The Mole Questions and Answers - Chemistry Form 3 Topical Revision

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 100 cm³ 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. Calculate the volume of oxygen gas used during the burning of magnesium (O = 16, molar gas volume = 24,000cm³ at room temperature)
10. A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%, sulphur 11.5%, water 45.3%
    1. Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11)
    2. 6.95g of the hydrated salt in (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)
11.  
    1. 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
    2. Identify the colour of the product formed in (i)
12.  
    1. 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;
       1. On the diagram, mark the positions of the other three sodium ions
       2. The melting and boiling points of sodium chloride are 801^oC and 1413^oC respectively. Explain why sodium chloride does not conduct electricity at 25^oC, but does not at temperatures between 801^oC and 1413^oC
    2. Give a reason why ammonia gas is highly soluble in water
    3. The structure of ammonium ion is shown below;
       
       Name the type of bond represented in the diagram by N → H
    4. Carbon exists in different crystalline forms. Some of these forms were recently discovered in soot and are called fullerenes
       1. What name is given to different crystalline forms of the same element
       2. 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
       3. The relative molecular mass of one of the fullerenes is 720. What is the molecular mass of this fullerene
13. Calculate the volume of oxygen gas used during the burning of magnesium (O = 16, molar gas volume = 24,000cm³ at room temperature)
14. 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	28
    3	42
    4	56

    1. Write the general formula of the hydrocarbons in the table
    2. Predict the relative atomic mass of the hydrocarbons with 5 carbon atoms
    3. Determine the relative atomic mass of the hydrocarbon in (ii) above and draw its structural formula (H=1.0, C=12.0)
15.  Galvanized iron sheets are made by dipping the sheets in molten Zinc.
    1. Explain how zinc protects iron from rusting
    2. Name the process applied in galvanization of iron with zinc
16. 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)
17. 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)
18. A sample of biogas contains 35.2% by mass of methane. A biogas cylinder contains 5.0kg of the gas. Calculate:
    1. Number of moles of methane in the cylinder (Molar mass of methane = 16)
    2. 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)
19. 0.84g of aluminium were reacted completely with chlorine gas. Calculate the volume of chlorine gas used. (Molar gas volume is 24dm³, Al = 27)
20. 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)
21. 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)
22. A weighed sample of crystallined sodium carbonate (Na₂CO₃.nH2O) 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)
23. 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.
24. 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, O=16]
25. 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) + 6½O_2(g) → 4CO_2(g) + 5H₂O_(l); ΔH^θ = -3,000KJmol^-1
    (Specific heat capacity of water = 4.2J g^-1oC, density of water 1gcm^-3 Molar gas volume 22.41dm³ at s.t.p)
26. 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)
27. 10.08g of ethanedioic acid (H₂C₂O₄.xH₂O) crystals were dissolved in water and made to 1dm³ solution. 25.0 cm³ of this solution was completely neutralized by 20cm³ of 0.2M sodium hydroxide solution.
    Calculate
    1. Molarity of the acid
    2. the value of x in H₂C₂O₄.xH₂O acid
28. 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)
29. 60 litres of sulphur(IV) oxide were made to react with 40 litres of oxygen.
    1. Which reactant was in excess and by how much?
    2. What is the volume of the product?
30. 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)
31. 0.84g of aluminium reacted completely with chlorine gas. Calculate the volume of chlorine gas used (Molar gas volume is 24dm³, Al = 27)

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 = ⁸/₄₀ = 0.2M ✓
   Moles of NaOH in 25cm³
   25 x 0.2 = 0.005
    1000
   Mole ratio 1:2
   Moles of acid = ^0.005/₂
   = 0.0025
   1 x 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 × 100 = 0.2moles
                                                   50
   Then D grams is 0.2 × 56
   = 11.2g
4. Number of moles of Q = 960 cm³ x 1 mole
                                           24000 cm³
   = 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
   - (3x24) 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 x 72 =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 24000 cm³ = 1mole
   150cm³ = ?
   150 x 1
   24000
   = 0.00625 moles 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 0.6625g RFM 106 Mole 0.6625/106 = 0.00625 Ratio      0.00625/0.00625              = 1  Na2CO3.9H2O	1.0125g 18 1.0125/18 = 0.5625 0.05625/0.00625 = 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
   1 mole of MgCl₂ = 2moles of Cl^- ions
   0.01789moles of MgCl₂ = 0.01789 x 2
   = 0.03478 moles of Cl^- ions
   1mole = 6.0 x 10²³ ions
   0.03578moles = 0.03578 x 6.0x 10²³
                                         1
   = 2.1468 x 10²² ions of Cl^-
9. 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)
10.  
    1. 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: FeSO₄.7H₂O
    2. 6.95g = ^6.95/₂₇₈ = 0.025
       ∴ 0.05 moles in 250cm³ = 0.025 x ¹⁰⁰⁰/₂₅₀ = 0.1
11.  
    1. R.F.M of PbI₂ = 207 + (127 × 2) = 461
       2 moles of I^- ions produces 1 mole of PbI₂
       Moles of I-ions = 0.1 × 300 = 0.03 mole
                                    1000
       Mole ratio PbI₂ : I^- mole of PbI₂ formed = ^0.03/₂ = 0.05
                         1  : 2 
       Mass of PbI₂ formed = 0.015 mole × 461
       = 6.915 g
    2. Yellow precipitateMocks Topical Analysis www.easyelimu.com 20
12.  
    1.  
       2. At 25^o C, sodium chloride is in solid form. Ions cannot move. Between 801^oC and 1413^oC sodium chloride is in liquid state, ions are mobile
    2. Both ammonia and water are polar moleculer and hydrogen bonds are formed
    3. N → H // co-ordinate bond / Dative bond
    4.  
       1. Allotrope
       2. 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
       3. ⁷²⁰/₁₂ = 60
13. 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)
14.  
    1. C_nH_2n, where n = No. of carbon atoms
    2. 70
    3. C₅H₁₀, CH₃CH=CHCH₂CH₃
       OR CH₃CH₂CHCH₂ = CH₂
15.  
    1. 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
    2. Sacrificial protection or cathodic protection
16. NH_3(g) + HNO_3(g) → NH₄NO_3(s)
    RMM of NH₄NO₃ = 80
    Moles of NH₄NO₃ = ⁴⁸⁰⁰/₈₀ = 60moles
    RMM of NH₃ = 17
    Mass of NH₃ = 60 x 17 = 1020KJ
    
    
17. From the equation of step 3
    SO_3(g) + H₂SO_4(l) → H₂S₂O_7(l)
    RFM of H₂S₂O₇ = 2 + (2 × 32) + (7 × 16) = 178 ✓ ½ mark
    178g of Oleum are produced by 22.4 liters of SO3 ✓ ½ mark
    178 kg of Oleum will produce
    178 × 1000 × 22.4l ✓ 1½ mark
              178g
    = 22,4000 liters ✓ ½ mark
    (Total 13 marks)
18.  
    1. 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
       = 110 Moles
    2. CH₄ + 2O₂ → CO₂ + 2H₂O – (ignore states)
       Volume = 110 x 24.0
       = 2640 dm³
       Mark consequential from equation and b(ii) (Without equation max *TZM*)
19. Volume of Cl₂ used
    = 0.047 x 24
    = 1.128dm³
20. 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 = ^0.156/₁ = 0.156
    Mole ratio C:H = 1:1.95
    E.F = CH₂
21. Na₂CO_3.xH2O → Na₂CO₃ + H₂O √1
    34.8g                15.9g        18.9g
                             106          18
                            0.15 √1     1.15 
                            0.15           0.15
    x = 7 √1
    Total 3 marks
22. % of H₂O lost = 14.5%^
    5 of anhydrous Na₂CO₃ = 85.5% (½mk)
    R.F.M of Na₂CO₃ = 106 (½mk)
    RMM of H₂O = 18 (½mk)
    

    Na2CO3	H2O
    85.5  106  0.8066  0.8055	14.5   18 0.8066  0.8055

    n = 1 (Na₂CO₃.H₂O) (½mk)
23. 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
24.  
    

    Element	C	H	O
    %	68.9	13.5	21.6
    Molar mass	12	1	16
    Moles	68.9/12  5.403	13.5/1  13.5	216/16  1.35
    MR	5.403/1.33  4	13.5/1.35  10	1.35/1.35  1
    Ratio	4	10	1

    
    h (C₄H₁₀O) = 74
    h (12 x 4) + (10 x 1) +16 = 74
    74h = 74
    h= 1
    Formula C₄H₁₀O
25. Moles C₄H₁₀ = ^1.12/_22.4 = 0.05 mol
    Heat produced + 0.05 × (3000) = 150 kJ
    Usefull heat = 75×150 = 112.5 kJ
                            100
    Let volume of water = V
    Room temperature = 25^oC
    Boiling point = 100^oC
    Change in temperature, ΔT = 100 − 25 = 75^oC ½ mk
    Q = ΔT × mass × C 
    = 75 × V × 4.2 =112500
    315V = 112500
    V = ¹¹²⁵⁰⁰/₃₁₅ ½ mk
    V = 357cm³ ½ mk
26. RFM Na₂CO₃ = 43 + 12 + 48 = 106
    Mol. Na₂CO₃ = ^19.6/₁₀₆ = 0.8149057
    Molarity of Na₂CO₃ = ^0.1849057/_0.25 = 0.73962M
    Na₂CO_3(aq) + MgCl_2(aq) → MgCO_3(s) + 2NaCl_(aq)
    Mole ratio NaCO₃ : MgCl₂ is 1:1
    ∴ mol. MgCl₂ Reacted = 0.1849
    If 2.0 mol. = 1000cm³ solution MgCl₂
    = 0.1849mol = 0.1849 × 1000
                                      2
    = 92.45 or 92.5 cm³
    
    
    
    
    
27.  
    1. ACID                            BASE
       1                                   2
       ½ ×0.004                   20cm³ × 0.2 moles
       = 0.002 moles  ✓½      1000cm³ = 0.004 moles
       25cm³ ________ 0.002 moles ✓ ½
       1000cm³ ______ ?
       1000cm³ × 0.002 moles = 0.08M✓ ½
    2. 0.08 moles ________________ 10.08g H₂C₂O₄.xH2O ✓ ½
       1 mole __________________ ?
       1 mole × 10.08 = 126 ✓ ½
       0.08 moles
       126 _______ H₂C₂O₄.xH₂O
       18x = 126 – 90 ✓ ½
       18x = 36
       x = 2 ✓ ½
28. Mg_(g) + 2HCl _(aq) → MgCl_{2 (aq)} + H_{2 (g)}
    24g ___________22.4dm³
    16g __________?
    1.6g x 22.4dm³ ✓ ½ = 1.4933 dm³
    24g
29.  
    1. 2SO_2(g) + O_2(g) → 2SO_3(g), SO₂ : O₂
       2              1              2         60 : 30 ✓½
       60l         40l                         
       Oxygen ✓½ by 10 litres
    2. 60 litres
30. Mass of Oxygen = 12 – 8.4 = 3.5g
    

    Element	Fe	O
    Mass	8.4	3.6
    RAM	56	16
    No. of moles	8.4/56  0.15	3.6/16  0.225
    Mole ration	0.15/0.15  1  2	0.225/0.15  1.5 × 2  3

    
    The empirical formula is Fe₂O₃ 

31. 2AL 3CL₂ = 2ALCL₃
    Mole of alumminium 0.84 = 0.0311.bt accrdin 2 d equation 2 mole of AL react with 3mole of Cl. 
                                       27
    1mole of Al required 1.5mole of cl
    0.0311mole of Al required 0.0311 ×1.5mole of Cl = 0.04665mole.
    1 mole of a gas occupied 22.4dm³,0.04665 mole will occupy 0.04665 × 22.4 = 1.0450dm³ of Cl