Doc
Brown's Advanced Level Chemistry
Quantitative redox reaction analysis
GCE
Advanced Level REDOX Volumetric Analysis Titration Revision Questions - ANSWERS
email doc
brown - comments - query?
All my advanced A level
inorganic
chemistry revision notes
Use your
mobile phone or ipad etc. in 'landscape' style
This is a BIG website, you need to
take time to explore it
[SEARCH
BOX]
ORIGINAL
REDOX TITRATION QUESTIONS
*
acidbase
& other nonredox titration Questions
Qualitative Analysis
*
REDOX REACTION THEORY
* If you spot a silly error please
EMAIL
query?comment?error
I
DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do, BUT I
AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE
EMAIL ME ASAP TO SORT IT OUT!
Most of the answers have been rounded up or
rounded down to three significant figures (3sf)
Studying
Constructing full inorganic redox
equations from halfequations and Redox titrations
first HELPS!
Question 1: ANSWERS
(a) MnO4(aq) +
8H+(aq) + 5Fe2+(aq) ===> Mn2+(aq)
+ 5Fe3+(aq) + 4H2O(l)
(b) (i) mol MnO4 = 0.0200 x
24.3 / 1000 = 0.000486
therefore mol Fe2+ = 5 x 0.000486 (1 : 5 in equation) =
0.00243 in 20 cm3
so scaling up to 1 dm3, the molarity
of Fe2+ = 0.00243 x 1000 / 20 = 0.122 mol dm3.
(ii) The end point is the first faint permanent pink due to a trace excess of KMnO4.
(c) mol MnO4 = 0.0200 x
25.45 / 1000 = 0.000509
mol Fe = 5 x 0.000509 = 0.002545,
mass Fe = 0.002545 x
55.9 = 0.1423 g
total Fe in wire = 0.1423 x 10 = 1.423 g (only
1/10th
of the made up solution used in titration)
so % Fe = 1.423 x 100 / 1.51 =
94.2 %
(d) The choice of acid is fully discussed in
Ex 6.4 of Advanced Redox Chemistry Part 2 but basically you should know
that ...
The titrations are done with (i) dilute sulfuric
acid present to prevent side reactions e.g. MnO2 formation
(brown colouration or black precipitate).
(ii) Nitric acid is an oxidising agent and
would oxidise iron(II) ions to iron(III) ions,
and (iii) chloride ions are oxidised to chlorine
by manganate(VII) ions (see half-cell potential data in Q3c),
therefore, using either acid (ii) or (iii)
would give a false
titration result.
(e) (i) 23.87 cm3, i.e.
the average titration value, which is statistically more accurate than an
individual titration result.
(ii) molarity = mol / volume in dm3
BUT, remember, 1 dm3 =
1000 cm3,
burettes and pipettes work in cm3
and molarity works in dm3,
a bit of pain I know, but live with
it and go with the flow!
therefore mol = molarity x vol in cm3
/ 1000
mol MnO4 =
0.0200 x 23.87/1000 = 4.774 x 104
(iii) From the balanced redox equation, 1 mol
MnO4 oxidises 5 mol of iron(II) ions
mol Fe2+ = 5 x 4.774 x
104 = 2.387 x 103
(iv) You can assume mass of Fe = mass of Fe2+
ion, (2 electrons don't weigh much!)
mol = mass/atomic mass, so mass = mol x
atomic mass
mass Fe = 2.387 x 103
x 55.8 = 0.1332 g
(v) Total mass of iron in original
sample = 10 x 0.1332 = 1.332 g (scaling up by factor of
250/25)
(vi) % iron in the original salt =
1.332 x 100 / 8.25 = 16.1% (1dp, 3sf)
Question 2 ANSWERS
:(a)
(i) I2(aq) + 2e ==> 2I(aq)
(iodine reduced by thiosulfate ion, reduction is electron gain) |
(ii) 2S2O32(aq)
==> S4O62(aq)
+ 2e (thiosulfate ion oxidised by iodine, oxidation is
electron loss) (Note
that you needed to reverse the half-cell equation as given in question
to arrive at the full ionic redox equation) |
(i) + (ii) gives: 2S2O32(aq) + I2(aq)
==> S4O62(aq)
+ 2I(aq) |
(b) mol S2O32
= 0.0120 x 23.5 / 1000 = 0.000282, mole iodine as I2 = mol S2O32
/ 2 (1 : 2 in equation) = 0.000141,
Ar(I) = 126.9, so Mr(I2)
= 2 x 126.9 = 253.8
therefore: mass of iodine = 0.000141 x 126.9 x 2 =
0.0358
g
(c) mol S2O32
= 0.095 x 26.5 / 1000 = 0.002518,
mol of iodine = mole 'thio' / 2 = 0.002518 / 2 =
0.001259 in 25.0
cm3,
scaling up to 1 dm3 gives 0.001259 x 1000 /25 =
0.0504
mol dm3 of molecular iodine I2.
mass concentration of
I2
= 0.0504 x 2 x 126.9 = 12.8 g dm3 of iodine
Question 3 ANSWERS:
(a)
(i) Sn2+(aq) + 2Fe3+(aq)
==> Sn4+(aq) + 2Fe2+(aq)
(ii) Cr2O72(aq)
+ 14H+(aq) + 6Fe2+(aq) ==> 2Cr3+(aq)
+ 6Fe3+(aq) + 7H2O(l)
(b) for a 25.0 cm3 aliquot titrated
mol
Cr2O72 = 0.0200 x 26.4 / 1000 = 0.000528,
mol Fe2+
titrated = 6 x Cr2O72 = 0.003168
(from 1 mol Cr2O72 : 6
mol Fe2+ in the redox
equation),
mol Fe2O3
= mol Fe2+ / 2 = mol Fe3+ / 2 = 0.003168 / 2 = 0.001584
(moles Fe2+ or Fe3+ halved, because of 2
moles of Fe3+ in in one mole of Fe2O3)
Mr(Fe2O3) =
159.8 (Fe = 55.9, 0 = 16.0)
so mass of Fe2O3 = 0.001584 x 159.8 = 0.2531 g.
Total mass
of Fe2O3 = 0.2531 x 10 = 2.531 g
Remember only 1/10th
titrated, so need to scale up by 10 to account for all the iron oxide in the
original sample.
Therefore % Fe2O3 = 2.531 x 100 / 2.83 =
89.4%
Note that that overall the ratio of mol Cr2O72
: mol Fe2O3 is 3 : 1.
Now its ok to solve this problem using the 3
: 1 ratio if you are very confident to use short cuts in this type of
calculation. However, I think from a teaching and learning point of view, I
prefer to show the full logic of each step in the calculation, particularly
when dealing with mole ratios and this is what I've tried to do on this
page!
(c) Potassium manganate(VII) isn't used for
this titration because it is strong enough to oxidise chloride ions (from the
hydrochloric acid) to form chlorine, giving a completely false titration.
(The half-cell potential proves the
feasibility of this reaction, with a greater positive potential than
chlorine/chloride, the manganate(VII) ion can oxidise the chloride ion to
chlorine. Further note, that the dichromate(VI) ion cannot oxides the
chloride ion, which is why it would used in this context.)
Note on the question design: There is actually a flaw in this question.
In order to ensure all the Fe3+ is reduced, you would need excess Sn2+
solution, BUT, any excess Sn2+ would be oxidised by the Cr2O72
giving a false titration value. Never-the-less, it is a legitimate problem to
solve!
Question 4 ANSWERS:
mol Fe2+ = 0.100 x
25.0 / 1000 = 0.0025,
mol MnO4 = mol Fe2+ / 5
(from equation 1 : 5) = 0.0005 in 24.15 cm3,
scaling up to 1 dm3,
molarity of MnO4 = 0.0005 x 1000 / 24.15 = 0.0207
mol
dm3.
Question 5 ANSWERS:
mol
Cr2O72 = 0.0200 x 21.25 / 1000 = 0.000425,
mol
of Fe salt = mol Fe2+
titrated = 6 x Cr2O72 = 6 x 0.000425 =
0.00255,
BUT only 1/10th of Fe2+
salt used in titration,
so 1 g of FeSO4.(NH4)2S04
.xH2O is equal to 0.00255 mol.
Scaling up to 1 mol gives a
molar mass for the salt in g mol1 of 1 x 1 /0.00255 = 392.2.
So the
formula mass for FeSO4.(NH4)2S04.xH2O
is 392.2
Now the formula mass of FeSO4.(NH4)2S04
= 284.1, this leaves 392.2 284.1 = 108.1 mass units.
Mr(H2O)
= 18, so 108.1 / 18 = 6.005 mol of water, so x = 6 in the salt formula,
FeSO4.(NH4)2S04.6H2O.
Question 6 ANSWERS:
(a) 2MnO4(aq) +
16H+(aq) + 5C2O42(aq)
==> 2Mn2+(aq) + 8H2O(l) +
10CO2(g)
or 2MnO4(aq) +
6H+(aq) + 5H2C2O4(aq)
==> 2Mn2+(aq) + 8H2O(l) +
10CO2(g)
(b) Mr(H2C2O4.2H2O)
= 126.0
total mol H2C2O4.2H2O (or C2O42)
= 1.52 / 126 = 0.0120635
but mol of C2O42 in
titration = 0.0120635/10 = 0.00120635 (1/10th used, 25 of 250 cm3),
mol MnO4 = mol of C2O42
/ 2.5 (2:5 or 1:2.5 ratio),
mol MnO4 = 0.00120634 / 2.5 =
0.00048254 (in 24.55 cm3),
scaling up to 1 dm3 the molarity
of MnO4 = 0.00048254 x 1000 / 24.55 = 0.0196554
= 0.01966 mol dm-3
(5dp, 4sf) OR 0.0197 mol dm3 (4dp, 3sf)
Mr(KMnO4) = 158.0, so in
terms of mass concentration = 0.01965523 x
158 = 3.10555
= 3.106 g dm-3 (3dp,
4sf) OR 3.11 g dm3 (2dp, 3sf)
[NOTE: The titration volume, mass and formula
masses are quoted to four significant figures (4sf), so it might be considered
legitimate to quote the answer to four significant figures]
Question 7 ANSWERS:
mol KHC2O4.H2C2O4.2H2O
(Mr = 254.2) = 0.150 / 254.2 = 0.000590087
ratio of tetroxalate to manganate(VII) is 2:2.5 or
1:1.25 (note equiv of 2 C2O42 in salt),
so mol MnO4
in titration = 0.000590087 / 1.25 = 0.000472069 in 23.2 cm3,
scaling up
to 1 dm3 gives for [MnO4] = 0.000472069 x
1000 / 23.2 = 0.02034781
= 0.02034 mol dm-3
(5dp, 4sf) OR 0.0203 mol dm3 (4dp, 3sf)
[NOTE: Quoting the concentration to 4dp, 3sf is
more appropriate here because the mass is only quoted to 3sf and the titration
is only likely to be accurate to the nearest 0.05 cm3]
Question 8 ANSWERS:
(a) 2MnO4(aq) +
6H+(aq) + 5H2O2(aq)
==>
2Mn2+(aq)
+ 8H2O(l) + 5O2(g)
(b) in titration mol MnO4
= 0.0200 x 20.25 / 1000 = 0.000405,
MnO4 : H2O2
ratio is 2:5 or 1 : 2.5, so mol H2O2 in titration = 0.000405
x 2.5 = 0.0010125,
scaling up for total mol H2O2
in diluted solution (of 1 dm3 or 1000 cm3) = 0.0010125 x 1000 / 25.0 =
0.0405 mol,
but in the original 50 cm3
solution,
therefore scaling up to 1 dm3, the
original
molarity of H2O2
is 0.0405 x 1000 / 50 = 0.810 mol dm3.
Question 9 ANSWERS:
(a) Zn(s) + 2Fe3+(aq) ===> Zn2+(aq)
+ 2Fe2+(aq)
(b) You need to know the redox equation to get the mole
ratio involved in this titration
MnO4(aq) +
8H+(aq) + 5Fe2+(aq) ==>
Mn2+(aq)
+ 5Fe3+(aq) + 4H2O(l)
Changes in oxidation state: Mn from +7 to +2 and
Fe from +2 to +3
Therefore, on electron/oxidation state change, 1
mole of MnO4- can oxidise 5 moles of Fe2+
Therefore:
mol MnO4 in
titration = 0.0100 x 26.5 / 1000 = 0.000265,
mol Fe (Fe2+) = mol MnO4
x 5 = 0.001325 in 20.0 cm3 of the alum solution,
scaling up gives
total mol Fe = 0.001325 x 500 / 20 = 0.033125,
total mass Fe in the 13.2 g of alum
= 0.033125 x 55.9 = 1.852,
so % Fe = 1.852 x 100 / 13.2 =
14.0%
Question 10 ANSWERS:
2MnO4(aq) +
16H+(aq) + 5C2O42(aq)
==> 2Mn2+(aq) + 8H2O(l) +
10CO2(g)
mol MnO4
in titration = 0.05 x 24.5 / 1000 = 0.001225,
ratio MnO4:Na2C2O4 is
2:5 or 1:2.5, so mol Na2C2O4 titrated =
0.001225 x 2.5 = 0.003063 in 5 cm3,
scaling up to 1 dm3,
molarity Na2C2O4 = 0.003063 x 1000 /
5 = 0.613 mol dm3
Mr(Na2C2O4)
= 134, so concentration = 0.613 x 134 = 82.1 g dm3
Question 11 ANSWERS:
MnO4(aq) +
8H+(aq) + 5Fe2+(aq) ===> Mn2+(aq)
+ 5Fe3+(aq) + 4H2O(l)
mol KMnO4 = 0.0100 x 43.85 / 1000 = 0.0004385,
mol Fe (Fe2+) = mol KMnO4 x 5,
mol Fe = 0.0004385 x 5
= 0.0021925, so mol FeSO4.xH2O is also
0.0021925,
in the titration 1/20th of the salt was used (25/500),
so 1/20th of 12.18 g = 0.0021925 mol of the salt = 0.609
g,
scaling up the mass of 1 mole of the salt is 0.609 x 1 / 0.0021925 =
277.8,
so formula mass of FeSO4.xH2O is 277.8, now
the formula mass of FeSO4 is 152.0,
so the formula mass of xH2O
= 277.8 152.0 = 125.8,
Mr(H2O) = 18, so x = 125.8 / 18 =
6.989,
so x = 7 and the formula of the salt is
FeSO4.7H2O,
i.e. seven molecules of water of crystallisation.
Question 12 ANSWERS:
(a) 2MnO4(aq) + 6H+(aq)
+
5NO2(aq) ==> Mn2+(aq)
+ 5NO3(aq) + 3H2O(l)
(b) mol KMnO4 in titration = 0.0250 x
25 / 1000 = 0.000625,
mol ratio MnO4:NO2
is 2:5 or 1:2.5, so mol NO2 in titration = 0.000625 x 2.5
= 0.0015625 in 24.2 cm3,
scaling up to 1 dm3 gives a
molar
concentration of NaNO2 of 0.0015625 x 1000 / 24.2 = 0.0646 mol dm3
Mr(NaNO2) = 69, so in
terms of mass concentration = 0.0646 x 69
= 4.46 g dm3
Question 13 ANSWERS:
Mr(FeC2O4)
= 143.9, mol FeC2O4 in
original solution = 2.68 / 143.9 = 0.01862,
scaling down the mol FeC2O4
in the titration = 0.01862 x 25 / 500 = 0.000931,
mol KMnO4 in
titration = 0.0200 x 28.0 / 1000 = 0.00056,
so ratio KMnO4:FeC2O4
is 0.00056:0.000931 = giving the 'not so easy to spot' 3:5 the reacting mole ratio.
FeC2O4
is made up of a Fe2+ ion and a C2O42
ion, and the full redox equation is:
3MnO4(aq)+
5FeC2O4(aq) + 24H+(aq)
===> 3Mn2+(aq)
+ 5Fe3+(aq)+ 12H2O(l) +
10CO2(g)
or 3MnO4(aq)+
5Fe2+(aq) + 5C2O42(aq) + 24H+(aq)
===> 3Mn2+(aq)
+ 5Fe3+(aq)+ 12H2O(l) +
10CO2(g)
Question 14 ANSWERS:
(a) IO3(aq)
+ 5I(aq) + 6H+(aq)
===> 3I2(aq) + 3H2O(l)
(b) mol I titrated = 0.100 x 20.0
/ 1000 = 0.002, mole ratio
IO3:I is 1:5,
so mole
IO3 reacted = 0.002 / 5 = 0.0004,
so 0.0004 = 0.012 x
(volume IO3 required) / 1000,
volume
IO3 required = 0.0004 x 1000 / 0.012 = 33.3 cm3
(c)(i) mole S2O32
('thio') = 0.0500 x 24.1 / 1000 = 0.001205,
2S2O32(aq) + I2(aq)
===> S4O62(aq)
+ 2I(aq)
I2:S2O32
ratio is 1:2 in the titration reaction, so mol I2 = mole S2O32
/ 2 = 0.001205 / 2 = 0.0006025,
now the IO3:I2
reaction ratio is 1:3,
so mol IO3 reacting to give iodine
= mole I2 formed / 3 = 0.0006025 / 3 = 0.000201 in 25 cm3,
so scaling up to 1 dm3 the
molarity of the KIO3 (IO3)
= 0.000201 x 1000 / 25 = 0.00804 mol dm3,
Mr(KIO3)
= 214.0, so in terms of mass, concentration = 0.00804 x 214 =
1.72 g dm3.
A quicker approach if confident! ratios from all equations involved are: 2S2O32
: I2 : 1/3IO3, means
that the overall mole iodate(V) = mole thiosulphate / 6, so you can 'jump' from line
'1' to the last 'few' lines. However in exams these days all the stages (i.e. , to
, !) are often 'broken down' for you and it might be best you work through the
problem thoroughly and methodically.
(ii) Starch indicator is used for the
titration, when the last of the iodine reacts with the thiosulphate, the blue
colour from the starchiodine 'complex' is discharged and the solution becomes
colourless.
Question 15 ANSWERS:
(i) mol KMnO4
= 0.0200 x 22.5 / 1000 = 0.00045,
mol Fe2+ = mol KMnO4 x
5 = 0.00225 in 25 cm3,
scaling up to 1 dm3, molarity of
the original Fe2+ = 0.00225 x 1000 / 25.0 = 0.090 mol dm3
(ii) the 2nd titration gives the total
concentration of Fe2+ + Fe3+ because any Fe3+
has been reduced to Fe2+,
mol KMnO4 = 0.0200 x 37.6 /
1000 = 0.000752, total mol Fe2+ titrated = mol KMnO4 x 5 =
0.00376 in 25 cm3,
scaling up to 1 dm3, total molarity
of Fe2+ + Fe3+ in original solution = 0.00376 x 1000 / 25.0 = 0.150 mol dm
3,
so using the result from (a) the
Fe3+ concentration = 'Fe'
total Fe2+ = 0.150 0.090 = 0.060 mol dm3.
Question 16 ANSWERS:
You can ignore the 25 cm3
of the solution because you use the same volume in each titration and you can
work on the ratio of the moles of 'Fe' out of the (a) and (b) titration calculations.
(a) mol Fe2+ = 5 x MnO4
= 5 x 0.0200 x 16.9 / 1000 = 0.00169 mol = unreacted iron
(Fe2+ formed from the
iron dissolving in the acid to form).
(b) mol Fe3+ = EDTA4
= 0.100 x 17.6 / 1000 = 0.00176 mol = total mol iron in the sample titrated.
(c) calculation (a) gives the relative moles of
unreacted iron Fe, as it dissolved to form the titratable Fe2+.
Calculation (b) gives the total unreacted Fe
plus the rust i.e. Fe3+,
because any Fe2+ formed from Fe has been oxidised to Fe3+.
So from the original mixture (in terms of the 25 cm3 sample), mol
unreacted Fe = 0.00169,
mol of rusted iron = 0.00176 0.00169 = 0.00007.
Therefore
the % rusted iron = 0.00007 x 100 / 0.00176 = ~4 % rusted iron.
Question 17
ANSWERS:
(a) I2(aq) + 2S2O32(aq)
==> S4O62(aq) + 2I(aq)
or I2(aq) +
2Na2S2O3(aq) ==> Na2S4O6(aq) + 2NaI(aq)
(b) Starch indicator is used, starch gives a blue/black colour with iodine, this colour disappears when the last of the iodine is titrated,
so a blue to colourless sharp endpoint is observed.
(c) mole 'thio' = 0.100 x 17.6/1000 = 0.00176,
mol I2 = 0.00176
χ 2 = 0.00088 in 25 cm3,
scaling up gives 0.00088 x 1000 χ 25 =
0.0352 mol dm3 for molarity of iodine,
formula mass I2 = 2 x 127 = 254, so
concentration = 0.0352 x 254 = 8.94 g dm3
Question 18 ANSWERS:
(a) Cr2O72(aq)
+ 14H+(aq) + 6I(aq)
===> 2Cr3+(aq)
+ 3I2(aq)
+ 7H2O(l)
(b) 2S2O32(aq) + I2(aq)
===> S4O62(aq)
+ 2I(aq)
(c) mol 'thio' = 20.0 x
0.100/1000 = 0.002,
therefore from equation (b),
mol
iodine = mol 'thio'/2 = 0.001
(d) From equation (a) mol
dichromate(VI) reacting = mol iodine liberated/3 = 0.000333
(3sf)
(e) Mr(K2Cr2O7)
= 294.2
mass K2Cr2O7
titrated = 0.000333 x 294.2 = 0.0980 g (3 sf)
(f) Since the aliquot of 25.0 cm3 is
1/10th of the total solution in the flask, the total mass of the
K2Cr2O7 in original sample
dissolved in the flask solution = 10 x 0.0980g = 0.98g
and the % purity of
the K2Cr2O7 = 0.98 x 100/1.01 = 97.0 % (3 sf)
Question 19 ANSWERS:
(a) H2C2O4
+ 2NaOH ===> Na2C2O4
+ 2H2O
or: (COOH)2
+ 2NaOH ===> (COONa)2 + 2H2O
(b) moles used in the NaOH in titration =
0.20 x 8.5/1000 = 0.0017 mol NaOH
from the equation mol H2C2O4
= mol NaOH/2
therefore mol H2C2O4
= 0.0017/2 = 0.00085 in 10 cm3 of the acid solution.
volume of H2C2O4
solution = 10/1000 = 0.01 dm3
therefore molarity of H2C2O4
= 0.00085/0.01 = 0.085 mol dm-3 (0.085M)
(c) (i) involves the gain of 5 electrons,
(ii) involves the loss of 2 electrons, so to balance the electron gain and
loss you need to add together 2 x (i) plus 5 x (ii)
2 x (i):
2MnO4(aq) +
16H+(aq) + 10e
===> 2Mn2+(aq)
+ 8H2O(l)
|
5 x (ii) 5H2C2O4(aq)
10e ===> 10CO2(g) +
10H+(aq)
|
adding gives: 2MnO4(aq) +
6H+(aq) + 5H2C2O4(aq)
===> 2Mn2+(aq)
+ 8H2O(l) + 10CO2(g) |
The electrons cancel out, as does 10 of the H+
ions, giving the correctly balanced redox equation for the titration.
Triple check: 1. balance out electron change (10
electrons change). 2. The total ion charge should be the same on both
sides of the equation (4+). 3. Final atom count is the same on both
sides of the equation (2 Mn, 28 O, 16H).
So you can't go wrong!
(d) from part (c): moles H2C2O4
titrated with KMnO4 = 0.0085 in 10 cm3 of the
ethanedioic solution.
the mole ratio is 5 : 2 for
H2C2O4 : KMnO4
therefore moles of KMnO4
in the redox titration = mol H2C2O4 x
2/5
mol KMnO4 = 0.00085 x 2/5
= 0.00034 (in 8.2 cm3 = 8.2/1000 = 0.0082 dm3)
molarity KMnO4 =
0.00034/0.0082 = 0.0415 mol dm-3 (0.0415 M,
4dp, 3sf)
Question 20
ANSWERS: Lawn sand - iron analysis
(a) MnO4(aq) +
8H+(aq) + 5Fe2+(aq) ===> Mn2+(aq)
+ 5Fe3+(aq) + 4H2O(l)
(b) mol KMnO4 used in titration =
(24.50/1000) x 0.02 = 4.90 x 10-4 mol
mol Fe2+ titrated = 5
x 4.90 x 10-4 = 2.45 x 10-3 mol
(from mole ratio in redox equation above)
mass Fe2+ = 2.45 x 10-3
x 55.8 = 0.1367 g
% Fe in lawn sand = 100 x 0.02744
/ 2.50 = 5.47% Fe2+ (2dp, 3sf)
Question Q21
ANSWERS: Iron tablet analysis
(a) Cr2O72(aq)
+ 14H+(aq) + 6Fe2+(aq)
===> 2Cr3+(aq)
+ 6Fe3+(aq)
+ 7H2O(l)
(b) mol Cr2O72-
used in titration = (23.80 / 1000) x 0.02 = 4.76 x 10-4 mol
From the equation: mol Fe2+
= mol Cr2O72- x 6 = 2.856 x 10-3
mol
Mass Fe2+ = 2.856
x 10-3 x 55.8 = 0.1594 g
% Fe2+ in tablet = 100
x 0.1594 / 1.02 = 15.6 % Fe2+ (2dp, 3sf)
Question 22 ?
?

I
DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do, BUT I
AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE
EMAIL ME
ASAP TO SORT IT OUT!
what A levels do I need to study chemistry at Cambridge
University what A levels do I need to study chemistry at
Oxford University what A levels do I need to study chemistry at
Durham University what A levels do I need to study chemistry at
York University what A levels do I need to study chemistry at
Edinburgh University what A levels do I need to study chemistry
at St Andrews University what A levels do I need to study
chemistry at Imperial College London University what A levels do
I need to study chemistry at Warwick University what A levels do
I need to study chemistry at Sussex University what A levels do
I need to study chemistry at Bath University what A levels do I
need to study chemistry at Nottingham University what A levels
do I need to study chemistry at Surrey University what A levels
do I need to study chemistry at Bristol University what A levels
do I need to study chemistry at Cardiff University what A levels
do I need to study chemistry at Birmingham University what A
levels do I need to study chemistry at Manchester University
what A levels do I need to study chemistry at University College
London University what A levels do I need to study chemistry at
Strathclyde University what A levels do I need to study
chemistry at Loughborough University what A levels do I need to
study chemistry at Southampton University what A levels do I
need to study chemistry at Sheffield University what A levels do
I need to study chemistry at Glasgow University what A levels do
I need to study chemistry at Liverpool University what A levels
do I need to study chemistry at Leeds University what A levels
do I need to study chemistry at Queens, Belfast University what
A levels do I need to study chemistry at Kings College, London
University what A levels do I need to study chemistry at
Heriot-Watt University what A levels do I need to study
chemistry at Lancaster University what A levels do I need to
study chemistry at East Anglia (UEA) University what A levels do
I need to study chemistry at Newcastle University what A levels
do I need to study chemistry at Keele University what A levels
do I need to study chemistry at Leicester University what
A levels do I need to study chemistry at Bangor University
what A levels do I need to study chemistry at Nottingham Trent
University what A levels do I need to study chemistry at Kent
University what A levels do I need to study chemistry at
Aberdeen University what A levels do I need to study chemistry
at Coventry University what A levels do I need to study
chemistry at Sheffield Hallam University what A levels do I need
to study chemistry at Aston University what A levels do I need
to study chemistry at Hull University what A levels do I need to
study chemistry at Bradford University what A levels do I need
to study chemistry at Huddersfield University what A levels do I
need to study chemistry at Queen Mary, University of London
University what A levels do I need to study chemistry at Reading
University what A levels do I need to study chemistry at Glyndwr
University what A levels do I need to study chemistry at
Brighton University what A levels do I need to study chemistry
at Manchester Metropoliten University what A levels do I need to
study chemistry at De Montfort University what A levels do I
need to study chemistry at Northumbria University what A levels
do I need to study chemistry at South Wales University what A
levels do I need to study chemistry at Liverpool John Moores
University what A levels do I need to study chemistry at Central
Lancashire University what A levels do I need to study chemistry
at Kingston University what A levels do I need to study
chemistry at West of Scotland University what A levels do I need
to study chemistry at Lincoln University what A levels do I need
to study chemistry at Plymouth University what A levels do I
need to study chemistry at Greenwich University what A levels do
I need to study chemistry at
Liverpool Metropolitan University
[SEARCH
BOX]
|
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
Website content © Dr
Phil Brown 2000+. All copyrights reserved on revision notes, images,
quizzes, worksheets etc. Copying of website material is NOT
permitted. Exam revision summaries & references to science course specifications
are unofficial. |
|