chem130 (f 03) review questions for midterm i page - 1 · chem130 (f 03) review questions for...

CHEM*130 (F 03) REVIEW QUESTIONS FOR MIDTERM I PAGE - 1

USEFUL FORMULAS AND CONVERSIONS

1. For solid:n(x) '

mass (x)M(x)

(units: gg mol &1

)

2. For solution:c(x) '

n(x)V(x)

(units: molL

)

3.ppm (w/w) '

1 g106 g

'1 g

1 million g'

1 µg1 g

'1 mg1 kg

ppm (w/v) '1 g

1 million mL'

1 g106 mL

'1 mg

103 mL'

1 mg1 L

ppb ' 1000 ppt ' 1 × 10!3 ppm '1 µg1 L

'1 × 10!6 g1 × 103 mL

'1 g

109 mL

4.% yield '

actual yieldtheoretical yield

× 100

5.∆Ho(reaction) = 3 n ∆Ho

f (products) ! 3 n ∆Hof (reactants)


PART AQUESTION 1

To find: Convert chemical formulas into chemical names

Chemical Formula Cation Anion Chemical Name

(a) K2Cr2O7 2 K+ Cr2O72! potassium dichromate

(b) NiSO4@6H2O Ni2+ SO42! nickel (II) sulfate hexahydrate

(c) Ag2S 2 Ag+ S2! silver (I) sulfide

(d) N2O5 molecule (no ions) dinitrogen pentaoxide

(e) (NH4)2SO3 2 NH4+ SO3

2! ammonium sulfite

(f) Fe(MnO4)2 Fe2+ 2 MnO4! iron (II) permanganate

or ferrous permanganate

(g) ClO2 molecule (no ions) chlorine dioxide

(h) Ba(HCO3)2 Ba2+ 2 HCO3! barium hydrogen carbonate

or barium bicarbonate

PART AQUESTION 2

To find: Write the chemical formulas from chemical names

Chemical Name Cation Anion Chemical Formula

(a) potassium hydrogen sulfate K+ HSO4! KHSO4

(b) lithium oxide 2 Li+ O2! Li2O

(c) magnesium chlorate Mg2+ 2 ClO3! Mg(ClO3)2

(d) manganese (II) nitrite Mn2+ 2 NO2! Mn(NO2)2

(e) aluminum carbonate 2 Al3+ 3 CO32! Al2(CO3)3

(f) copper (I) chromate 2 Cu+ CrO42! Cu2CrO4

(g) mercury (II) iodide Hg2+ 2 I! HgI2

(h) dibromine trioxide molecule (no ions) Br2O3


Avogadro’s number = 6.022 × 1023 atoms mol!1

301 mg325 g

'301 mg325 mL

'301 mg

325 × 10!3 L' 926.1538 ppm ' 926 ppm (3 s.f.)

Combustion: a complete OXIDATION of a compound via the use of air (or O2)the final products of organic compounds are usually CO2(g) and H2O(R)

C8H10N4O2(s) + ? O2(g) ÷ 8 CO2(g) + 5 H2O(R) + ...

PART AQUESTION 3

Given: 325 g coffee solution301 mg caffeine (C8H10N4O2, 194 g mol!1)density of coffee = 1 g mL!1

(a) the number of moles of caffeine present

n(x) 'mass (x)

M(x)n(caffeine) '

mass (caffeine)M(caffeine)

'301 × 10!3g194 g mol &1

= 1.5515 × 10!3 mol = 1.55 × 10!3 mol (3 s.f.)

(b) the number of N atoms in the cup from the caffeine (C8H10N4O2, 194 g mol!1)

1 mole caffeine contains 4 moles of N atoms in caffeinen (N atoms in caffeine) = n(caffeine) × (4 moles of N atoms in caffeine ÷ 1 mole caffeine)

= (1.5515 × 10!3 mol) × (4)

ˆ # N atoms in the cup from the caffeine= n(N atoms in caffeine) × (6.022 × 1023 atoms mol!1)= [(1.5515 × 10!3 mol) × (4)] × (6.022 × 1023 atoms mol!1)= 3.7373 × 1021 atoms= 3.74 × 1021 atoms (3 s.f.)

(c) the concentration of the caffeine in the cup in ppm1 ppm ' 1 mg L !1

(d) the number of moles of H2O produced on complete combustion of 301 mg caffeine

1 mole C8H10N4O2(s) produces 5 moles of H2O(R)ˆ n(H2O) produced = n(caffeine) × (5 moles H2O ÷ 1 mole caffeine)

= (1.5515 × 10!3 mol) × (5)= 7.7575 × 10!3 mol= 7.76 × 10!3 mol (3 s.f.)


PART AQUESTION 4

Given: FeCl3 (ferric chloride or iron (III) chloride)c(FeCl3) = 0.108 mol L!1

(a) mass of FeCl3 in 1.0 L solution

c(x) 'n(x)V(x)

(units: molL

) n(x) 'mass (x)

M(x)

n(FeCl3) = c(FeCl3) × V(FeCl3) = (0.108 mol L!1) × (1.0 L) = 0.108 mol

M(FeCl3) = [55.847 + 3 (35.453)] g mol!1 = 162.206 g mol!1

ˆ mass (FeCl3) = n(FeCl3) × M(FeCl3)= (0.108 mol) × (162.206 g mol!1)= 17.518 g= 18 g (2 s.f.)

(b) mmol of Cl! in 242 mL of FeCl3 solution

n(FeCl3) = c(FeCl3) × V(FeCl3) = (0.108 mol L!1) × (242 × 10!3 L) = 2.6136 × 10!2 mol

1 mole of FeCl3 contains 3 moles of Cl!

ˆ n(Cl!) = n(FeCl3) × (3 moles Cl! ÷ 1 mole FeCl3)= (2.6136 × 10!2 mol) × (3)= 7.8408 × 10!2 mol= 78.4 × 10!3 mol (3 s.f.)

= 78.4 mmol

(c) To prepare 1.50 L of a 2.50 mmol L!1 FeCl3 solution

n(FeCl3) = c(FeCl3) × V(FeCl3) = (2.50 mmol L!1) × (1.50 L) = 3.75 mmol

ˆ volume of 0.108 mol L!1 FeCl3 required = n(FeCl3) ÷ c(FeCl3)= (3.75 mmol) ÷ (0.108 mol L!1)= 34.722 mL= 34.7 mL (3 s.f.)

34.7 mL 0.108 mol L!1 FeCl3

+) 1465.3 mL water 1500.0 mL Total volume (1.50 L)


(d) To prepare 0.500 L of a 0.268 mol L!1 Cl! solution

c(x) 'n(x)V(x)

(units: molL

)

n(Cl!) = c(Cl!) × V(Cl!) = (0.268 mol L!1) × (0.500 L) = 0.134 mol

1 mole of FeCl3 contains 3 moles of Cl!

n(FeCl3) = n(Cl!) × (1 mole FeCl3 ÷ 3 moles Cl!)= (0.134 mol) × (a)

ˆ volume of 0.108 mol L!1 FeCl3 required = n(FeCl3) ÷ c(FeCl3)= [(0.134 mol) × (a)] ÷ (0.108 mol L!1)= 0.41358 L= 414 mL (3 s.f.)

414 mL 0.108 mol L!1 FeCl3

+) 86 mL water 500.0 mL Total volume (0.500 L)

PART AQUESTION 5

Given: 3.5 mL of 2.4 ppm Pb2+ diluted to 105 mL total volume(i.e. 3.5 mL ÿ 2.4 ppm Pb2+

105 mL ÿ ? ppm Pb2+)

(a) the concentration of Pb2+ in ppm

(2.4 ppm) (3.5 mL ÷ 105 mL) = 0.080 ppm (2 s.f.)

(b) the concentration of Pb2+ in ppb1 ppm = 103 ppb

(0.080 ppm) (103 ppb ÷ 1 ppm) = 0.080 × 103 ppb = 80 ppb (2 s.f.)


PART AQUESTION 6

Given: 1.234 g CaF2 (calcium fluoride)in 234 mL pure water

M(CaF2) = [40.08 + 2 (18.998)] g mol!1 = 78.076 g mol!1

n(x) 'mass (x)

M(x)

n(CaF2) present = mass(CaF2) ÷ M(CaF2)= (1.234 g) ÷ (78.076 g mol!1)= 1.58051 × 10!2 mol

(a) the fluoride concentration in mol L!1

1 mole of CaF2 contains 2 moles of F! ions

n(F!) present = n(CaF2) × (2 moles F! ÷ 1 mole CaF2)= (1.58051 × 10!2 mol) × (2)= 3.16102 × 10!2 mol

ˆ c(F!) in mol L!1 = n(F!) ÷ V(F!) = (3.16102 × 10!2 mol) ÷ (234 × 10!3 L) = 0.13509 mol L!1

= 0.135 mol L!1 (3 s.f.)

(b) the fluoride concentration in ppm

In aqueous solution, 1 ppm = 1 mg L!1

M(F!) = 18.998 g mol!1 = 18.998 × 103 mg mol!1

ˆ c(F!) in ppm = (0.13509 mol L!1) × (18.998 × 103 mg mol!1)= 2566.372 mg L!1

= 2.57 × 103 ppm (3 s.f.)


Residence time (τ) 'amount of substance in the reservoir

rate of inflow to (outflow from) reservoir

units: τ 'g, moles, mol L !1, t

g s !1, moles min!1, mol L !1 s !1, t year !1

109.5 days 'the total mass of Hg in all components of the body

6.0 mg day !1

residence time (τ) of Hg in the body 'the total mass of Hg in all components of the body

constant intake of Hg in the diet

PART AQUESTION 7

Given: residence time (τ) of Hg in the body= 0.30 years= (0.30 years) (365 days / year)= 109.5 days

constant intake of Hg in the diet = 6.0 mg day!1 (i.e. the rate)

To find: the total mass of Hg in all compartments of the body in grams

ˆ the total mass of Hg in all compartments of the body= (residence time (τ) of Hg in the body) × (constant intake of Hg in the diet)= (109.5 days) × (6.0 mg day!1)= 657 mg= 657 × 10!3 g= 0.657 g= 0.66 g (2 s.f.)


volume conversion: 1 L = 1 dm3 = 10!3 m3 = 103 cm3

1 kmol = 103 mol

PART AQUESTION 8

Given: 5.0 × 102 m3 of 1.0 M solution of Fe2(SO4)3, iron (III) sulfate

To find: the number of kmol of Fe3+

1 mole of Fe2(SO4)3 contains 2 moles of Fe3+

c(Fe3+) = c(Fe2(SO4)3) × (2 moles Fe3+ ÷ 1 mole Fe2(SO4)3)= (1.0 mol L!1) × (2)= 2.0 mol L!1

V(Fe3+) = (5.0 × 102 m3)= (5.0 × 102 m3) (103 L / m3)= 5.0 × 105 L

c(x) 'n(x)V(x)

(units: molL

)

In 5.0 × 102 m3 (i.e. 5.0 × 105 L) of 1.0 M Fe2(SO4)3 solution,

ˆ n(Fe3+) present = c(Fe3+) × V(Fe3+)= (2.0 mol L!1) × (5.0 × 105 L)= 1.0 × 106 mol= (1.0 × 106 mol) (10!3 kmol / mol)= 1.0 × 103 kmol (2 s.f.)


PART AQUESTION 9

Given: C 32.43%H 8.163%N 37.81%O 21.60%M(NDMA) = 2.2 × 102 g mol!1

(a) the simplest molecular formula

Assuming the percentages are equal to the masses of C, H, N and O in a 100 g sample of thecompound,

n(C) = mass(C) ÷ M(C) = 32.43 g ÷ 12.011 g mol!1 = 2.700 moln(H) = mass(H) ÷ M(H) = 8.163 g ÷ 1.0079 g mol!1 = 8.099 moln(N) = mass(N) ÷ M(N) = 37.81 g ÷ 14.007 g mol!1 = 2.699 moln(O) = mass(O) ÷ M(O) = 21.60 g ÷ 15.999 g mol!1 = 1.350 mol

C : H : N : O = 2.700 : 8.099 : 2.699 : 1.350Dividing through by the smallest of these numbers (i.e. 1.350) gives:

C : H : N : O = 2 : 6 : 2 : 1

ˆ The most likely simplest molecular formula is: C2H6N2O

(b) the true molecular formula

M(C2H6N2O) = [(2 × 12.011) + (6 × 1.0079) + (2 × 14.007) + (15.999)] g mol!1 = 74.0824 g mol!1

M(NDMA) = 2.2 × 102 g mol!1

the number of units of the simplest formula in the molecular formula= M(NDMA) ÷ M(C2H6N2O)= (2.2 × 102 g mol!1) ÷ (74.0824 g mol!1)= 2.97 (. 3)

ˆ The molecular formula of NDMA is: (C2H6N2O)3 ! C6H18N6O3

(c) the precise molar mass for NDMA

M(NDMA) [or M(C6H18N6O3)] (molar masses obtained from the Periodic Table)= [(6 × 12.011) + (18 × 1.0079) + (6 × 14.007) + (3 × 15.999)] g mol!1

= 222.2472 g mol!1

= 222.25 g mol!1


PART AQUESTION 10

To find: Balance the chemical equations

(a) Mg3N2(s) + H2O(R) ÷ Mg(OH)2(s) + NH3(g)Step 1: Mg3N2(s) + H2O(R) ÷ 3 Mg(OH)2(s) + 2 NH3(g)Step 2: Mg3N2(s) + 6 H2O(R) ÷ 3 Mg(OH)2(s) + 2 NH3(g)

(b) NH3(g) + O2(g) ÷ N2(g) + H2O(R)Step 1: 2 NH3(g) + O2(g) ÷ N2(g) + H2O(R)Step 2: 2 NH3(g) + O2(g) ÷ N2(g) + 3 H2O(R)Step 3: 2 NH3(g) + 3/2 O2(g) ÷ N2(g) + 3 H2O(R)Step 4: 4 NH3(g) + 3 O2(g) ÷ 2 N2(g) + 6 H2O(R)

(c) Cu(s) + HNO3(aq) ÷ 3 Cu(NO3)2(aq) + NO(g) + H2O(R)Step 1: 3 Cu(s) + HNO3(aq) ÷ 3 Cu(NO3)2(aq) + NO(g) + H2O(R)Step 2: 3 Cu(s) + HNO3(aq) ÷ 3 Cu(NO3)2(aq) + 2 NO(g) + H2O(R)Step 3: 3 Cu(s) + 8 HNO3(aq) ÷ 3 Cu(NO3)2(aq) + 2 NO(g) + H2O(R)Step 4: 3 Cu(s) + 8 HNO3(aq) ÷ 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(R)

PART AQUESTION 11

Given: reactants: aluminum metal Al(s)perchloric acid HClO4(aq)

products: aluminum (III) perchlorate Al(ClO4)3(aq)hydrogen gas H2(g)

Step 1: Al(s) + HClO4(aq) ÷ Al(ClO4)3(aq) + H2(g)Step 2: Al(s) + 3 HClO4(aq) ÷ Al(ClO4)3(aq) + H2(g)Step 3: Al(s) + 3 HClO4(aq) ÷ Al(ClO4)3(aq) + 3/2 H2(g)Step 4: 2 Al(s) + 6 HClO4(aq) ÷ 2 Al(ClO4)3(aq) + 3 H2(g)

ˆ The complete reaction is:2 Al(s) + 6 HClO4(aq) ÷ 2 Al(ClO4)3(aq) + 3 H2(g)

The coefficients of H2(g) and HClO4(aq) are 3 and 6 , respectively.


PART AQUESTION 12

Given: 9.37 g of C2H6O(R), limiting reactantexcess O2(g)

To write: combustion equation for C2H6O(R)

To find: grams of H2O(R) produced upon complete combustion

The combustion reaction is:

C2H6O(R) + 3 O2(g) ÷ 2 CO2(g) + 3 H2O(R) ethanol (in excess)

9.37 g (L.R.)

M(C2H6O) = [(2 × 12.011) + (6 × 1.0079) + (15.999)] g mol!1 = 46.0684 g mol!1

n(x) 'mass (x)

M(x)

n(C2H6O) oxidized = mass(C2H6O) ÷ M(C2H6O)= 9.37 g ÷ 46.0684 g mol!1

= 0.20339 mol= n(limiting reactant)

1 mole C2H6O(R) produces 3 moles of H2O(R)

n(H2O) produced = n(C2H6O) × (3 moles H2O ÷ 1 mole C2H6O)= (0.20339 mol) × (3)= 0.61018 mol

M(H2O) = [(2 × 1.0079) + (15.999)] g mol!1 = 18.0148 g mol!1

ˆ mass(H2O) produced = n(H2O) × M(H2O)= (0.61018 mol) × (18.0148 g mol!1)= 10.9923 g= 11.0 g (3 s.f.)


% yield 'actual yield

theoretical yield× 100%

PART AQUESTION 13

Given: 300 g of Cr2O3(s) (limiting reactant)excess C(s)191 g Cr(s) obtained

To find: % yield of Cr(s)

Cr2O3(s) + 3 C(s) ÷ 2 Cr(s) + 3 CO(g) 300 g (in excess) 191 g (L.R.)

n(x) 'mass (x)

M(x)

M(Cr2O3) = [(2 × 51.996) + (3 × 15.999)] g mol!1 = 151.989 g mol!1

n(Cr2O3) = mass(Cr2O3) ÷ M(Cr2O3)= (300 g) ÷ (151.989 g mol!1)= 1.9738 mol= n(limiting reactant)

1 mole of Cr2O3(s) produces 2 moles of Cr(s)

(1) n(Cr) produced theoretically = n(Cr2O3) × (2 moles Cr ÷ 1 mole Cr2O3)= (1.9738 mol) × (2)= 3.9476 mol

(2) M(Cr) = 51.996 g mol!1

actual n(Cr) obtained = mass(Cr) ÷ M(Cr)= (191 g) ÷ (51.996 g mol!1)= 3.67336 mol

ˆ % yield = [(3.67336 mol) ÷ (3.9476 mol)] × 100%= 93.053%= 93.1% (3 s.f.)


1 t = 106 g

PART AQUESTION 14

Given: 1.00 Mg (1.00 × 106 g = 1.00 ton) ore containing Cu2O(s)76.0% purity of Cu2O(s)

To find: mass of Cu(s) produced in tonnes (t)

mass of pure Cu2O in ore = (1.00 × 106 g) × (76.0%) = 7.60 × 105 g

M(Cu2O) = [2 (63.546) + 15.999] g mol!1 = 143.091 g mol!1

n(x) 'mass (x)

M(x)

n(Cu2O) = mass(Cu2O) ÷ M(Cu2O)= (7.60 × 105 g) ÷ 143.091 g mol!1

= 5.3113 × 103 mol

Based on the reaction,Cu2O(s) + H2(g) ÷ 2 Cu(s) + H2O(R)

1 mole of Cu2O(s) produces 2 moles of Cu(s)

n(Cu) produced = n(Cu2O) × (2 moles Cu ÷ 1 mole Cu2O)= (5.3113 × 103 mol) × (2)= 1.0623 × 104 mol

M(Cu) = 63.546 g mol!1

ˆ mass(Cu) produced = n(Cu) × M(Cu)= (1.0623 × 104 mol) × (63.546 g mol!1)= 6.7502 × 105 g= 0.67502 × 106 g= 0.675 t (3 s.f.)


PART AQUESTION 15

Given: The following three unbalanced sequential reactions: Reaction Î: N2(g) + H2(g) ÷ NH3(g) Reaction Ï: NH3(g) + O2(g) ÷ NO(g) + H2O(R) Reaction Ð: NO(g) + H2O(R) + O2(g) ÷ HNO3(aq)start with 2.4 mol of H2(g)

To find: the theoretical yield of HNO3(aq) in moles

The balanced sequential reactions are:Reaction Î: N2(g) + 3 H2(g) ÷ 2 NH3(g)Reaction Ï: 4 NH3(g) + 5 O2(g) ÷ 4 NO(g) + 6 H2O(R)Reaction Ð: 4 NO(g) + 2 H2O(R) + 3 O2(g) ÷ 4 HNO3(aq)

From Reaction Î: 3 moles of H2(g) produces 2 moles of NH3(g)

n(NH3) = n(H2) × (2 moles NH3 ÷ 3 moles H2)= (2.4 mol) × (2/3)= 1.6 mol

From Reaction Ï: 4 moles of NH3(g) produces 4 moles of NO(g)

n(NO) = n(NH3) × (4 moles NO ÷ 4 moles NH3)= (1.6 mol) × (4/4)= 1.6 mol

From Reaction Ð: 4 moles of NO(g) produces 4 moles of HNO3(aq)

n(HNO3) = n(NO) × (4 moles HNO3 ÷ 4 moles NO)= (1.6 mol) × (4/4)= 1.6 mol= 1.6 mol (2 s.f.)


PART AQUESTION 16

Given: The following three sequential reactions: Reaction Î: Cl2(g) + 2 KOH ÷ KCl + KClO + H2O Reaction Ï: 3 KClO ÷ 2 KCl + KClO3

Reaction Ð: 4 KClO3 ÷ 3 KClO4 + KClstart with 1.00 kg (1.00 × 103 g) of KClO4 product

To find: mass of Cl2(g) in kg

n(x) 'mass (x)

M(x)

(a) assume 100% efficient in all the reactions

M(KClO4) = [39.098 + 35.453 + 4 (15.999)] g mol!1 = 138.547 g mol!1

n(KClO4) = mass(KClO4) ÷ M(KClO4)= (1.00 × 103 g) ÷ (138.547 g mol!1)= 7.21777 mol

From Reaction Ð: n(KClO3) = n(KClO4) × (4 moles KClO3 ÷ 3 moles KClO4)= (7.21777 mol) × (4/3)= 9.62369 mol

From Reaction Ï: n(KClO) = n(KClO3) × (3 moles KClO ÷ 1 mole KClO3)= (9.62369 mol) × (3)= 28.8711 mol

From Reaction Î: n(Cl2) = n(KClO) × (1 mole Cl2 ÷ 1 mole KClO)= (28.8711 mol) × (1)= 28.8711 mol

M(Cl2) = [2 (35.453)] g mol!1 = 70.906 g mol!1

ˆ mass(Cl2) required = n(Cl2) × M(Cl2)= (28.8711 mol) × (70.906 g mol!1)= 2.04713 × 103 g= 2.05 kg (3 s.f.)

(b) assume only 75% efficient in the last stepThe starting material of Cl2(g) must be 4/3 (i.e. 1/75%) the amount in order to yield 75%efficient, thus the mass of Cl2(g) required is: (2.05 kg ÷ 75%) = 2.73 kg.


PART AQUESTION 17

Given: 4.21 g of H2(g)26.7 g of O2(g)

To find: mass of all substances left after the reaction

M(H2) = [2 (1.0079)] g mol!1 = 2.0158 g mol!1

M(O2) = [2 (15.999)] g mol!1 = 31.998 g mol!1

M(H2O) = [2 (1.0079) + 15.999] g mol!1 = 18.0148 g mol!1

n(x) 'mass (x)

M(x)

2 H2(g) + O2(g) ÷ 2 H2O(R)Initial mass(x): 4.21 g 26.7 g !

M(x): 2.0158 g mol!1 31.998 g mol!1 18.0148 g mol!1

n(x): 2.0885 mol 0.8344 mol !

ratio: 2.0885 mol ÷ 2 moles 0.8344 mol ÷ 1 mol != 1.04425 = 0.8344

(in excess) (ˆ L.R.)

Change s 2× 0.8344 mol s 0.8344 mol r 2× 0.8344 mol

At equilibrium

n(x): [2.0885 ! (2× 0.8344)] mol (0.8344 ! 0.8344) mol (2× 0.8344) mol= 0.4197 mol = 0 = 1.6688 mol

M(x): 2.0158 g mol!1 31.998 g mol!1 18.0148 g mol!1

mass(x): 0.8460 g 30.063 g0.85 g (2 s.f.) ZERO 30.1 g (3 s.f.)

H2(g) leftover H2O(R) produced


PART AQUESTION 18

1. All salts in aqueous solution dissociate into ions (cations and anions).2. Insoluble salts (solids) are written as the electrically neutral formula.3. Covalent substances (e.g. CO2(g), H2O(R)) do not dissociate in aqueous solution.

(a) AgNO3(aq) + Na2S(aq) ÷ Ag2S(s) + NaNO3(aq)

Step 1: Balance the reaction2 AgNO3(aq) + Na2S(aq) ÷ Ag2S(s) + 2 NaNO3(aq)

Step 2: Cancel out the spectator ions2 Ag+(aq) + 2 NO3

!(aq) + 2 Na+(aq) + S2!(aq) ÷ Ag2S(s) + 2 Na+(aq) + 2 NO3!(aq)

Step 3: Write the balanced NIE2 Ag+(aq) + S2!(aq) ÷ Ag2S(s)

(b) Ba(ClO4)2(aq) + (NH4)2SO4(aq) ÷ NH4ClO4(aq) + BaSO4(s)

Step 1: Balance the reactionBa(ClO4)2(aq) + (NH4)2SO4(aq) ÷ 2 NH4ClO4(aq) + BaSO4(s)

Step 2: Cancel out the spectator ionsBa2+(aq) + 2 ClO4

!(aq) + 2 NH4+(aq) + SO4

2!(aq) ÷ 2 NH4+(aq) + 2 ClO4

!(aq) + BaSO4(s)Step 3: Write the balanced NIE

Ba2+(aq) + SO42!(aq) ÷ BaSO4(s)

(c) Al2(SO3)3(s) + HNO3(aq) ÷ Al(NO3)3(aq) + SO2(g) + H2O(R)

Step 1: Balance the reactionAl2(SO3)3(s) + 6 HNO3(aq) ÷ 2 Al(NO3)3(aq) + 3 SO2(g) + 3 H2O(R)

Step 2: Cancel out the spectator ionsAl2(SO3)3(s) + 6 H+(aq) + 6 NO3

!(aq) ÷ 2 Al3+(aq) + 6 NO3!(aq) + 3 SO2(g) + 3 H2O(R)

Step 3: Write the balanced NIEAl2(SO3)3(s) + 6 H+(aq) ÷ 2 Al3+(aq) + 3 SO2(g) + 3 H2O(R)

(d) HF(aq) + Na2CO3(aq) ÷ NaF(aq) + H2O(R) + CO2(g)

Step 1: Balance the reaction2 HF(aq) + Na2CO3(aq) ÷ 2 NaF(aq) + H2O(R) + CO2(g)

Step 2: Cancel out the spectator ions2 HF(aq) + 2 Na+(aq) + CO3

2!(aq) ÷ 2 Na+(aq) + 2 F!(aq) + H2O(R) + CO2(g)Step 3: Write the balanced NIE

2 HF(aq) + CO32!(aq) ÷ 2 F!(aq) + H2O(R) + CO2(g)


OR

Note that the net ionic equation (NIE) for all neutralization reactions of astrong acid (e.g. HCl, HBr, HI, H2SO4, HNO3, and HClO4) and a strong base(e.g. NaOH, KOH, Ca(OH)2 etc.) with production of water is written as:

H+(aq) + OH!(aq) ÷ H2O(R)

H3O+(aq) + OH!(aq) ÷ 2 H2O(R)

PART AQUESTION 19

Given: c(KOH) = 0.2532 M c(H2SO4) = 0.1076 MV(KOH) = 50.00 mL

To find: V(H2SO4) in mL

c(x) 'n(x)V(x)

(units: molL

)

n(KOH) present = c(KOH) × V(KOH)= (0.2532 mol L!1) × (50.00 × 10!3 L)= 1.266 × 10!2 mol

The complete neutralization reaction is:H2SO4(aq) + 2 KOH(aq) ÷ K2SO4(aq) + 2 H2O(R)

1 mole of H2SO4(aq) reacts with 2 moles of KOH(aq)

n(H2SO4) required = n(KOH) × (1 mole H2SO4 ÷ 2 moles KOH)= (1.266 × 10!2 mol) × (1/2)= 6.330 × 10!3 mol

ˆ V(H2SO4) needed = n(H2SO4) ÷ c(H2SO4)= (6.330 × 10!3 mol) ÷ (0.1076 mol L!1)= 5.8829 × 10!2 L= 58.83 mL (4 s.f.)


Note that the net ionic equation (NIE) for all neutralization reaction of astrong base (e.g. NaOH, KOH, Ba(OH)2 etc.) and a strong acid (e.g. HCl, HBr,HI, HNO3, H2SO4 and HClO4) with production of water is written as:

H+(aq) + OH!(aq) ÷ H2O(R)

PART AQUESTION 20

Given: c(HCl) = 0.3200 MV(HCl) = 31.50 mL V(Ba(OH)2) = 25.00 mL

To find: c(Ba(OH)2)

c(x) 'n(x)V(x)

(units: molL

)

n(HCl) present = c(HCl) × V(HCl)= (0.3200 mol L!1) × (31.50 × 10!3 L)= 1.008 × 10!2 mol

The complete titration reaction is:Ba(OH)2(aq) + 2 HCl(aq) ÷ BaCl2(aq) + 2 H2O(R)

1 mole of Ba(OH)2(aq) reacts with 2 moles of HCl(aq)

n(Ba(OH)2) titrated = n(HCl) × (1 mole Ba(OH)2 ÷ 2 moles HCl)= (1.008 × 10!2 mol) × (½)= 5.040 × 10!3 mol

ˆ c(Ba(OH)2) required = n(Ba(OH)2) ÷ V(Ba(OH)2)= (5.040 × 10!3 mol) ÷ (25.00 × 10!3 L)= 0.2016 mol L!1

= 0.2016 M (4 s.f.)


PART AQUESTION 21

Given: 0.6120 g of a strong acid, HX (monotropic acid)c(NaOH) = 0.3000 MV(NaOH) = 20.30 mL

To find: the molar mass and the chemical name of HX

c(x) 'n(x)V(x)

(units: molL

) n(x) 'mass (x)

M(x)

n(NaOH) titrated = c(NaOH) × V(NaOH)= (0.3000 mol L!1) × (20.30 × 10!3 L)= 6.090 × 10!3 mol

The complete titration reaction is:HX(aq) + NaOH(aq) ÷ NaX(aq) + H2O(R)

1 mole of HX(aq) reacts with 1 mole of NaOH(aq)

n(HX) present = n(NaOH) × (1 mole HX ÷ 1 mole NaOH)= (6.090 × 10!3 mol) × (1)= 6.090 × 10!3 mol

ˆ the calculated M(HX) = mass(HX) ÷ n(HX)= (0.6120 g) ÷ (6.090 × 10!3 mol)= 100.49 g mol!1

= 100.5 g mol!1 (4 s.f.)

There are only six strong acids: 1. HCl: M(HCl) = [1.0079 + 35.453] g mol!1 = 36.4609 g mol!1

2. HBr: M(HBr) = [1.0079 + 79.904] g mol!1 = 80.9119 g mol!1

3. HI: M(HI) = [1.0079 + 126.904] g mol!1 = 127.9119 g mol!1

4. H2SO4: M(H2SO4) = [2 (1.0079) + 32.06 + 4 (15.999)] g mol!1 = 98.0718 g mol!1

5. HNO3: M(HNO3) = [1.0079 + 14.007 + 3 (15.999)] g mol!1 = 63.0119 g mol!1

6. HClO4: M(HClO4) = [1.0079 + 35.453 + 4 (15.999)] g mol!1 = 100.4569 g mol!1

ˆ The strong acid is HClO4, perchloric acid.


PART AQUESTION 22

Given: c(Na2C2O4) = 2.84 MV(Na2C2O4) = 500.0 mL

To find: mass (g) of rust, Fe2O3(s), removed

c(x) 'n(x)V(x)

(units: molL

)

n(Na2C2O4) used = c(Na2C2O4) × V(Na2C2O4)= (2.84 mol L!1) × (500.0 × 10!3 L)= 1.42 mol

The complete titration reaction is:

Fe2O3(s) + 6 Na2C2O4(aq) + 6 HCl(aq)÷ 2 Na3Fe(C2O4)3(aq) + 6 NaCl(aq) + 3 H2O(R)

1 mole of Fe2O3(s) reacts with 6 moles of Na2C2O4(aq)

n(Fe2O3) removed = n(Na2C2O4) × (1 mole Fe2O3 ÷ 6 moles Na2C2O4)= (1.42 mol) × (1/6)= 0.23667 mol

M(Fe2O3) = [2 (55.847) + 3 (15.999)] g mol!1 = 159.691 g mol!1

n(x) 'mass (x)

M(x)

ˆ mass(Fe2O3) removed = n(Fe2O3) × M(Fe2O3)= (0.23667 mol) × (159.691 g mol!1)= 37.794 g= 37.8 g (3 s.f.)


PART AQUESTION 23

Given: c(Pb(NO3)2) = 0.200 M = c(Pb2+)V(Pb(NO3)2) = 50.0 mL = V(Pb2+)c(NaCl) = 0.126 M = c(Cl!)V(NaCl) = 100.0 mL = = V(Cl!)

To find: maximum mass(PbCl2) produced

The precipitation reaction is: Pb(NO3)2(aq) + 2 NaCl(aq) ÷ PbCl2(s) + 2 NaNO3(aq)

c(x) 'n(x)V(x)

(units: molL

) n(x) 'mass (x)

M(x)

NIE: Pb2+(aq) + 2 Cl!(aq) ÷ PbCl2(s) c(x): V(x):

0.200 M 0.126 M !50.0 mL 100.0 mL !

n(x): (0.200 mol L!1) × (50.0 × 10!3 L) (0.126 mol L!1) × (100.0 × 10!3 L)= 0.0100 mol = 0.0126 mol

ratio: (0.0100 mol ÷ 1 mol) (0.0126 mol ÷ 2 mol)

= 0.0100 = 0.00630(ˆ L.R.)

2 moles of Cl!(aq) produces 1 mole of PbCl2(s)

n(PbCl2) produced = n(Cl!) × (1 mole PbCl2 ÷ 2 moles Cl!)= (0.0126 mol) × (½)= 0.00630 mol

M(PbCl2) = [207.2 + 2 (35.453)] g mol!1 = 278.106 g mol!1

ˆ mass(PbCl2) produced = n(PbCl2) × M(PbCl2)= (0.00630 mol) × (278.106 g mol!1)= 1.75206 g= 1.75 g (3 s.f.)


PART AQUESTION 24

Back Titration1. Back Titration is used when the analyte

(a) is unstable in solution (e.g. it volatilizes easily from the solution)(b) is a solid (‡ It is difficult to determine exactly whether all the solids

dissolve in aqueous solution.)2. The analyte with unknown concentration (e.g. CaCO3(s) dissolved in acidic

solution) is the limiting reactant, some of the other reagent (e.g. HCl) is leftover(in excess).

3. The amount of HCl remaining is determined by acid-base titration with NaOH(as titrant) BACK to equivalence point.

4. The amount of analyte (CaCO3) is then calculated from the amount of reagent(HCl) that was consumed, which is not available to take part in the backtitration.

Given: 1.058 g impure sample of CaCO3(s) (analyte, limiting reagent)50.0 mL of 1.08 M HCl(aq) (in excess)28.3 mL of 1.26 M NaOH(aq) (titrant)

(a) net ionic equations for the reaction in each titration

Reaction Î: CaCO3(s) + 2 HCl(aq) ÷ CaCl2(aq) + CO2(g) + H2O(R)NIE Î: CaCO3(s) + 2 H+(aq) ÷ Ca2+(aq) + CO2(g) + H2O(R)

Reaction Ï: HCl(aq) + NaOH(aq) ÷ NaCl(aq) + H2O(R)NIE Ï: H+(aq) + OH!(aq) ÷ H2O(R)

(b) the percent purity of the CaCO3 samplec(x) '

n(x)V(x)

(units: molL

)

total n(HCl) used = c(HCl) × V(HCl)= (1.08 mol L!1) × (50.0 × 10!3 L)= 0.0540 mol

From Reaction Ï: 1 mole HCl(aq) reacts with 1 mole NaOH(aq)

n(NaOH) required to neutralize excess HCl = c(NaOH) × V(NaOH)= (1.26 mol L!1) × (28.3 × 10!3 L)= 0.035658 mol= n(HCl) leftover


% purity by mass 'calculated mass

sample mass× 100%

ˆ n(HCl) consumed up by the reaction with CaCO3(s)= (0.0540 mol) ! (0.035658 mol)= 0.018342 mol

From Reaction Î: 1 mole CaCO3(s) reacts with 2 moles HCl(aq)

n(CaCO3) present = n(HCl) consumed up × (1 mole CaCO3 ÷ 2 moles HCl)= (1.8342 × 10!2 mol) × (½)= 9.171 × 10!3 mol

M(CaCO3) = [40.08 + 12.011 + 3 (15.999)] g mol!1 = 100.088 g mol!1

n(x) 'mass (x)

M(x)

ˆ mass(CaCO3) present = n(CaCO3) × M(CaCO3) = (9.171 × 10!3 mol) × (100.088 g mol!1) = 0.9179 g

ˆ % purity of CaCO3(s) sample = [mass(CaCO3) ÷ sample mass] × 100%= [0.9179 g ÷ 1.058 g] × 100%= 86.759%= 86.8% (3 s.f.)


Iodometric Titration• useful when the analyte (unknown concentration) has oxidizing properties1. adding a source of iodide ion (I!) in acidic solution

÷ I!(aq) is oxidized quantitatively to I2(s) by the analyte÷ I!(aq) is added in excess, so that Cu2+(aq) is the limiting reactant

2. titration of the I2(s) against Na2S2O3(aq) of known concentration

PART AQUESTION 25

Given: 2.00 g mineral sample containing Cu2+(aq) (analyte, limiting reagent)I!(aq) (in excess)31.4 mL of 0.0500 M Na2S2O3(aq) (titrant)

To find: the percent of Cu by mass in the mineral

Reaction Î: 2 Cu2+(aq) + 4 I!(aq) ÷ 2 CuI(aq) + I2(s)Reaction Ï: I2(s) + 2 S2O3

2!(aq) ÷ S4O62!(aq) + 2 I!(aq)

c(x) 'n(x)V(x)

(units: molL

) n(x) 'mass (x)

M(x)

n(S2O32!) required for titration = c(S2O3

2!) × V(S2O32!)

= (0.0500 mol L!1) × (31.4 × 10!3 L)= 1.57 × 10!3 mol

From Reaction Ï: 1 mole I2(s) reacts with 2 moles S2O32!(aq)

n(I2) reacted = n(S2O32!) × (1 mole I2 ÷ 2 moles S2O3

2!)= (1.57 × 10!3 mol) × (½)= 7.85 × 10!4 mol

From Reaction Î: 2 moles Cu2+(aq) yield 1 mole I2(s)n(Cu2+) present = n(I2) × (2 moles Cu2+ ÷ 1 mole I2)

= (7.85 × 10!4 mol) × (2)= 1.57 × 10!3 mol

M(Cu) = 63.546 g mol!1

mass(Cu) present = n(Cu) × M(Cu) = (1.57 × 10!3 mol) × (63.546 g mol!1)= 9.97672 × 10!2 g

ˆ % of Cu by mass = [mass(Cu) ÷ sample mass] × 100%= [9.97672 × 10!2 g ÷ 2.00 g] × 100%= 4.98836%= 4.99% (3 s.f.)


PART AQUESTION 26

Given: 4.2852 g ferrous sulfate, [email protected] g Fe(s) obtained

To find: the value of x (i.e. the number of water hydrate)

The reactions involved in the gravimetric analysis of Fe(s) are:Î FeSO4@xH2O(s) ÷ Fe2+(aq) + SO4

2!(aq) + x H2O(R)Ï Fe2+(aq) + Mg(s) ÷ Mg2+(aq) + Fe(s)

M(Fe) = 55.847 g mol!1

n(x) 'mass (x)

M(x)

n(Fe) obtained = mass(Fe) ÷ M(Fe)= (0.8607 g) ÷ (55.847 g mol!1)= 1.54117 × 10!2 mol

From Reactions Î and Ï: 1 mole FeSO4@xH2O(s) yields 1 mole of Fe(s)n(FeSO4@xH2O) used = n(Fe) × (1 mole FeSO4@xH2O ÷ 1 mole Fe)

= (1.54117× 10!2 mol) × (1)= 1.54117× 10!2 mol

M(FeSO4@xH2O) = mass(FeSO4@xH2O) ÷ n(FeSO4@xH2O)= (4.2852 g) ÷ (1.54117 × 10!2 mol)= 278.048 g mol!1

M(FeSO4) = [55.847 + 32.06 + 4 (15.999)] g mol!1 = 151.903 g mol!1

M(xH2O) = M(FeSO4@xH2O) ! M(FeSO4)= 278.048 g mol!1 ! 151.903 g mol!1

= 126.145 g mol!1

M(H2O) = [2 (1.0079) + 15.999] g mol!1 = 18.0148 g mol!1

ˆ x = M(xH2O) ÷ M(H2O)= (126.145 g mol!1) ÷ 18.0148 g mol!1

= 7.002 (4 s.f.)


PART AQUESTION 27

Given: Before heating, mass(BaCl2@2H2O) + mass(NaCl) = 2.4103 gAfter heating, mass(BaCl2) + mass(NaCl) = 2.1058 g remained

ˆ mass(H2O) = 2.4103 g ! 2.1058 g = 0.3045 g

(a) the mass of BaCl2@2H2O in the original samplen(x) '

mass (x)M(x)

M(H2O) = [2 (1.0079) + 15.999] g mol!1 = 18.0148 g mol!1

n(H2O) produced = mass(H2O) ÷ M(H2O)= (0.3045 g) ÷ (18.0148 g mol!1)= 1.69028 × 10!2 mol

The heating reaction is: BaCl2@2H2O(s) + NaCl(s) ÷ BaCl2(s) + NaCl(s) + 2 H2O(g)

1 mole BaCl2@2H2O(s) contains 2 moles H2O(R)n(BaCl2@2H2O) used = n(H2O) × (1 mole BaCl2@2H2O ÷ 2 moles H2O)

= (1.690276× 10!2 mol) × (½)= 8.45138 × 10!3 mol

M(BaCl2@2H2O) = [137.33 + 2 (35.453) + 2 (18.0148)] g mol!1 = 244.2656 g mol!1

mass(BaCl2@2H2O) = n(BaCl2@2H2O) × M(BaCl2@2H2O)= (8.45138 × 10!3 mol) × (244.2656 g mol!1)= 2.064 g (4 s.f.)

(b) the % Ba in the original sample

1 mole BaCl2@2H2O(s) contains 1 mole Ban(Ba) present = n(BaCl2@2H2O) × (1 mole Ba ÷ 1 mole BaCl2@2H2O)

= (8.45138 × 10!3 mol) × (1)= 8.45138 × 10!3 mol

M(Ba) = 137.33 g mol!1

mass(Ba) = n(Ba) × M(Ba)= (8.45138 × 10!3 mol) × (137.33 g mol!1)= 1.160628 g

ˆ % Ba = [mass(Ba) ÷ sample mass] × 100%= [1.160628 g ÷ 2.4103 g] × 100%= 48.15% (4 s.f.)


PART BQUESTION 1

Standard States (25oC and 1 atm) of Elements

Solids Gases Liquids

carbon C(s, graphite) sodium Na(s) potassium K(s) iron Fe(s) sulfur S(s) chromium Cr(s) iodine I2(s)

hydrogen H2(g) nitrogen N2(g) oxygen O2(g) chlorine Cl2(g) fluorine F2(g)

bromine Br2(R) mercury Hg(R)

(a) Heat of formation of isoamyl alcohol C5H12O(R)

5 C(s, graphite) + 6 H2(g) + ½ O2(g) ÷ C5H12O(R)

(b) Heat of formation of ammonium iodide

½ N2(g) + 2 H2(g) + ½ I2(s) ÷ NH4I(s)

(c) ∆H = ! [the enthalpy of formation of CrO2Cl2(R)]

CrO2Cl2(R) ÷ Cr(s) + O2(g) + Cl2(g)

(d) ∆H = !2 [the enthalpy of formation of mercury (I) bromodate, HgBrO3(s)]

2 HgBrO3(s) ÷ 2 Hg(R) + Br2(R) + 3 O2(g)

(e) Heat of combustion of glucose C6H12O6(s)

C6H12O6(s) + 6 O2(g) ÷ 6 CO2(g) + 6 H2O(R)

(f) Heat of combustion of benzoic acid C7H6O2(s)

C7H6O2(s) + 15/2 O2(g) ÷ 7 CO2(g) + 3 H2O(R)


∆Hof of all elements in their standard states (1 atm, 25oC) are ZERO.

∆Hof [H2(g)] = 0 ∆Ho

f [K(s)] = 0 ∆Hof [Br2(R)] = 0

∆Hof [O2(g)] = 0 ∆Ho

f [Na(s)] = 0 ∆Hof [Hg(R)] = 0

∆Hof [N2(g)] = 0 ∆Ho

f [I2(s)] = 0 ∆Hof [H+(aq)] = 0

PART BQUESTION 2

To find: ∆Hof = 0 at 298 K (i.e. elements in their standard states)

(a) I2(s): ∆Hof [I2(s)] = 0 at standard states

(b) O3(g): 3/2 O2(g) ÷ O3(g) ˆ ∆Horxn = ∆Ho

f [O3(g)]

(c) KBr(s): K(s) + ½ Br2(R) ÷ KBr(s) ˆ ∆Horxn = ∆Ho

f [KBr(s)]

(d) H2O(R): H2(g) + ½ O2(g) ÷ H2O(R) ˆ ∆Horxn = ∆Ho

f [H2O(R)]

(e) NH3(g): ½ N2(g) + 3/2 H2(g) ÷ NH3(g) ˆ ∆Horxn = ∆Ho

f [NH3(g)]

PART BQUESTION 3

Given: 2 Al2O3(s) ÷ 4 Al(s) + 3 O2(g) ∆H = +3340 kJ∆Hrxn = +960 kJ

To find: mass of Al(s) produced

∆Hrxn = +960 kJ = (+3340 kJ ÷ 4 mol Al) × n(Al)n(Al) = 1.1497 mol

M(Al) = 26.982 g mol!1

ˆ mass(Al) produced = n(Al) × M(Al)= (1.1497 mol) × (26.982 g mol!1)= 31.021 g= 31.0 g (3 s.f.)


PART BQUESTION 4

Given: 2 NaN3(s) ÷ 2 Na(s) + 3 N2(g)

To find: the enthalpy change for the reaction (∆Horxn) at 1 atm

∆Hof of all elements in their standard states (1 atm, 25oC) are ZERO.

∆Hof [Na(s)] = 0

∆Hof [N2(g)] = 0

∆Ho(reaction) = 3 n ∆Hof (products) ! 3 n ∆Ho

f (reactants)

∆Ho(reaction) = 3 n ∆Hof (products) ! 3 n ∆Ho

f (reactants)ˆ ∆Ho

rxn = {2× ∆Hof [Na(s)] + 3× ∆Ho

f [N2(g)]} ! {2× ∆Hof [NaN3(s)]}

= {0 + 0} ! {2× ∆Hof [NaN3(s)]}

= !2 ∆Hof [NaN3(s)]

PART BQUESTION 5

Given: Î CH4(g) + 2 O2(g) ÷ CO2(g) + 2 H2O(R) ∆Ho = !890 kJ mol!1

Ï C2H4(g) + 3 O2(g) ÷ 2 CO2(g) + 2 H2O(R) ∆Ho = !1410 kJ mol!1

To find: ∆Ho for the reaction: 2 CH4(g) + O2(g) ÷ C2H4(g) + 2 H2O(R)

Using Hess’s Law,

2× Î: 2 CH4(g) + 4 O2(g) ÷ 2 CO2(g) + 4 H2O(R) (2×) ∆Ho = !890 kJ mol!1

(!1)× Ï: 2 CO2(g) + 2 H2O(R) ÷ C2H4(g) + 3 O2(g) (!1×) ∆Ho = !1410 kJ mol!1

2 CH4(g) + O2(g) ÷ C2H4(g) + 2 H2O(R) ∆Ho

rxn = !370 kJ mol!1


PART BQUESTION 6

Given: ∆Ho combustion (kJ mol!1)Î C(s, graphite) !395.5Ï H2(g) !285.8Ð CH4(g) !890.2

To find: the standard molar enthalpy of formation of CH4(g)(i.e. C(s) + 2 H2(g) ÷ CH4(g) ∆Ho

f [CH4(g)] = ? )

The data can be represented by the following thermochemical equations: Î C(s, graphite) + O2(g) ÷ CO2(g) ∆Ho

comb = !395.5 kJ mol!1

Ï H2(g) + ½ O2(g) ÷ H2O(g) ∆Hocomb = !285.8 kJ mol!1

Ð CH4(g) + 2 O2(g) ÷ CO2(g) + 2 H2O(g) ∆Hocomb = !890.2 kJ mol!1

Using Hess’s Law:∆Hocombustion (kJ mol!1)

1× Î: C(s, graphite) + O2(g) ÷ CO2(g) (1×) !395.5

2× Ï: 2 H2(g) + O2(g) ÷ 2 H2O(g) (2×) !285.8

!1× Ð: CO2(g) + 2 H2O(g) ÷ CH4(g) + 2 O2(g) (!1×) !890.2

C(s, graphite) + 2 H2(g) ÷ CH4(g) ∆Hof [CH4(g)] = !76.9 kJ mol!1

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chem130 (f 03) review questions for midterm i page - 1 · chem130 (f 03) review questions for...