MARK
PUNT
/25
CHM 172
Tutorial test A4 / Tutoriaaltoets A4
Surname
Van
Student number
Studentenommer
Initials
Voorletters
Memorandum
ia
Signature
Handtekening
22 September 2014
et
VRAAG 1
[9]
In die teenwoordigheid van soutsuur reageer UO(NO3)2 met
kaliumdichromaat om UO2(NO3)2 en CrCℓ3 te produseer.
Balanseer hierdie redoksreaksie deur gebruik te maak van
die half-reaksiemetode. Toon al die elemente se
okisdasietoestande aan in die ioniese vergelyking,
identifiseer die oksidasie- en reduksie-halfreaksies en sluit
al die toeskouerione in die finale gebalanseerde vergelyking
in.
Pr
QUESTION 1
[9]
In the presence of hydrochloric acid UO(NO3)2 reacts with
potassium dichromate to produce UO2(NO3)2 and CrCℓ3.
Balance this redox reaction using the half-reaction method.
Show the oxidation states of all the elements in the ionic
equation, clearly identify the oxidation- and reduction halfreactions and return all the spectator ions in the final
balanced equation.
or
Date / Datum
UO(NO3)2 (𝑎𝑎) + K2Cr2O7 (𝑎𝑎) → CrCℓ3 (𝑎𝑎) + UO2(NO3) 2 (𝑎𝑎)
Unbalanced net-ionic reaction (without spectator ions) / Ongebalanseerde net-ioniese reaksie (sonder toeskouer-ione):
Oxidation:
2+
+6
UO (𝑎𝑎) +
Cr2O72-
+3
3+
+6

(𝑎𝑎) → Cr (𝑎𝑎) + UO2
2+
(𝑎𝑎)
UO2+ (𝑎𝑎) + H2O (ℓ) → UO22+ (𝑎𝑎) + 2H+ (𝑎𝑎) + 2e¯ 
ni
ve

+4
rs
i
Balancing / Balansering:
ty
of
Total ionic (unbalanced):
UO2+ (𝑎𝑎) + 2NO3-(aq) + 2K+(aq) + Cr2O72- (𝑎𝑎) → Cr3+ (𝑎𝑎) + 3Cℓ¯(aq) + UO22+(𝑎𝑎) + 2NO3-(aq)
i.e. the spectator ions are NO3- and K+ and Cℓ¯
(did not change or does not appear on other side of equation)
Net ionic (unbalanced): UO2+ (𝑎𝑎) + Cr2O72- (𝑎𝑎) → Cr3+ (𝑎𝑎) + UO22+ (𝑎𝑎)
Reduction:

Cr2O72- (𝑎𝑎) + 14H+ (𝑎𝑎) + 6e¯ → 2Cr3+ (𝑎𝑎) + 7H2O (ℓ)


All or nothing
×3
×1
3UO2+ (𝑎𝑎) + 3H2O (ℓ) → 3UO22+ (𝑎𝑎) + 6H+ (𝑎𝑎) + 6e¯
©
U
Cr2O72- (𝑎𝑎) + 14H+ (𝑎𝑎) + 6e¯ → 2Cr3+ (𝑎𝑎) + 7H2O (ℓ)
3UO2+ (𝑎𝑎) + Cr2O72- (𝑎𝑎) + 8H+ (𝑎𝑎) → 3UO22+ (𝑎𝑎) + 2Cr3+ (𝑎𝑎) + 4H2O (ℓ)
Add 6 NO3-
Add 2 K+ -
Add 8 Cℓ -
Add 6 NO3-
Add 6 Cℓ -

Add 2 K+ 2Cℓ -
3UO(NO3)2 (𝑎𝑎) + K2Cr2O7 (𝑎𝑎) + 8HCℓ (𝑎𝑎) → 3UO2(NO3)2 (𝑎𝑎) + 2CrCℓ3 (𝑎𝑎) + 2KCℓ+ 4H2O (ℓ)

CHM 172
A4 Monday

1/3
No phases -1
© University of Pretoria
2014
36.3 𝑔 O
C (12.01 𝑢)
0.2407
0.02004
1.998 ≃ 2

∴ Empirical formula: C2H4O
OR by mass percentage:
% of O: Given as 36.3%
:
:
:
:
% of C:
0.882 𝑔 𝐶𝑂2
C (12.01 𝑢)
54.5
4.537
2.00
:
:
:
:
ni
ve
QUESTION 3
[5]
A piece of nickel foil, 0.550 𝑚𝑚 thick and with an area of
1.25 𝑐𝑚2 , was allowed to react with fluorine gas to give
1.009 𝑔 of a nickel fluoride, Nix Fz (The density of nickel is
8.908 𝑔. 𝑐𝑚 −3 ). Determine the formula of this compound.
U
Mass of Ni:
1.25 𝑐𝑚2 × 0.0550 𝑐𝑚 ×
= Volume of metal
©
Ni (58.69 𝑢)
0.6124
0.01043
1
∴ Empirical formula: NiF2
CHM 172
A4 Maandag

:
:
:
:
:
O (16.00 𝑢)
0.1604
0.01003
1


2
H (1.01 𝑢)
9.20
9.109
4.02
:
:
:
:
O (16.00 𝑢)
36.3
2.268
1
VRAAG 3
[5]
ʼn Stuk nikkelfoelie, 0.550 𝑚𝑚 dik met ʼn oppervlakte van
1.25 𝑐𝑚2 is toegelaat om met fluoorgas te reageer om
1.009 𝑔 van ʼn nikkel fluoried, Nix Fz, te vorm. (Die digtheid
van nikkel is 8.908 𝑔. 𝑐𝑚−3 ). Bepaal die formule van hierdie
verbinding.
8.90836.3 𝑔 Ni
∴ Mass of F: 1.009 𝑔 − 0.6124 𝑔 = 03967 𝑔 𝐹
Mass (𝑔)
Mol
÷smallest
12.01 𝑔 𝐶
ty
Mass (𝑔)
Mol
÷smallest
:
× 44.01 𝑔 𝐶𝑂 × 100 = 54.5 % 𝐶
rs
i
∴In a 100 𝑔 sample
H (1.01 𝑢)
0.0401
0.03970
3.958 ≃ 4
0.442 𝑔 𝑠𝑎𝑚𝑝𝑙𝑒
% of H: 100 − (36.3 + 54.5) = 9.20 % 𝐻


or
0.442 𝑔 − (0.2407 + 0.1604)𝑔 = 0.0401 𝑔 H
2
Pr
Mass (𝑔)
Mol
÷smallest
12.01 𝑔 C
0.882 𝑔CO2 × 44.01 𝑔 CO = 0.2407𝑔 𝐶 (g)
et
From the mass of CO2, we get the mass of C:
So mass of H:

0.442 𝑔 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 × 100 𝑔 compound = 0.1604 𝑔 𝑂
of
Mass of O:
VRAAG 2
[6]
ʼn Onbekende verbinding bevat C, H en O. Dit is bekend dat
die persentasie suurstof 36.3% is. Wanneer 0.442 𝑔 van ʼn
suiwer monster van hierdie verbinding ontbrand, word
0.882 𝑔 CO2 opgevang saam met ʼn onbekende hoeveelheid
H2O. Bepaal die empiriese formule van hierdie verbinding.
ia
QUESTION 2
[6]
An unknown compound contains C, H and O. It is known
that the percentage of oxygen is 36.3%. When 0.442 𝑔 of a
pure sample of this compound combusts, 0.882 𝑔 of CO2 is
collected along with an unknown amount of H2O.
Determine the empirical formula of this compound.
𝑐𝑚3
= 0.6124 𝑔 𝑁𝑖


F (19.00 𝑢)
0.3967
0.02088
2
2/3


©Universiteit van Pretoria
2014
STUDENT NUMBER
STUDENTENOMMER
1 𝑚𝑜𝑙
Mol N2O4 (92.02 u) required for all the N2H4 to react:
or
1.00 × 102 𝑔 × 32.05 𝑔 = 3.12 𝑚𝑜𝑙 N2H4
Mol N2H4 (32.05 𝑢) available:
no mark for the mass to mole calculation

1 𝑚𝑜𝑙 N O
3.12 𝑚𝑜𝑙 N2 H4 × 2 𝑚𝑜𝑙 N2H4 = 1.56 𝑚𝑜𝑙 N2O4 required
1 𝑚𝑜𝑙
2.00 × 102 𝑔 × 92.02 𝑔 = 2.17 𝑚𝑜𝑙 N2O4 available
et
2 4
But, mol N2O4 available:
ia
QUESTION 4
[5] VRAAG 4
[5]
In the early days of the study of rockets a fuel mixture was In die vroeë dae van vuurpylnavorsing is ‘n
used that contained two liquids, hydrazine (N2H4) and brandstofmengsel gebruik wat bestaan het uit twee
dinitrogen tetroxide (N2O4).
vloeistowwe, hidrasien (N2H4) en distikstoftetroksied (N2O4).
2
2
2
2
1.00 × 10 𝑔 of N2H4 is mixed with 2.00 × 10 𝑔 N2O4 and 1.00 × 10 𝑔 N2H4 word gemeng met 2.00 × 10 g N2O4 en
allowed to react. Determine the mass of nitrogen gas that toegelaat om te reageer. Bepaal die massa van stikstofgas
can be formed during the reaction.
wat gevorm kan word tydens die reaksie.
2N2H4(ℓ) + N2O4(ℓ) → 3N2 (𝑔) + 4H2O(𝑔)
Pr
no mark for the mass to mole calculation
But 2.17 mol N2O4 available > 1.56 mol N2O4 required

[2] if motivation is clear
∴There is too much N2O4 to react with all the N2H4 ⇒ N2H4 is the limiting reagent.
OR:
3 𝑚𝑜𝑙 N2
28.02 𝑔 N2
of
∴Mass of N2 produced = 3.12 𝑚𝑜𝑙 N2 H4 ×
2 𝑚𝑜𝑙 N2 H4
×
𝑚𝑜𝑙 N2
= 131 𝑔 𝑁2
3 sig. figs.

ty
Mol N2H4 (32.05 u) required for all the N2O4 to react:
2 𝑚𝑜𝑙 N H
2.17 𝑚𝑜𝑙 N2 O4 × 1 𝑚𝑜𝑙 N2 O4 = 4.34 𝑚𝑜𝑙 N2O4 required
2 4
rs
i
.... 3.12 mol N2H4 available < 4.34 mol N2H4 required ...... N2H4 is limiting and N2O4 is in excess.
OR - Compare the theoretical yield for the complete conversion of each reagent:
3 𝑚𝑜𝑙 N2
2 H4
×
28.02 𝑔 N2
𝑚𝑜𝑙 N2
ni
ve
3.12 𝑚𝑜𝑙 N2 H4 × 2 𝑚𝑜𝑙 N
3 𝑚𝑜𝑙 N2
2.17 𝑚𝑜𝑙 N2 O4 × 1 𝑚𝑜𝑙 N
2 O4
×
28.02 𝑔 N2
𝑚𝑜𝑙 N2
= 131 𝑔 𝑁2 can be produced if all the N2H4 reacts
= 182 𝑔 𝑁2 can be produced if all the N2O4 reacts.
The one that theoretically produces the least product is the limiting reagent.
U
This means that N2H4 is the limiting reagent.
Partial Periodic Table of the Elements / Gegeeltelike periodieke tabel
©
1
H
1.01
3
Li
6.94
11
Na
22.99
19
K
39.10
4
Be
9.01
12
Mg
24.31
20
Ca
40.08
CHM 172
A4 Monday
21
Sc
44.96
22
Ti
47.87
23
V
50.95
24
Cr
52.00
25
Mn
54.94
26
Fe
55.85
27
Co
58.93
3/3
28
Ni
58.69
29
Cu
63.55
30
Zn
65.39
5
B
10.81
13
Al
26.98
31
Ga
69.72
6
C
12.01
14
Si
28.09
32
Ge
72.61
7
N
14.01
15
P
30.97
33
As
74.92
8
O
16.00
16
S
32.07
34
Se
78.96
9
F
19.00
17
Cl
35.45
35
Br
79.90
2
He
4.00
10
Ne
20.18
18
Ar
39.95
36
Kr
83.80
© University of Pretoria
2014