Solution to Exercises for Experiment 7
Δ
2 KClO3 (s ) ⎯MnO
⎯⎯
→ 2 KCl (s ) + 3 O2 (g)
2
MM
122.6
74.6
32.0
Balanced chemical equation for the decomposition of KClO3
(i.e., the reaction being studied in this experiment).
Using Dalton’s Law of Partial Pressures
Obtained from App VI
Barometric pressure
PO 2
=
PT
− PH
P V M O2
!g$
PO2 V = # & R T ⇒ R = O2
" M %O2
gO2 T
O
VO = VH
2
2
2O
that was
displaced during
reaction (in L)
How R was determined
experimentally.
32.0 g/mol
T O (in K) à K = °C + 273
2
gO2 = gunk - gresidue
The Law of Conservation of Mass is being used. After heating,
the O2 (g) being produced in the reaction escapes. The mass of
the residue differs from the mass on the unknown by the amount
of O2 that was produced in the reaction.
If the mass of the unknown and/or residue are not given - nO (or gO ) can be solved for by using the ideal gas
2
2
equation. P, V and T can be obtained as shown above, while the known value for the gas constant, R, can be used.
PO2 V = nO2 R T
Obtained from gO = gunk - gresidue, which can be converted to mol O2 (g O / 32.0 g/mol) or
2
it can also be obtained by using the ideal gas equation.
⎛ 2 mol KClO3 ⎞ ⎛ 122.6 g KClO3 ⎞
gKClO3 = # mol O2 ⎜
⎝ 3 mol O2 ⎟⎠ ⎜⎝ 1 mol KClO3 ⎟⎠
% KClO3 (in unk) =
g KClO3
• 100
g unk
2
Once mol of O2 is known, the grams of
KClO3 and % KClO3 in the unknown
mixture can be calculated as shown.
NOTE : nO2
α
gKClO3
α % KClO
3
1. More moles of gas are being produced, which would lead to more water being displaced (i.e., Vgas
would also be greater). However - since n ∝V - R would not be affected.
PO V ↑
PO2 V = nO2 R T ⇒ R = 2
nO2 ↑ T
2. Since more mol of O2 are being produced (coming from a source other than from the decomposition
of KClO3), this would erroneously indicate that more KClO3 was present in the unknown mixture,
which would lead to an erroneously higher calculated % KClO3 in the unknown mixture.
3.
L • torr ⎞
PO2 V = nO2 R T ⇒ (765 − 35.7 torr) (0.350 L) = nO2 ⎛⎜ 62.4
⎟ (32 + 273 K )
⎝
mol • K ⎠
nO2 =
729.3 • 0.350
= 0.0134 mol
62.4 • 305
80% KClO3
⎛ 2 mol KClO3 ⎞ ⎛ 122.6 g KClO3 ⎞ ⎛ 100 g mixture ⎞
gmixture = 0.0134 mol O2 ⎜
= 1.36 g
⎝ 3 mol O2 ⎟⎠ ⎜⎝ 1 mol KClO3 ⎟⎠ ⎜⎝ 80.0 g KClO3 ⎟⎠
L • torr ⎞
4. PO2 V = nO2 R T ⇒ (760 − 23.8 torr) (0.2500 L) = nO2 ⎛⎜ 50.0
⎟ (25.0 + 273 K )
⎝
mol • K ⎠
nO2 =
736.2 • 0.2500
= 0.0124mol
50.0 • 298
⎛ 2 mol KClO3 ⎞
gKClO3 = 0.0124 mol O2 ⎜
⎝ 3 mol O2 ⎟⎠
% KClO3 (in unk) =
⎛ 122.6 g KClO3 ⎞
⎜⎝ 1 mol KClO ⎟⎠ = 1.01 g
3
g KClO3
1.01 g
• 100 =
• 100 = 67.3 %
g unk
1.500 g
L • torr ⎞
⎛
5. PO2 V = nO2 R T ⇒ (740.0 − 31.8 torr) (0.2000 L) = nO2 ⎜ 70.0
⎟ (30.0 + 273 K )
⎝
mol • K ⎠
nO2 =
708.2 • 0.2000
= 0.00668mol
70.0 • 303
⎛ 2 mol KClO3 ⎞ ⎛ 122.6 g KClO3 ⎞
gKClO3 = 0.00668 mol O2 ⎜
= 0.546 g
⎝ 3 mol O2 ⎟⎠ ⎜⎝ 1 mol KClO3 ⎟⎠
% KClO3 (in unk) =
g KClO3
0.546 g
0.546
• 100 ⇒ 50.0% =
• 100 ⇒ g unk =
• 100 = 1.092 g
g unk
g unk
50.0
gresidue = gunk − gO2 = 1.092 − 0.214 = 0.878 g
⎛ 32.0 g O2 ⎞
? gO2 = 0.00668 mol O2 ⎜
= 0.214 g
⎝ 1 mol O2 ⎟⎠
6. It doesn’t matter what gas producing reaction is being studied, the calculated value that would be
obtained for R is the same since it is a constant.
7. Less moles of gas are being produced, which would lead to less water being displaced (i.e., Vgas
would also be smaller). However - since n ∝V - R would not be affected.
PO2 V = nO2 R T ⇒ R =
PO2 V ↓
nO2 ↓ T
8. Since less mol of O2 are being produced because all the KClO3 did not decompose, this would
erroneously indicate that less KClO3 was present in the unknown mixture, which would lead to an
erroneously lower calculated % KClO3 in the unknown mixture.