13. The mole Questions and Answers

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) + HNO_3(aq) KNO_3(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) + 3Cl_2(g) 2FeCl₃

 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) PbI_2(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

 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 0^o and 1atm. Given that

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

  boiled to 100^oC in order to make some coffee?

 C₄H_10(g) + 6 ½ O_2(g) 4CO_2(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)

The Mole Answers

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:

Na₂SO_3(s) + 2HCL_(aq) 2NaCl_(aq) + SO_2(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

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 Cl^–ions

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 I^–ions produces I mole of pbI₂

Moles of I^–ions = 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) C_nH_2n, where n = No. of carbon atoms

ii) 70

iii) C_sH₁₀, 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) NH₄NO_3(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

SO_3(g) + H₂SO_4(L) ____________ H₂S₂O_7(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 + H₂SO_4(aq) _______ Na₂SO_4(aq) + H₂O_(L) + CO_2(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.

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 = 25^oC  

 Boiling point = 100^oC

  Change in temperature, T = 100-25 = 75^oC 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 Co_3(aq) + Mg Cl_2(aq) + MgCo_3(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 ________________ 10.08_g 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) ___________ MgCl_{2 (aq)} + H_{2 (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

The empirical formula is Fe₂O₃