol Cl
0.0408
S1.00Cl2.01
SCl2
4)
Given:
0.5000 g citric acid
0.6871 g CO2
0.1874 g H2O
molar mass compound = 192 g mol−1
mass element
molar mass element
=
mass compound molar mass compound
mass C 
x
12.01 g C
=
0.6871 g CO2
44.01 g CO2
y
2.02 g H
mass H 
=
0.1874 g H2 O 18.02 g H2 O
x = 0.1875 g C = 0.01561 mol C
0.01561
0.01561
0.01561
C1.000H1.332O1.167
y = 0.0210 g H = 0.0208 mol H
emp. formula: C6H8O7
mass O  0.5000 g – (0.1875 g + 0.0210 g) = 0.2915 g O = 0.01822 mol O
192 g mol-1
= 1
192
C0.01561 H 0.0208 O0.01822
molecular formula: C6H8O7
1
AP CHEM WKST KEY: STOICHIOMETRY
5)
Given:
30.0 g CH2(NH2)COOH
mass element
molar mass element
=
mass compound molar mass compound
x
14.01 g N
=
30.0 g CH2 (NH2 )COOH 75.08 g CH2 (NH2 )COOH
x = 5.60 g N
6)
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
20.0 g
a)
115 g
1 mol Al
3 mol H2
20.0 g Al (
)(
) = 1.11 mol H2
26.98 g Al 2 mol Al
115 g H2 SO4 (
1 mol H2 SO4
3 mol H2
)(
) = 1.17 mol H2
98.09 g H2 SO4 3 mol H2 SO4
b)
Al
c)
1 mol Al2 (SO4 )3 342.17 g Al2 (SO4 )3
1 mol Al
20.0 g Al (
)(
)(
) = 127 g Al2 (SO4 )3
26.98 g Al
2 mol Al
1 mol Al2 (SO4 )3
d)
1 mol Al
3 mol H2 SO4 98.09 g H2 SO4
20.0 g Al (
)(
)(
) = 109 g H2 SO4 reacted
26.98 g Al
2 mol Al
1 mol H2 SO4
mass excess reactant = 115 g – 109 g = 6 g H2SO4
7)
3CCl4 + 2SbF3 → 3CCl2F2 + 2 SbCl3
excess
14.6 g
a)
14.6 g SbF3 (
b)
% yield= (
1 mol SbF3
3 mol CCl2 F2 120.91 g CCl2 F2
)(
)(
) = 14.8 g CCl2 F2
178.76 g SbF3
2 mol SbF3
1 mol CCl2 F2
actual yield
8.62 g
) 100 = (
) 100 = 58.2% yield
theoretical yield
14.8 g
2
AP CHEM WKST KEY: STOICHIOMETRY
8)
9)
3Zn(s) + 2H3PO4(aq) → Zn3(PO4)2(s) + 3H2(g)
a)
1 mol Zn
2 mol H3 PO4 1000 mL H3 PO4
50.0 g Zn (
)(
)(
) = 408 mL H3 PO4
65.38 g Zn
3 mol Zn
1.25 mol H3 PO4
b)
250.0 mL H3 PO4 (
0.725 mol H3 PO4 1 mol Zn3 (PO4 )2 386.08 g Zn3 (PO4 )2
)(
)(
) = 35.0 g Zn3 (PO4 )2
1000 mL H3 PO4
2 mol H3 PO4
1 mol Zn3 (PO4 )2
2NaOH(aq) + MgCl2(aq) → Mg(OH)2(s) + 2NaCl(aq)
150.0 mL
1.825 M
a)
250.0 mL
1.450 M
150.0 mL NaOH (
250.0 mL MgCl2 (
b)
NaOH
c)
M1V1 = M2V2
1.825 mol NaOH 1 mol Mg(OH)2 58.33 g Mg(OH)2
)(
)(
) = 7.984 g Mg(OH)2
1000 mL NaOH
2 mol NaOH
1 mol Mg(OH)2
1.450 mol MgCl2 1 mol Mg(OH)2 58.33 g Mg(OH)2
)(
)(
) = 21.14 g Mg(OH)2
1000 mL MgCl2
2 mol MgCl2
1 mol Mg(OH)2
M2 =
M1 V1
V2
[NaOH] =
(1.825 M)(150.0 mL)
= 0.6844 M NaOH
400.0 mL
[MgCl2 ] =
(1.450 M)(250.0 mL)
= 0.9062 M MgCl2
400.0 mL
Note: NaOH is the limiting reactant. Therefore, OH− is in both the limiting reactant and the precipitate. The Na+
and Cl− are spectator ions and Mg2+ is the second ion found in the precipitate.
concentration of Mg2+ lost from solution 
-
0.6844 M OH (
1 mol Mg2+
2+
in precipitate
- ) = 0.3422 M Mg
2 mol OH
[ ]i
Δ[ ]
[ ]e
Na+
0.6844
0
0.6844
[Na+] = 0.6844 M
OH−
0.6844
– 0.6844
0
[Mg2+] = 0.5640 M
Mg2+
0.9062
– 0.3422
0.5640
Cl−
1.812
0
1.812
[Cl−] = 1.812 M
3