Chemistry 2018 Past Questions | WAEC

Study the following Chemistry past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.

1. (a)(i) Sketch a graphical representation of Charles’ law.

(ii) Calculate the volume of oxygen that would be required for the complete combustion of 2.5 moles of ethanol at s.t.p. [molar volume at s.t.p. = 22.4 dm $^{3}$

(b)(i) State the collision theory of reaction rates.

(ii) Using the collision theory, explain briefly how temperature can affect the rate of a chemical reaction.

(c)(i) Define esterification.

(ii) Give two uses of alkanoates.

(iii) Give the products of the alkaline hydrolysis of ethyl ethanoate.

(d) A tin coated plate and a galvanized plate were exposed for the same length of time.

(i) Which of the two plates corrodes faster

(ii) Explain briefly your answer in 2(d)(i)

Solutions & Explanation:

(a)(i) Sketch of Graphical Representation of Charles’ Law: Volume (cm$^3$)

(ii) Volume of O$_{2(g)}$ required for complete combustion of 2.5 moles of ethanol at s.t.p.

C$_{2}$H$_{5}$OH$_{(aq)}$ + 3O2$_{2}$ -> 3H$_{2}$O + 2CO$_{2(g)}$

1 mole of ethanol = 3 x 22.4 dm$^{2}$. of O$_{2(g)}$

.-. 2.5 moles of ethanol = 3 x 22.4 dm$^{3}$ x 2.5

= 168.0 dm$^{3}$.

(b)(i) Collision theory of reactants rates states that atoms/molecules/ions/particles/reactions involved in a chemical reaction must collide with effective collision before reaction can take place.

(ii) Using collision theory to explain effect of temperature on the rate of a chemical reaction: Increase in temperature makes the reacting molecules to gain kinetic energy and move faster. Where the temperature decreases, the reacting molecules lose kinetice energy and move slower. Frequency of collision and effective collision increase or decrease, therefore rate of reaction increases or decreases.

(c)(i) Esterification is the reaction between an alkanol and alkanoic acid to produce an Alkanoate (ester) and water in the presence of a mineral acid.

(ii) Two uses of alkanoates are: (1) Used in perfumes/cosmetics. (2) Used in artificial flavouring for foods. (3) Used in production of soap. (4) Used as plasticizers. (Any 2)

(iii) Products of the alkaline hydrolysis of ethyl ethanoate are: ethanol (C$_{2}$H$_{5}$OH) and sodium ethanoate (CH$_{3}$COONa) or potassium ethanoate CH$_{3}$COOK.

(d)(i) Tin-coated plate corrodes faster.

(ii) Zinc is more reactive than tin, higher in the electrochemical series than tin. When the tin coated plate is exposed, the plate goes into solution and corrodes. In zinc coated (galvanized) plate, it is zinc that dissolves and corrodes hence protecting the plate better and zinc coated plate does not corrode.

2. (a)(i) Draw the structure of the sixth member of the alkenes.

(ii) Calculate the relative molecular mass of the sixth member of the alkene.

(iii) State one difference between cracking and reforming in the petroleum industry. [H = 1, C = 12]

(b)(i) Define the term enthalpy of neutralization.

(ii) Describe briefly how the enthalpy of neutralization of the reaction of dilute hydrochloric acid and aqueous potassium hydroxide could be determined.

(i) State which of the electrodes will be the: 1. anode; II. cathode.

(ii) Give the reason for your answer in 3(c)(i).

(iii) State the type of reaction occurring at each electrode.

(iv) Write a balanced equation for the overall cell reaction.

(d)(i) Name the compound formed when iron is exposed to moist air for a long time.

(ii) Write a balanced chemical equation for the reaction in 3(d)(i).

(iii) Name one ore of iron.

Solutions & Explanation:

(a)(i) Structure of the sixth member of the alkenes: C$_n$H$_{2n}$; n = 7, C$_7$H$_{2 \times 7}$ = C$_7$H$_14$

(Double can be positioned within any of the carbon chain)

(ii) Relative molecular mass of the sixth member of alkene

(C$_7$H$_14$ $\begin {bmatrix} C & = 12 \\ H & = 1 \end{bmatrix}$

(C$_7$H$_14$ = (12 x 7) + (1 x 14)

= (12 x 7) + (1 x 14)

= 84 + 14

= 98 (no units)

(iii) One differences between cracking and reforming in the petroleum indutry:

CRACKING	REFORMING
Involves breaking large molecules of petroleumfractions into smaller molecules	Involves re-arrangement of atoms in molecules of petroleum fractio’n to obtain branched and cyclic hydrocarbons
Usect to improve the quantity of petrol.	Usect to improve the quality of petrol.
Can be achieved thermally or catalytically	Occurs in presence of a catalyst

(b)(i) Enthalpy of neuttalization is the beat evolved when an acid reacts with a base to form one mole of water or when one mole of OH$^{+}$ from an acid reacts with one mole of OH$^{-}$ from a base to form one mole of water.

(ii) Determination of enthalpy of neutralization of the reaction of dilute hydrochloric acid and aqueous potassium hydroxide

HCI$_{(aq)}$+ KOH$_{(aq)}$ -> KC1 + H$_{2}$O

Equimolar solutions of hydrochloric acid and potassium hydroxide are prepared sepa’rately. A known volume of the acid is placed in a calorimeter/polystyrene beaker and the temperature recorded.

The same volume of the potassium hydroxide solution at the same temperature is poured into the calorimeter and the mixture stirred gently. The maximum temperature of the mixture is recorded.

Heat lost in the reaction = heat gained by the solution formed

OR $\bigtriangleup$ H neutralization = mass of solution x specific heat capacity of water x change in temperature.

(ii) Cathode : silver electrode reasons in 3(c)(i)

(iii) Silver has a more positive Standard electrode potential than copper. Copper is more electropositive than silver or copper is higher in the electrochemical series

(iv) Type of reaction occuring at each electrode.

Anode : oxidation (occurs at the copper electrode).

Cathode : reduction (occurs at the siliver electrode).

(v) Balanced chemical equation for the overall cell reaction.

Cu$_s$ + 2Ag$^+$ $\to$ CU$^{2+}_{(aq)}$ + 2Ag$_(s)$

(d) Compound formed when iron is exposed to moist air for a long time; (i) Hydrated Iron(III) oxide. (ii) Balanced chemical equation for the reaction in 3(d)(i).

4Fe$_s$ + 3O$_{2(g)}$ + 2 x H$_2$O$_l$ $\to$ 2Fe$_2$O$_3$.XH$_2$O

where X is 1

4Fe$_s$ + 3O$_{2(g)}$ + 2H$_2$O$_l$ $\to$ 2Fe$_2$O$_3$.H$_2$O

(iii) One ore of iron: haematite, magnetite, siderite/spathic iron ore, limonite, iron pyrites

3. (a)(i) Draw and label a diagram for the laboratory preparation of a dry sample of sulphur(IV)oxide.

(ii) Write a balanced chemical equation for the reaction in (a)(i).

(iii) State the precaution that must be taken in the preparation of the gas stated in (a)(i).

(iv) Give a reason why the precaution stated in (a)(ii) must be taken.

(b)(i) State Dalton’s law of partial pressures.

(ii) The volume of a sample of methane collected over-water at a temperature of 12°C and a pressure of 700 mmHg was 30cm3

. Calculate the volume of the dry gas at s.t.p. [Saturated vapour pressure of water at 12°C is 10 mmHg]

(c)(i) Write an equation for the reaction between chlorine and water.

(ii) Why does litmus paper turn red when put in the resulting solution in (c)(i)?

(d)(i) State the trend in the boiling points of chlorine, bromine and iodine.

(ii) Explain briefly why water has a higher boiling point than ammonia.

Solutions & Explanation:

(a)(i) A labelled diagram for the laboratory preparation of a dry, sample of sulphur (iv) oxide.

(Any one situable but must correspond to reactants in the diagram)

(ii) Balanced chemical equation for the reaction in 4(a)(i)

NaSO$_{3(aq)}$ + 2HCl$_{(aq)}$ —> 2NaCl$_{3(aq)}$ H2O$_{(l)}$ + SO$_{2(g)}$

K$_{2}$SO$_{3(aq)}$ + 2HCl$_{(aq)}$ —> H$_2$O$_{l}$ + SO$_{2(g)}$

NaSO$_{3(aq)}$ + H$_2$SO$_{4(aq)}$ —> 2Na$_{2}$SO$_4$ + H$_2$O$_{(l)}$ + SO$_{2(g)}$

(iii) Precaution in the preparation of the gas in 4(a)(i). The gas should be prepared in a fume cupboard (chamber).

(iv) Reason for the precaution. It is because the gas is poisonous.

(b)(i) Dalton’s law of partial pressures states that for a mixture of gases which do not react, the total pressure is equal to the sum of the partial pressures of the individual gases.

(ii) Volume of methane (CH$_4$) collected over water at temperature of 12°C and a pressure of 700mmHg was 30cm$_3$.

Volume of the gas (dry) at s.t.p.

$\frac{P_1V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$P_1$ = (700 — 10) = 690 mmHg (dry gas)

$T_1$ = 12 + 273 = 285 K

$V_2$ = $\frac{P_1V_1T_2}{P_2T_1}$

P$_2$ = 760mmHg, T$_2$ = 273k

$V_2$ = $\frac{690 \times 30 \times 273}{760 \times 285}$

$V_2$ = 26.09cm$^3$

$V_2$ = 26.10cm$^3$

(c)(i) Equation for the reaction between chlorine and water. Cl$_{2(g)}$ + H$_2$O $\to$ HCI + HCIO

(ii) Litmus paper turned red (from blue to red) because H$^+$ ions are released or HCI produced is acidic. Also HCIO produced is acidic too. The solution is acidic.

(d)(i) Trend in the boiling points of chlorine, bromine and iodine. Boiling point of iodine is greater than that of bromine which is greater than that of chlorine. Boiling point increases from chlorine to iodine.

(ii) Water has a higher boiling point than ammonia because water has stronger hydrogen bonds than ammonia. Also because oxygen atom has two lone pairs of electrons and nitrogen atom has one pair of electron. Also oxygen is more electronegative than nitrogen.

4. (a)(i) State two industrial uses of hdrogen.
(ii) Consider the equation below.
1. State the type of hardness of water being removed as shown by the above equation.
2. Give two disadvantages of hardness of water.
(b)(i) In the extraction of aluminium by electrolysis, graphite electrodes are used. State the disadvantages of using this type of electrode.
(ii) Calcuim oxide reacts with water to form slaked line: I. Write a balanced equation for this reaction; II. State one use of slaked line.
(c)(i) What is meant by saponification?
(ii) List the raw materials needed for the manufacture of soap.
(iii) Name the main by-product obtained from the manufacture of soap.
(d) With the aid of chemical equations explain briefly how iron is extracted in the blast furnace using iron ore, coke and limestone as raw materials at the:
(i) bottom of the furnace; (ii) middle of the furnace (iii) top of the furnace.

Solutions & Explanation:

(a)(i) Two industrial uses of hydrogen: —filling of weather balloons. —manufacture of plastics. —manufacture of ammonia by the Haber process. —manufacture of methanol. —oil refinery process/hydrocracking —used in fuel cells. —manufacture of margarine. —in oxy-hydrogen flame (for welding and cutting metals). —as reducing agent in production of metals, e.g. Cu and.Pb from their oxides. —conversion of coal to crude oil. —(liquid) hydrogen is used as rocket fuel/gaseous fuel (Any 2).

(ii) Mg(HCO$_3$)$_{2(aq)}$ —> MgCO$_3$$_{(s)}$ + H$_2$O$_{(l)}$ + Cu$_3$.

In the equation above: (1) the type of hardness of water being removed is temporary hardness of water. (2) two disadvantges of hardness of water are: —scales/furrinab of kettles/pipes causing blockage leading to power wastage. —soap is wasted not suitable for laundry. —not suitable for use in tanning, textiles and paper production (Any 2).

(b)(i) Disadvatnages of using graphite electrodes in the extraction of aluminium by electrolysis: At high temperature oxygen gas produced at the anode reacts with the graphite electrodes to form carbon(IV) oxide.

(ii) Balanced equation of reaction between calcium oxide and water to form slaked lime: CaO + H$_2$O $\to$ Ca(OH)$_2$

(iii) One use of slaked lime: —used to treat acidic soil. —used to make mortar/plasters/cement/hold bricks. —in manufacture of glass. —purification of water/removing hardness of water. —manufacture of white wash. —to test for CO$_2$. —production of bleaching powder: —to recover or produce ammonia from ammonium chloride used in solvay process.

(c)(i) Saponification is the alkaline hydrolysis of fats/oils to produce soap and propane —1,2,3-triol

(ii) Raw materials needed for the manufacture of soap: —oil/fat. —alkali (1) (sodium hydroxide/potassium hydroxide

(iii) Main by-product obtained from the manufacture of soap is propane —1,2,3-triol.

(d) Extraction of iron in the blast furnace using iron-ore, coke and limestone as raw materials with the aid of chemical equations..

(i) Bottom of furnace: The O$_2$ in the hot air combines with carbon(coke) to produce carbon (IV) oxide which is reduced to carbon (II) oxide.

C + O$_2$ $\to$ CO$_{2(g)}$

CO$_{2(g)}$ + C $\to$ 2CO$_{(g)}$

(ii) Middle of the furnace: The calcium carbonate CaCO$_3$ decomposes under great heat to form CO$_2$ and CaO and CaCO$_3$ decomposes to remove the impurities SiO$_2$

CaCO$_3$ —> CaO$_s$ + CO$_{2(g)}$

CaO + SiO$_2$ –> CaSiO$_3$.

(iii) Top of furnace: Iron (III) oxide is reduced to metallic iron Fe$_2$O$_2$ + 3CO 2Fe + 3CO$_2$ OR Fe$_3$O$_4$ + 4CO —> 3Fe + 4CO$_2$.

5. (a) (i) Define the term fermentation

(ii) Name the catalyst that can be used for this process

(b) Name two factors which determine the choice of an indicator for an acid-base titration

(c) Consider the following reaction equation: Fe + H$_2$SO$_4$ $\to$ FeSO$_4$ + H$_2$. Calculate the mass of unreacted iron when 5.0g of iron reacts with 10cm$^3$ of 1.0 moldm$^3$ H$_2$SO$_4$, [Fe = 56.0]

(d) Name one:

(i) Heavy chemical used in electrolytic cells

(ii) Fine chemical used in textile industries

(e) Explain briefly how a catalyst increases the rate of a chemical reaction.

(f) (i) Write the chemical formula for the product formed when ethanoic acid reacts with ammonia

(ii) Give the name of the product formed in (f)(i)

(g) List three properties of aluminum that makes it suitable for the manufacture of drinks cans

(h) State two industrial uses of alkylalkanoates

(i) Name two steps involved in the crystallization of a salt from its solution

(j) List two effects of global warming

Solution & Explanation:

(ai) Fermentation is the conversation of aqueous glucose or sugar to ethanol/alcohol and release of carbon (iv) oxide as a by product. Alternatively, it is the decomposition or breakdown of glucose/starch/carbohydrate to ethanol/alcohol and release of carbon (iv) oxide

(aii)
Zymase

(b)

-The strength of an acid
-The PH of the solution

No. of moles of H$_2$SO$_4$ = 10cm$^3$

Concentration = 1mol/dm$^3$

Amount = Vol x Concentration

= $\frac{10}{100}$ x 1

= 0.01mol

1 mole of H$_2$SO$_4$ = 1 mole of Fe

Mass of Fe reacted = 56 x 0.01g = 0.56g

Mass of unreacted Fe = (5.0 – 0.056)g = 4.44g

(di) tetraoxsulphate (iv)

(dii). dyes

(e) A catalyst increases the rate of reaction by lowering the activation energy of the reaction

(fi) CH$_3$COOH$_s$ + NH3$_g$ → CH$_3$COONH$_4$

(fii) Ammonium ethanoate

(g)

(a) It doesn’t corrode easily.

(b) it doesn’t react with contain inside.

(h) (a) Sweet production industries

(b) perfume industries.

(i)
(a) Evaporation

(b)seedling

(j)
(a) it increases the volume water in ocean and seas.

(b) It causes increase in rainfall and also corrode painted surfaces.

Chemistry 2018 Past Questions | WAEC

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