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Empirical formula: CH (b) Concentrated nitric acid and concentrated sulfuric
acid. (c) (i) NO2
+
(ii) HNO3 + H2SO4 → NO2+ + HSO4
– + H2O (iii)
�
NO2O2N
� H�
NO2H�
2 (a) (i) I: Br2, II: NaOH (ii) Dye, colouring or indicator. (iii) Add phenylamine to sodium nitrite and
hydrochloric acid below 10 °C. Add this product to an alkaline solution of phenol.
(b)
C C
C
CH
H H
H
H
C
C
H
C C
C
CH
H H
H
H
C
C
H
C C
CCC CCC
• Theporbitalsoverlapaboveandbelowtheplaneof the ring structure.
• Theelectronsaredelocalisedandspreadacrossall six carbon atoms.
• Inbenzeneallthebondsarethesamelengthbetween the lengths of the single and double bonds.
Phenolismorereactivebecausetheringisactivated.Thelonepairfromtheoxygenontheringis delocalised into the ring so electrophiles are more attractedtotheringthaninbenzene.
3 (a) (i) FeBr3 or AlBr3
(ii) Halogen carrier.
(iii)
� Br2 � HBr
Br
(iv) Bromobenzene (b)
Step 1 Step 2
Br
� H�
Br� H
Intermediate
Br
�
(c) (i)
H
HH
H H
HH
H H
H H
H H
H H
H
p orbitalsoverlap
Cyclohexene
Benzene
p orbitalsoverlap togive a� bondabove andbelow thetwo carbonatoms
See diagram00.00.00.00
H
HH
H H
HH
H H
H
(ii) Inbenzenetheπelectronsarespreadoverthewhole ring structure, as they are delocalised. When cyclohexene is reacted with bromine the bromine is polarised by the double bond. However,whenbenzeneisreactedwithbromine, the bromine cannot be polarised by theringastheelectronsarespreadoverthemolecule. Electrophiles are less attracted to the benzene.Greaterenergyisneededtobreaktheπcloudinbenzenethanincyclohexene.
4 (a) (i) D = Propanone (ii) E = Propanal (b) (i) Reagent(s):2,4-dinitrophenylhydrazine.
Observation:orangeprecipitate. (ii) Reagents:Tollens’reagent.ObservationforD:
(ii) Aldehydes can be oxidised but ketones cannot. (c) (i) CH3CH=CHCH2OH (ii) Redox reaction/reduction or addition. (d) C4H6O + 5O2 → 4CO2 + 3H2O 7 (a) (i) Reagent:Tollens’reagent.Observation:silver
mirror formed. Organic product: butanoic acid. (ii) Reagent:bromine.Observation:decolourises.
Typeofreaction:electrophilicaddition. (b) Recrystallise the product in order to purify the
orange precipitate. Measure the melting points of the crystals. Compare known melting points with data tables.
8 (a)
C
H
H
C
H
H
C OH
Butan-1-ol
H C
H
H H
H
C
H
H
C
CH3
H
C OH
2-methylpropan-1-ol
H
H
H
C
H
C
OH
C H
2-methylpropan-2-ol
H
H
CH HH H
H
C
H
H
C
H
O
C H
Butan-2-ol
H
H C
H
H H
H
(b) (i) Butan-1-ol can be oxidised to produce butanal andfurtheroxidisedtogivebutanoicacid.Thisoccurs as primary alcohols can be oxidised. Butan-2-ol can be oxidised to produce butanone. No further oxidation occurs as ketones cannot be oxidised. 2-Methylpropan-2-ol cannot be oxidised as it is a tertiary alcohol and is not oxidised. 2-Methylpropan-1-ol can be oxidised to produce 2-methylpropanal and further oxidised togive2-methylpropanoicacid.
D: 2-methylpropan-1-ol. E: 2-methylpropanoic acid. (ii) (CH3)2CHCOOH + C2H5OH →
(CH3)2CHCOOC2H5 + H2O (c)
CC
H
H
C
H
H
CHO C
H
H H
H
O
O�Na�
9 (a) (i) Methyl butanoate. (ii) Heat/boil/warm/reflux with aqueous HCl or
(ii) C18H34O2 + C2H6O → C20H38O2 + H2O (c) Hydrolysis using hot aqueous HCl.
1.2 Polymers and synthesis 1 (a) (i) Hydrochloric acid. Aqueous acid heated under
reflux. (ii) Amino acids. (iii)
CC
H
H
H N
H
H
(or equivalent)
O
O H
�
(iv) A reaction with water which breaks down a compoundintosmallersub-units.Inthereactionabovethepeptidebondisbroken.
(b) (i) For stereoisomerism in compound A, there must be a chiral centre, which is a carbon atom with four different groups attached to it. Stereoisomerism occurs when there is a differentspatialarrangementofthegroups.Twomirror images that are non-superimposable are observed.
(ii) Themoleculecontainstwochiralcentresandeach chiral centre has two isomers.
(iii) Use a naturally occurring amino acid. (iv) Ahigherdosewouldberequired.Theother
stereoisomermayhavesideeffects. 2 (a) (i) RCH(NH2)COOH (ii)
(c) Itisacondensationreaction.Whenthereactionoccurs a small molecule such as water is lost.
(d) Fibres/clothing. 4 (a) (i)
H HHHHH
H3C
CC
CCC C
CH3 H3C CH3 H3C CH3
(ii) CH2
(iii)
H
C C
CH3
H
H
(b) (i) Name of functional group: peptide.
C
O
N
H
(ii) Condensation. (iii) Displayed formula of H:
O
C CC
H
H
NC
H
H
N
H O
O HH
H
Skeletal formula of H:
N
H
OH
O
H2N
O
(iv) No. of moles of glycine = 1.40/75.0 = 0.0187 moles. Expected no. of moles of dipeptide = 0.0187/2 = 9.33 × 10–3 mol. Expected mass = 9.33 × 10–3 × 132 = 1.232 g. % = (1.10/1.232) × 100. Percentage yield 89.3%
(v)
H CC
HH
H
NCl�
H O
OH
�
5 (a) Inadditionpolymerisation,theC=Cdoublebondinan unsaturated molecule (monomer), such as ethene, breaks open and the monomer adds on to agrowingpolymerchain.Themonomersaddoneat a time to form a long polymer such as poly(ethene), HDPE.
Incondensationpolymerisation,apolymersuchasPETisformedwhenmonomers–inthiscaseethane-1,2-diolandbenzene-1,4-dicarboxylicacid– join together and lose a small molecule for each bondthatforms.Inthecaseofthispolyester,awater molecule forms for each ester linkage that forms between –OH and HOOC–.
C
O
HO
H
H
H
C C
H
H
Cn C
H
H
Addition: Monomer ethene
H n
H
H
H
C C
H
H
H
O C
n
H
C O
H
Condensation
C
H
H
C OH �
H
H
C
O
C
O
C
O
OHHO
� H2O
(b) (i)
O
C C
H
H
H
C O
CH3
(ii) Reagents: HCl (aq). Conditions: Heat and reflux (iii) CH3COOH (c)
10 (a) HOOC(CH2)4COOH and H2N(CH2)6NH2. (b) condensation polymerisation – small molecule, H2O,
is eliminated. (c) Structural similarity Bothhavepeptideoramidelinks Both form H-bonds between molecules. Chemical similarity Both can be hydrolysed back to monomers by
reaction with hot aqueous acid or alkali: e.g. –CO(CH2)4CONH(CH2)6NH– →
HOOC(CH2)4COOH + H2N(CH2)6NH2 Differences Protein can be water-soluble, nylon-6,6 not. Proteins are biodegradable, nylon-6,6 is not. Nylon-6,6 has a regular chain of alternating
monomers. Proteins are made from 20 different amino acids.
2 (a) Adsorption at 1700 cm–1 shows the presence of C=O. Adsorption at 1200 cm–1 shows the presence of C–O. No broad adsorption at 2500–3500 cm–1 so does not
contain O–H. (b) ThemassspectrumshowsanM+ peak at 116 therefore 116 – 32 (2 oxygen atoms) = 84 therefore 6 carbon atoms maximum. Molecular formula of X = C6H12O2. Base peak = 57 CH3CH2C=O. NMR. peak areas show 12H atoms and four different
protonenvironments. δ = 0.9 ppm suggests CH3 next to CH2 (triplet) δ = 1.2 ppm suggests 2 × CH2 next to CH (as split
into a doublet). δ = 2.3 ppm CH2 next to CH3 (quartet) Other CH2 next to C=O δ = 4.1 ppm is CH next to 2 × CH3.
C
C
H
CH3
CH3
C
O
O
C
H
H
H
H
H
3 (a) Compound has a molecular mass of 74. Relativeempiricalformulaemass=74
(3 × 12 + 6 + 2 × 32) therefore empirical formula = molecular formula
= C3H6O2.
100
80
60
40
20
010 15 20 25 30 35 40 45 50 55 60 65 70 75
m/e
Rela
tive
inte
nsity
(b) (i) Compound X is not an aldehyde or a ketone. (ii) Compound X does not contain C=C and is not a
phenol. (c) Structure 1: ethyl methanoate. Structure 3: propanoic acid. (d) Peak at 1750 cm–1 indicates presence of C=O. Peak at 1250 cm–1 indicates presence of C–O. No peak at 2500–3500 cm–1 means it does not
contain O–H group. Itcannotbestructure3.
(e) Correct structure of compound Y:
O
C HC
H
H
OC
H
H
H
Peak at δ = 2.1 ppm is due to –CO–CH3
Peak at δ = 3.7 ppm is due to –O–CH3
Relativepeakareas1:1asbothgroupshavethesame number of carbon atoms.
Peaks show no splitting as there are no protons on the neighbouring carbon atoms.
aldehyde or ketone. Purify or recrystallise the orange precipitate. Measure the melting point of the crystals. Comparetheresultswithknownvaluesfromadatatable.
(c) (i)
H
HC
C
C
C
H
H
H
HH
OH
O
Abso
rptio
n of
ene
rgy
11 10 9 8 7 6 5 4 3 2 1 0Chemical shift/�
n.m.r. spectrum of 3-hydroxybutanone
1 1 3 3
(ii) Re-runthespectrumhavingaddedD2O to the sample.Thepeakdueto–O–Hwilldisappear.
(iii) Peak at δ = 1.4 ppm (doublet): (1:1) due to one proton on adjacent carbon atom. peak at δ = 4.3 ppm (quartet): (1:3:3:1) due to three protons on the adjacent
carbon atom. (iv) See graph answer to (c)(i). (v) Thenumberofprotonsinthesamechemical
Alcohol: 2-methylpropan-1-ol Carboxylic acid : propanoic acid. Heat the alcohol and the carboxylic acid with some
concentrated sulfuric acid in a water bath. (c) Mass spectrometry. Use the molecular ion peak, the m/zvalueforthe
peak at highest mass. (d) (i) Theδvalueorchemicalshiftindicatesthetypeof
protonorchemicalenvironment.Thenumberofpeaksgivesthenumberofdifferenttypesofproton or the different numbers of chemical environments.Therelativeratioofpeakareasshows the number of protons of each type. Splittingpatternsgiveinformationaboutthenumber of neighbouring hydrogen atoms
(protons)onthecarbonunderinvestigation:thesplitting pattern (singlet, doublet, triplet or quartet) is one greater than the number of adjacent protons.
D2O is used to identify the presence of O–H groups.
(ii)
CH3CH2COOCH2CH(CH3)2
S Q P R T
7 (a) (i) Find the m/zvalueofthepeakfurthesttotherightwith highest m/zvalue.
(ii) C2H3O2 has an empirical mass of 59. therefore 118/59 = 2 2 × empirical formula = C4H6O4. (b) (i) OH peak disappears. (ii) Peak at δ=3.3ppmidentifiesCHgroup.It
shows as a quartet splitting pattern showing presence of three protons on the adjacent carbon atom.Thereisonlyonehydrogeninthisenvironment.
Peak at δ = 1.2 ppm identifies CH3group.Itshows doublet splitting showing the presence of oneprotonontheadjacentcarbonatom.Therearethreeprotonsinthisenvironment.
8 (a) Infrared: X and YbothhaveC=Oat1680–1750cm–1. X and YbothhaveC–Oat1000–1300cm–1. X and YbothhaveO–Hinrange2500–3300cm–1. Only Y has an absorption at 3100–3500 cm–1 for N–H. Bothhavedifferentfingerprintregions. Mass spectrum: X and YhavedifferentM peaks, X = 88 and Y = 89. X and Yhavesomesimilarfragments,e.g.CH3CH+
at m/z = 28, COOH+ at 45, CHCOOH+ at 58, CH3CHCOOH+ at 73.
X and Yhavesomedifferentfragments,e.g.X has CH3CHCH3
+ at m/z = 43 or (CH3)2CHCO+ at 71; Y has H2NNCH+ at 29, H2NCHCH3
(ii) Increasingthetemperaturemovestheequilibriumto the left to counteract the effect of the increase inenergy.Theforwardreactionisexothermicandthereversereactionisendothermic.
4 (a) Kc = [Hl]2
______ [H2] [l2]
(b) (i) H2,0.14mol;I2,0.04mol;HI,0.32mol
(ii) Kc = 0.322
___________ 0.14 × 0.04
= 18.285 714 29
= 18 (to 2 sig. figs), no units (c) Kcisconstantbecauseallvolumescancelandthe
temperature is constant. Thecompositionisthesameastherearethesame
number of gas moles on either side of the equation. (d) Theforwardreactionisexothermic.Theequilibrium
pH = –log 4.59 = –0.665 (a) (i) A Brønsted–Lowry acid is a proton donor. (ii) A weak acid is partially dissociated. (iii) pH = –log[H+] (iv) A buffer solution minimises changes in pH after
addition of small amounts of both acids and alkalis.
(b) IntheequilibriumH2CO3(aq) Y H+(aq) + HCO3–(aq)
(e) 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O (f) A conjugate acid–base pair is two species differing
by H+
acid 1: HNO3; base 2: NO3–
acid 2: HCOOH2+; base 2: HCOOH
(g) (i) 6HNO3 + S → H2SO4 + 6NO2 + 2H2O or 4HNO3 + S → H2SO4 + 4NO2 + H2
(ii) TheoxidationnumberofNdecreasesfrom+5inHNO3 to +4 in NO2.
2.2 Energy 1 (a) (i) Energyisneededtoovercometheforceofattractionbetweenouterelectronsandnucleus (ii) Electronaffinityinvolvesaneutralatomgainingofanelectronbyattractionfromthenucleus (b) (i)
(iii) LatticeenthalpyofFeOismorenegativeandmore exothermic than lattice enthalpy of CaO. IonicradiusofFe2+ must be less than that of a Ca2+ion.Thisleadstogreaterattraction between Fe2+ ions and O2– ions than between Ca2+ ions and O2– ions.
2 (a) Lattice enthalpy is the enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions under standard conditions.
(c) Na+ has a larger radius than Mg2+ and Br– has a larger radius than Cl–
Na+ has a smaller charge than Mg2+
Br– experiences less attraction than Cl–
Na+ experiences less attraction than Mg2+
ThereforethereismuchlessattractionbetweenNa+ ions and Br– ions than between Mg2+ ions and Cl– ions. One mark for correct spelling, punctuation and grammar in at least two sentences.
3 (a) (i) ∆H1 = Enthalpy change of formation of magnesium oxide
∆H2 = Enthalpy change of atomisation of magnesium
∆H3 = First ionisation energy of magnesium (ii) Mg2+(g) + O2−(g) (iii) Theelectronbeinggainedisrepelledbythe
(ii) Lattice enthalpy of barium oxide is less exothermicthanthatofmagnesiumoxide.Thisis because Mg2+ has a smaller ionic radius than Ba2+ so there is a stronger attraction between thepositiveandnegativeions.
(c) Magnesium oxide has a high melting point because it has a highly exothermic lattice enthalpy resulting fromstrongattractiveforcesbetweenMg2+ and O2– ions.
8 (a) Thestandardelectrodepotentialofahalfcell,EØ, is the e.m.f. of a half cell compared with a standard hydrogen half cell, measured at 298 K with solution concentrations of 1 mol dm–3 and a gas pressure of 100 kPa (1 atmosphere).
(b) (i)
Cu(s)
Ag�(aq)Cu2�(aq)
Ag(s)
V
Salt bridge
Electrons
(ii) Direction from Cu(s) to Ag(s) shown near to wire (iii) Standard cell potential = 0.80 – 0.34 = 0.46 V
(iv) Cu + 2Ag+ → Cu2+ + 2Ag (c) EØ for Fe3+/Fe2+islesspositivethanforCl2/Cl–. ThereforetheFe2+ has a greater tendency to lose
electrons than Cl–. EØforI2/I–islesspositivethanforFe3+/Fe2+. ThereforetheI– has a greater tendency to lose
electrons than Fe2+. 9 (a) Thestandardelectrodepotentialofahalfcell,EØ,
is the e.m.f. of a half cell compared with a standard hydrogen half cell, measured at 298 K with solution concentrations of 1 mol dm–3 and a gas pressure of 100 kPa (1 atmosphere).
(b) Standard cell potential = 1.52 – 1.36 = 0.16 V (c) (i) 2MnO4
– + 10Cl– + 16H+ → 2Mn2+ + 5Cl2 + 8H2O
(ii) Cl from –1 to 0 Mn from +7 to +2 (iii) Cl– has been oxidised Mn in MnO4
– has been reduced (d) Standard electrode potential applies only to a
concentration of 1 mol dm–3,andpotentialsgivenoinformation about rates. When concentration of Cl–decreases,equilibriummovestowardsCl–, and Cl2/Cl–halfcellpotentialbecomeslessnegative.Thedifference between the electrode potentials becomes smaller and the Cl2/Cl–potentialmayevenbecomemorepositivethantheMnO4/Mn2+ potential.Thereactionwillthennotbefeasible.
10 (a) (i)
Cr(s)
Cd2�(aq) (1 mol dm�3)Cr3�(aq) (1 mol dm�3)
Cd(s)
T � 298 KV
Salt bridge
Electrons
(ii) Direction from Cr(s) to Cd(s) shown near to wire (iii) Cr half cell EØvalueismorenegative. (b) 2Cr(s) + 3Cd2+(aq) → 2Cr3+ (aq) + 3Cd(s) (c) (i) Standard cell potential = –0.40 – (–0.74) = 0.34 V (ii) Concentration of Cr3+ decreases so equilibrium
movestowardsCr3+ and Cr half-cell becomes morenegative.Thereforethereisabiggerdifference between the electrode potentials and the cell potential will increase.
11 (a) Thestandardelectrodepotentialofahalf-cell,EØ, is the e.m.f. of a half-cell compared with a standard hydrogen half-cell, measured at 298 K with solution concentrations of 1 mol dm–3 and a gas pressure of 100 kPa (1 atmosphere).
solid B = platinum electrode (ii) Direction from hydrogen half-cell along wire to
S2O82– half-cell
(iii) S2O82– + H2 → 2SO4
2– + 2H+
(c) molar mass of Na2S2O8 = 238.2 g mol–1
molar mass of Na2SO4 = 142.1 g mol–1
∴ 23.82 g of Na2S2O8 and 14.21g of Na2SO4
(d) Concentration of SO42– decreases so equilibrium
movestowardsSO42– and S2O8
2–/SO42– half-cell
becomesmorepositive.Thereforetherewillbeabigger difference between electrode potentials and the cell potential will increase.
12 (a) (i) moles of Cu = 0.68 × 5/1000 = 0.0034 mass of Cu = 0.0034 × 63.5 = 0.216 g % Cu = 0.216/0.28 = 77% (ii) Ratios Cu : N : O : H = 26.29/63.5 : 11.6/14 : 59.63/16 : 2.48/1 = 0.414 : 0.829 : 3.73 : 2.48 = 1 : 2 : 9 : 6 Empirical formula of A = CuN2O9H6 (iii) Formula usually shown as: Cu(NO3)2·3H2O (b) Cu → Cu2+: Cu from 0 to +2 NO3
– → NO: N from +5 to +2 ratio of Cu : N in equation is 3 : 2 3Cu + 8H+ + 2NO3
– → 3Cu2+ + 2NO + 4H2O (c) (i) moles of A = 90/24000 = 3.75 × 10–3
Mr of A = 0.24/ 3.75 × 10–3 = 64 GasisSO2
(ii) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
2.3 Transition elements 1 (a) Addaqueoussodiumhydroxidewhichgivesarusty
∴ mass FeSo4·7H2O = 2.1168 × 10–3 × 277.9 = 0.588 g
∴ % purity = 0.588 ______ 0.655
× 100 = 89.8%
2 (a) (i) Blue to green to yellow solution. (ii) Thelonepaironthechlorideionisdonatedto
thecopper(II)ion.Aco-ordinate(dativecovalent)bond is formed.
(b) A light blue precipitate forms, which is soluble in excess ammonia forming a deep blue solution.
(c) Ammonia molecule has one lone pair of electrons andthreebondingpairs.Thelonepairrepelsthebonding pairs more than they repel each other, reducing the bond angle. Each ammonia ligand bonded to copper has four bond pairs. So equal repulsion between bonding pairs. Angle larger, at 109.5.
3 Transitionelementsaredblockelementsthathaveanion with an incomplete d sub-shell.
Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9
Note that the d sub-shell is incomplete.
2�
Cu
H2O
H2O OH2
OH2
OH2
OH2
Theshapeistetrahedral. Water molecules are ligands (electron-pair donors) and
theseformco-ordinatebondstothecentralmetal.TheCu2+ ion is an electron-pair acceptor.
• Transitionmetalshavevariableoxidationstates,forexample Cu2+ and Cu+.
• Transitionmetalsformcolouredcompounds:CuCl42– is yellow. [Cu(H2O)6]2+ is blue.
• Transitionmetalsandtheirionsactascatalysts.Ironis a catalyst for the Haber process.
4 (a) 1s2 2s2 2p6 3s2 3p6 3d5
Ironhasanincompletedsub-shellintheFe3+ion.Itis a transition metal.
(b) 1 Variable oxidation state. 2 Forms coloured compounds. (c) Fe2+ reacts with hydroxide ions to form a green
precipitate whereas Fe3+ reacts with hydroxide to form an orange-brown precipitate.
(d) 4Fe2+ + O2 + 4H+ → 4Fe3+ + 2H2O (e) (i) Copper may react with MnO4
–. Copper reacts with Fe3+ reducing it back to Fe2+. (ii) Mn in MnO4
– changes from oxidation state +7 to +2 in Mn2+ (it is reduced). Fe changes from Fe2+ (+2) to Fe3+ (+3) and is oxidised.
(c) Waterhas2lonepairsand2bondingpairs.Thetwo lone pairs repel the bonding pairs more than the bond pairs repel each other. Bond angle is 104.5º but in the ligand a lone pair is used to form a co-ordinate bond to the metal. Water ligand has 3 bondpairsandonelonepair.Thereislessrepulsion. So the bond angle is greater than in a water molecule.
12 (a) (i) From pink to blue (ii) Tetrahedral (iii) Ligand substitution (b)
N
H
C
H
H
C
H
H
N
H
HH
Co
NH2
NH2
NH2
NH2
H2N
H2NCo
H2N
H2N
H2N
NH2
NH2
NH2
3� 3� (c) (i) Optical isomerism (ii)
N
H
C
H
H
C
H
H
N
H
HH
Co
NH2
NH2
NH2
NH2
H2N
H2NCo
H2N
H2N
H2N
NH2
NH2
NH2
3� 3�
13 (a) 1s2 2s2 2p6 3s2 3p6 3d6; Fe2+ has an incompletely filled d subshell.
(b) (i) [Fe(H2O)6]2+ (ii)
2�
Fe
H2O
H2O OH290°
OH2
OH2
OH2
(iii) Each ligand donates an electron pair to the metaliontoformacoordinate(dativecovalent)bond.
(c) [Fe(H2O)6]3+ + 4Cl– Y [FeCl4]2+ + 6H2O (d) FeCl2givesagreenprecipitateandFeCl3givesa
brown-red precipitate. Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) (e) Cr changes from +3 to +6, which is oxidation; Fe changes from +6 to +3, which is reduction. (f) 4FeO4
2– + 20H+ → 4Fe3+ + 3O2 + 10H2O
14 (a) MoO3 + 2Al → Mo + Al2O3
(b) [Kr]4d3, which has an incompletely filled d sub-shell, meaning it is a transition metal ion.
(c) Cr2O72–(aq) + 2H+(aq) + 3MoO2(s) →
3MoO42–(aq) + 2Cr3+(aq) + H2O(l)
(d) (i) K2FeO4
(ii) Moles of KOH = 4.00 × 10.0/1000 = 0.0400 mol Moles of Fe2O3 = 1.00/159.6 = 0.00627 mol From the equation, 1 mol Fe2O3 reacts with