## The mole Questions

1.  In an experiment magnesium ribbon was heated in air. The product formed was found to be

heavier than the original ribbon. Potassium manganate (VII) was on the other hand, heated in

air and product formed was found to be lighter. Explain the differences on the observation made

2. In a filtration experiment 25cm3 of a solution of Sodium Hydroxide containing 8g per

litre was required for complete neutralization of 0.245g of a dibasic acid. Calculate

the relative molecular mass of the acid (Na = 23.0, O = 16, H= 1)

3. D grams of Potassium hydroxide were dissolved is distilled water to make 100cm3 of solution.

50cm3 of the solution required 50cm3 of 2.0M nitric acid for complete neutralization.

Calculate the mass D of Potassium hydroxide (RFM of KOH = 56)

KOH(aq) + HNO3(aq) KNO3(aq) + H2O(l)

4.  When excess dilute hydrochloric acid was added to sodium sulphite, 960cm3 of sulphuric

(IV) Oxide gas was produced. Calculate the mass of sodium sulphate that was used.

(Molar gas volume = 24000cm3 and Molar mass of sulphite = 126g)

5.  The equation of the formation of iron (III) chloride is

2Fe(s) + 3Cl2(g) 2FeCl3

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Calculate the volume of chlorine which will react with iron to form 0.5g of Iron (III) chloride.

(Fe = 56 Cl=35.5). Molar gas volume at 298K = 24dm3)

6.  15.0cm3 of ethanoic acid (CH3COOH) was dissolved in water to make 500cm3 of solution.

Calculate the concentration of the solution in moles per litre

[C=12, H = 1, O = 16, density of ethanoic acid is 1.05g/cm3]

7.  When 1.675g of hydrated sodium carbonate was reacted with excess hydrochloric acid,

the volume carbon (IV) oxide gas obtained at room temperature and pressure was 150cm3.

Calculate the number of moles of water of crystallization in one mole of hydrated sodium

carbonate:- (Na=23, H =1, C=12, O=16, MGV at R.T.P = 24000cm3)

8.  How many chloride ions are present in 1.7g of magnesium chloride crystals?

(Avogadro’s constant = 6.0 x 1023, Mg = 24, Cl = 35.5)

9.  0.84g of aluminium reacted completely with chlorine gas. Calculate the volume of chlorine

gas used  (Molar gas volume is 24dm3, Al = 27)

10.  6.4g of a mixture of sodium carbonate and sodium chloride was dissolved in water to make

50cm3 solution. 25cm3 of the solution was neutralized by 40cm3 of 0.1M HCl(aq). What is

he percentage of sodium chloride in the solid mixture?

11  An unknown mass, x, of anhydrous potassium carbonate was dissolved in water and the solution made up to 200cm3. 25cm3 of this solution required 18cm3 of 0.22M nitric (V) acid for complete neutralization. Determine the value of x. (K=39.0, C =12.0, O =16.0)

12.   Calculate the volume of oxygen gas used during the burning of magnesium (O = 16, molar

gas volume = 24,000cm3 at room temperature)

13.   A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%,

sulphur 11.5%, water 45.3%

i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11)

ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total

volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution

in moles per litre. (Given that the molecula mass of the salt is 278)

14.   (i) Lead (II) ions react with iodide ions according to the equation;

Pb2+(aq) + 2I(aq) PbI2(s)

300cm3 of a 0.1m solution of iodide ions was added to a solution containing excess lead II ions.

Calculate the mass in grams of lead II iodide formed

(ii) Identify the colour of the product formed in (d) (i)

15.  a) The diagram below represents part of the structure of sodium chloride crystal

The position of one of the sodium ions in the crystal is shown as;

i) On the diagram, mark the positions of the other three sodium ions

ii) The melting and boiling points of sodium chloride are 801C and 1413C respectively. Explain

why sodium chloride does not conduct electricity at 25C, but does not at temperatures

between 801C and 1413C

b) Give a reason why ammonia gas is highly soluble in water

c) The structure of ammonium ion is shown below;

Name the type of bond represented in the diagram by N H

d) Carbon exists in different crystalline forms. Some of these forms were recently discovered

in soot and are called fullerenes

i) What name is given to different crystalline forms of the same element

ii) Fullerenes dissolve in methylbenzene while the other forms of carbon do not. Given that soot is

a mixture of fullerenes and other solid forms of carbon, describe how crystals of fullerenes can

be obtained from soot

iii) The relative molecular mass of one of the fullerenes is 720. What is the molecular mass of

this fullerene

16.  Calculate the volume of oxygen gas used during the burning of magnesium (O = 16, molar

gas volume = 24,000cm3 at room temperature)

17.   Study the information in the table below and answer the questions that follow

 Number of carbon atoms per molecule Relative molecular mass of the hydrocarbon 2 3 4 28 42 56

i) Write the general formula of the hydrocarbons in the table

ii) Predict the relative atomic mass of the hydrocarbons with 5 carbon atoms

iii) Determine the relative atomic mass of the hydrocarbon in (ii) above and draw its

structural formula (H=1.0, C=12.0)

18.  A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%,

sulphur 11.5%, water 45.3%

i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11) (3 mks)

ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total

volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution

in moles per litre. (Given that the molecula mass of the salt is 278)

19.  a) Galvanized iron sheets are made by dipping the sheets in molten Zinc.

i) Explain how zinc protects iron from rusting

ii) Name the process applied in galvanization of iron with zinc

20.  Calculate the percentage of copper in 1.0g of the alloy

(Cu = 63.5 Mg = 24)

21.   A factory uses nitric acid and ammonia gas as the only reactant for the preparation of the

fertilizer if the daily production of the fertilizer is 4800kg. Calculate the mass of ammonia

gas used daily

(N = 14.0, O= 16.0, H = 1.0)

22.  Calculate the volume of sulphur (VI) oxide gas that would be required to produce 178kg of

oleum in step 3 molar gas volume at s.t.p = 22.4 litres H = 1 O = 16 S = 32

23.  Using the answer in d (ii) above, determine:

i) The volume of 1M nitric acid that would react completely with one mole of copper

(Cu = 63.5)

ii) The volume of Nitrogen (IV) oxide gas produced when one mole of copper reacts

with excess 1M nitric acid at room temperature

24.  A sample of biogas contains 35.2% by mass of methane. A biogas cylinder contains 5.0kg

of the gas. Calculate:

(i) Number of moles of methane in the cylinder (Molar mass of methane = 16)

(ii) Total volume of carbon (IV) oxide produced by the combustion of methane in the cylinder

(Molar gas volume = 24.0dm3 at room temperature and pressure)

25. 0.84g of aluminium were reacted completely with chlorine gas. Calculate the volume

of chlorine gas used. (Molar gas volume is 24dm3, Al = 27)

26.  3.52g of Carbon (IV) Oxide and 1.40g of water are produced when a mass of a hydrocarbon

is completely burnt in oxygen. Determine the empirical formula of the hydrocarbon;

(H = 1 , C= 12, O = 16)

27.  Calculate the number of water molecules when 34.8g Na2CO3 xH2O is heated and 15.9g of

anhydrous Na2CO3 obtained (H=1, O=16, Na= 23, C = 12)

28.  A weighed sample of crystallined sodium carbonate (Na2CO3nH2O) was heated in a crucible

until there was no further change in mass. The mass of the sample reduced by 14.5%. Calculate

the number of moles (n) of water of crystallization (Na = 23, O = 16, C = 12, H = 1)

29. In a reaction 20cm3 of 0.1 M Sodium Carbonate completely reacted with 13cm3 of dilute

sulphuric acid. Find the molarity of the sulphuric acid used.

30. An organic compound P contains 68.9% carbon, 13.5% hydrogen and 21.6% oxygen.

The relative formula mass of p is 74. Determine its molecular formula. [C=12, H=1, 0=16]

31.  Campers GAZ cylinder contains about 1.12dm3 of butane measured at 0o and 1atm. Given that

25% of heat is lost, what is the maximum volume of water at room temperature which can be

boiled to 100oC in order to make some coffee?

C4H10(g) + 6 ½ O2(g) 4CO2(g) + 5H2O(l) H
= -3,000KJmol-1

(Specific heat capacity of water = 4.2J g-1C-0c, density of water 1gcm-3 Molar gas
volume 22.41 at s.t.p)

32.  An aqueous solution containing anhydrous sodium carbonate was prepared by dissolving

19.6g of the salt in 250cm3 of distilled. Calculate the volume of 2M of magnesium chloride

solution required to precipitate all the carbonate ions in the solution.
(Na=23, C= 12; O = 16; Mg = 24; Cl =35.5)

33.  10.08g of ethanedioic acid (H2C2O4.xH2O) crystals were dissolved in water and made to

1dm3 solution. 25.0cm3 of this solution was completely neutralized by 20cm3 of 0.2M

sodium hydroxide solution.

Calculate

i) Molarity of the acid

ii)the value of x in H2C2O4xH2O acid

34. 1.6g of magnesium metal is reacted with excess hydrochloric acid. Calculate the volume

of hydrogen gas produced

(Molar gas volume at stp = 22.4dm3 Mg=24)

35.  60 litres of sulphur(IV) oxide were made to react with 40 litres of oxygen.

a) Which reactant was in excess and by how much?

b) What is the volume of the product?

36.  During welding of cracked railway lines by thermite 12.0g of oxide of iron is reduced by

aluminium to 8.40g of iron. Determine the empirical formula of the oxide

(Fe= 56.0, O= 16.0)

1.  When a magnesium ribbon is heated in air it combines with oxygen forming magnesium oxide.

When potassium manganate (VII) is heated it decomposes giving off oxygen which escapes in air

2. RFM of NaOH = 40

Moles of NaOH = 8 = 0.2M

40

Moles of NaOH in 25cm3

25 x 0.2 = 0.005

1000

Mole ratio 1:2

Moles of acid = 0.005

2

= 0.0025

1x 0.245 = 98

0.0025

3.  No. Of moles of HNO3 acid

50 x 2 = 0.1moles

1000

Mole ratio 1:1

The KOH will have 0.1moles; 0.1 X 100 = 0.2moles

50

Then D grams is 0.2 X 56

= 11.2g

4.  Number of moles of Q = 960cm3 x 1mole

24000cm3

= 0.04moles

Equation:

Na2SO3(s) + 2HCL(aq) 2NaCl(aq) + SO2(g) + H2O(l)

Mole ratio Na2SO3 : SO2 is 1:1

No. of moles of Na2SO3 = 0.04moles

Mass of Na2SO3 = 126gmol-1 x 0.04

= 5.04g

5.   From the equation

– ( 3×24) litres of chlorine react with iron to produce [(56 x 2) + (35.5 X3)] g of Fecl3.

325 g of Fecl3 is produced by 72 litres of cl2

Then 0.5g of fecl3 is produced by:

0.5 x72 =0.11078 litres

325

= 110.78 cm3

6.  RMM (CH3OOH) = 60

Mass of 15cm3 and = 1.05 x 15 = 15.75g

Moles in 500cm3 solution = 15.75 = 0.2625

60

Molarity = 1000 x 0.2625

5000 = 0.525M

7.   If 24000cm3 = 1mole

150cm3 = ?

150 x 1

24000 = 0.00625moles of CO2

Since the ratio of Na2CO3 O2 produced is 1:1 the mass of Na2CO3 = 0.00625 x 106 = 0.6625g

 Na2Co3 H2O Mass 0.6625gRFM 106Mole 0.6625 = 0.00625 106Ratio 0.00625 0.00625 = 1Na2CO3.9H2O 1.0125g181.0125 = 0.5625 180.056250.0.00625= 9

8.  MgCl2 Mg2+(s) 2Cl

R.F.M of MgCl2 = 24 + 71

= 95

Moles of Mass = 1.7

R.F.M 95

= 0.01789moles

I mole of MgCl2 = 2moles of Cl-ions

0.01789moles of MgCl2 = 0.01789 x 2

= 0.03478moles of Clions

1mole = 6.0 x 1023ions

0.03578moles = 0.03578 x 6.0x 1023

1

= 2.1468 x 1022 ions of Cl

12.   Mass of O2 = (4.0 – 2.4)= 1.6g

Moles of O2 = 1.6/16 = 0.1

If 1 mol O2 ________ 24000cm3

0.1 Mol Mg = 0.5 mol O2 = 1200cm3

OR

2mg : O2

2(24)  24000

2.4/2(24) = x/240000

X = 2.4 x 24000  = 1200cm3

2(2.4)

13.  i)   Fe S O H2O

20.2/56 11.5/32 23.0/16 45.3/18

0.36/0.36    0.36/0.36    1.44/0.36 2.52/0.36

1 1 4 7

Empirical formula: FeSO4 + H2O

ii) 6.95g  = 6.95/278  = 0.025

 0.05 moles in 250cm3 = 0.025 x 1000/250   = 0.1

14. R.F.M of pbI2 = 207 + (127X2) = 461

2 moles of Iions produces I mole of pbI2

Moles of Iions = 0.1 X 300 = 0.03 mole

1000

Mole ratio PbI2: I mole of PbI2 formed = 0.03 = 0.05

I : 2 2

Mass of pbI2 formed = 0.015 mole X 461

= 6.915 g

d(i) Yellow precipitate

15.  a)  i)

ii) At 25C, sodium chloride is in solid form. Ions cannot move. Between 801 and

1413C sodium chloride is in liquid state, ions are mobile

b) Both ammonia and water are polar moleculer and hydrogen bonds are formed

c) N _________ H // co-ordinate bond / Dative bond

d)  i) Allotrope

ii) Add methylbenzene to soot in a beaker. Shake and filter. Warm the filtrate to

concentrate it. Allow the concentrate to cool for crystals to form. Filter to obtain

crystals of fullerene

iii) 720/12  = 60

16.   Mass of O2 = (4.0 – 2.4)= 1.6g

Moles of O2 = 1.6/16 = 0.1

If 1 mol O2 ________ 24000cm3

0.1 Mol Mg = 0.5 mol O2 = 1200cm3

OR

2mg : O2

2(24)  24000

2.4/2(24) = x/240000

X = 2.4 x 24000  = 1200cm3

2(2.4)

17.  i) CnH2n, where n = No. of carbon atoms

ii) 70

iii) CsH10, CH3CH=CHCH2CH3

OR CH3CH2CHCH2= CH2

18.  i)   Fe S O H2O

20.2/56 11.5/32 23.0/16 45.3/18

0.36/0.36    0.36/0.36    1.44/0.36 2.52/0.36

1 1 4 7

Empirical formula: FeSO4 + H2O

ii) 6.95g  = 6.95/278  = 0.025

 0.05 moles in 250cm3 = 0.025 x 1000/250   = 0.1

Concentration   = 6.95/278 x 1000/250   = 0.1

19.  a)  Zinc is more reactive// higher reduction potential than copper it will react with//

get oxidized in preference to iron oxygen to form Zinc Oxide coat which protects iron

from rusting

ii) Sacrificial protection or cathodic protection

20.   Mole of Mg that reacted = Answer in (c) (ii) x 2

1000   2

= 26 = 0.026 √½

1000

Mass of Mg in the alloy = 0.026 x 24

= 0.624g √½

Mass Cu in the alloy = (1.0 – 0.624)

= 0.376g √½

% of Cu = 0.376 x 100

1.0

= 37.6%√½

21.  NH(g) + HNO(g) NH4NO3(s)

RMM of NH4NO3 = 80

Moles of NH4NO3 = 4800 = 60moles

80

RMM of NH3 = 17

Mass of NH3 = 60 x 17 = 1020KJ

22.   From the equation of step 3

SO3(g) + H2SO4(L) ____________ H2S2O7(L)

RFM of H2S2O7 = 2 + (2 X 32) + (7 X 16) = 178  ½ mark

178g of Oleum are produced by 22.4 liters of SO3  ½ mark

178 kg ”  ”  ”  ”  ”  ” 178 X 1000 X 22.4L  1 ½ mark

178g

= 22,4000 liters  ½ mark

(Total 13 marks)

23.  i) Moles of copper = 0.635 = 0.01 moles

63.5

Volume of 1M Nitric acid 40 = 4000cm3  ½ mark

0.01

– Use value in d(ii) above

ii) 480cm3
 ½ mark = 48,000 cm3
 ½ mark

0.01

OR 4000 X 480 = 48,000cm3  ½ mark

40cm3

i.e. Answer in e(i) X 480cm3

Answer in d(i) [Total = 11 marks]

24.  (i)   35.2 x 1000

100 x 16

= 10Moles

Or mass of CH4 = 35.2 x 5 = 1.76g

1000

Mass in g = 1.76 x 1000 = 1760kg

Moles of methane = 1760

16

= 110Moles

(ii) CH4 + 2O2 CO2 + 2H2O – (ignore states)

Volume = 110 x 24.0

= 2640dm3

Mark consequential from equation and b(ii) (Without equation max *TZM*)

25.   Volume of Cl2 used

= 0.047 x 24

= 1.128dm3

26.  Mass due Carbon in CO2 = 12/4 x 35.2

= 0.96

Moles carbon = 0.96/12 = 0.08

Mass due Hydrogen in H2O = 2/18 x 1.40

= 0.156

Moles hydrogen = 0.156 = 0.156

1

Mole ratio C:H = 1: 1.95

E.F = CH2

27.  Na2CO3 x H2O Na2CO3 + H2O √1

34.8g  15.9g 18.9g

106 18

0.15
√1  1.15 3

1. 0.15

x  =  7 √1

28.  % of H2O lost = 14.5%^

5 of anhydrous Na2CO3 = 85.5%  (½mk)

R.F.M of Na2CO3 = 106  (½mk)

RMM of H2O = 18  (½mk)

NaCO3 H2O

85.5 14.5

106 18  (½mk)

0.8066  0.8055

0.8055   0.8055  (½mk)

n = 1 (Na2CO3.H2O)  (½mk)

29.  Moles of Na2CO3 = 20 x 0.1 = 0.002 moles

1000

Na2CO3 + H2SO4(aq) _______ Na2SO4(aq) + H2O(L) + CO2(g)

Mole ratio 1 : 1

Moles of H2SO4 = Moles of Na2CO3

= 0.002 moles

Molarity of H2SO4 = 10000 x 0.002 = 0.154 moles

13

30.

 Element C H O % 68.9 13.5 21.6 Molar mass 12 1 16 Moles 68.9/125.403 13.5/1 13.5 216/161.35 MR 5.43/1.334 13.5/1.35 10 1.35/1.351 Ratio 4 10 1

h (C4H10O) = 74

h (12×4) + (10×1) +16 = 74

74h = 74

H= 1

Formula C4H10O

31.  Moles C4H10 = 1.12  = 0.05 mol

22.4

Heat produced + 0.05 X (3000) = 150 kj

Usefull heat = 75X150 = 112.5 kj

100

Let volume of water = V

Room temperature = 25oC

Boiling point = 100oC

Change in temperature, T = 100-25 = 75oC ½ mk

T X mass X C   Q 315V = 112500

75 X V X 4.2 =112.5 V = 112500 ½ mk

1000 1 315

V = 357.km3 ½ mk

32.  RFM Na2CO3 = 43 + 12 + 48 = 106

Mol. Na2 CO3 = 19.6 = 0.8149057

106

Molarity of Na2 Co3 = 0.1849057 = 0.73962m

0.25

Na2 Co3(aq) + Mg Cl2(aq) + MgCo3(s)

Mole ratio Na CO3 : Mg Cl2 is 1:1

 mol. Mg Cl2 Reacted = 0.1849

If 2.0 mol. = 1000cm3 solution mg cl2

= 0.1849mol = 0.1849 X 1000

2

= 92.45 or 92.5 cm3

33.  i)  ACID BASE

1 2

½ 0.004 20cm3
X 0.2 moles

= 0.002 moles  ½ 1000cm3 = 0.004 moles

25cm3 ________ 0.002 moles  ½

1000cm3 ______ ?

1000cm3 X 0.002 moles = 0.08 M ½

ii) 0.08 moles ________________ 10.08g H2C2O4xH2O  ½

1 mole __________________ ?

1 mole  X 10.08  = 126  ½

0.08 moles

126 _______ H2C2O4xH2O

18x = 126 – 90  ½

18x = 36

X = 2  ½

34.  Mg (g) + 2HCL (aq) ___________ MgCl2 (aq) + H2 (g)

24g _________________________________ 22.4dm3

16g _____________________________________?

1.6 gx 22.4dm3
 ½ = 1.4933 dm3

35.  a) 2SO2(g) + O2(g) 2SO3(g), SO2  :  O2

2 1 2  60  :  30 ½

60 l 40 l  Oxygen ½ by 10 litres

36.   Mass of Oxygen = 12 – 8.4 = 3.5g

 Element Fe O Mass 8.4 3.6 R.A.M 56 16 No. of moles 8.4560.15 3.6160.225 Mole ration 0.150.1512 0.225 0.151.5 x23

The empirical formula is Fe2O3

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## 1 Comment

• ### Joash, March 18, 2023 @ 6:33 pmReply

Thank you for assistance…I like your channel

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