Alkanes: Properties, Reactions, and Petroleum Refining Practice Questions — JAMB UTME

Question 1Question →

What volume of oxygen at STP is required for the complete combustion of $10\text{ dm}^3$ of propane gas ( $C_3H_8$ ) measured at the same temperature and pressure?

10\text{ dm}^3

30\text{ dm}^3

50\text{ dm}^3

22.4\text{ dm}^3

Show answer & explanation

Answer: $50\text{ dm}^3$

Answer

The volume of oxygen required at STP for the complete combustion of

10\text{ dm}^3

of propane is

50\text{ dm}^3

.

According to the balanced chemical equation

C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)

,

1\text{ dm}^3

of propane requires

5\text{ dm}^3

of oxygen for complete combustion at the same temperature and pressure. Therefore,

10\text{ dm}^3

of propane requires

10 \times 5 = 50\text{ dm}^3

of oxygen.

Step-by-Step Solution

1

Write and balance the chemical equation for the complete combustion of propane gas.

C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)

A balanced chemical equation provides the correct mole and volume ratios of reactants and products.

2

Apply Gay-Lussac's Law of Combining Volumes for gases at the same temperature and pressure.

1\text{ volume of } C_3H_8 \text{ reacts with } 5\text{ volumes of } O_2

The volume ratio of gaseous reactants equals the ratio of their stoichiometric coefficients.

3

Multiply the given volume of propane by the volume ratio coefficient.

\text{Volume of } O_2 = 10\text{ dm}^3 \times 5 = 50\text{ dm}^3

Calculating

10 \times 5

gives the total volume of oxygen required.

Key Concept

Combustion Stoichiometry and Gay-Lussac's Law of Combining Volumes

Question 2Question →

Three isomeric alkanes have the molecular formula $C_5H_{12}$ : pentane, 2-methylbutane, and 2,2-dimethylpropane. Which of the following statements correctly accounts for the trend in their boiling points?

Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

2,2-Dimethylpropane has the highest boiling point because its compact spherical shape maximizes molecular interactions.

2-Methylbutane has a higher boiling point than pentane because branching increases the strength of intermolecular forces.

All three isomers have identical boiling points because they possess the same molecular formula and molar mass.

Show answer & explanation

Answer: Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

Answer

Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

Pentane possesses an unbranched, straight-chain hydrocarbon structure. This spatial arrangement allows adjacent molecules to lie close together with maximum surface contact. As a result, intermolecular van der Waals forces are strongest in pentane, requiring the highest temperature to transition from liquid to gas.

Step-by-Step Solution

1

Analyze the molecular structures of the three

C_5H_{12}

isomers.

Pentane is unbranched (straight-chain), 2-methylbutane is monobranched, and 2,2-dimethylpropane is highly branched (spherical).

Structural branching determines molecular shape and the overall contact area available between molecules.

2

Relate molecular shape to intermolecular forces.

Alkanes are non-polar and held together by weak London dispersion (van der Waals) forces, which scale with molecular surface contact area.

Greater surface contact area leads to stronger attractive forces that require more thermal energy to overcome.

3

Determine the boiling point trend based on surface area.

Pentane has the largest surface area of contact, giving it the strongest intermolecular forces and the highest boiling point, while 2,2-dimethylpropane has the lowest.

Increased branching compacts the molecule into a sphere, minimizing contact area and lowering the boiling point.

Key Concept

Effect of structural branching on alkane boiling points and intermolecular forces

Estimated Time:1m 0s

Question 3Question →

During the fractional distillation of crude oil in a refining column, components separate based on differences in their boiling point ranges. Which of the following petroleum fractions is collected at the very top of the fractionating column?

Refinery gas

Bitumen

Kerosene

Diesel oil

Show answer & explanation

Answer: Refinery gas

Answer

Refinery gas is collected at the very top of the fractionating column.

Refinery gas comprises short-chain alkanes containing 1 to 4 carbon atoms. Because these small molecules experience weak intermolecular forces, they possess the lowest boiling points among all petroleum fractions and rise to the coolest region at the very top of the fractionating tower.

Step-by-Step Solution

1

Analyze how fractional distillation separates petroleum fractions.

Fractions separate according to their boiling point ranges, which depend on hydrocarbon chain length and molar mass.

Smaller alkane molecules have weaker van der Waals forces and lower boiling points.

2

Determine the temperature gradient in a fractionating column.

The bottom of the column is the hottest, while the top is the coolest.

Vapors rise through the column and condense when the temperature drops below their respective boiling points.

3

Identify the fraction with the lowest boiling point.

Refinery gas (

C_1

–

C_4

alkanes like methane, ethane, propane, and butane) has the lowest boiling point range (

< 20^\circ\text{C}

) and remains gaseous at the top.

Components with the lowest boiling points travel to the coolest section at the top before exiting.

Key Concept

Fractional Distillation of Crude Oil and Boiling Point Trends

Estimated Time:45s

Question 4Question →

Arrange the following crude oil (petroleum) fractions in order of increasing boiling point range, starting from the fraction with the lowest boiling point to the one with the highest boiling point.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest boiling point range is Refinery gas, followed by Petrol (Gasoline), Kerosene (Paraffin), and Diesel oil (Gas oil).

In the fractional distillation of petroleum, fractions condense and exit the fractionating column at different levels depending on their boiling point ranges. Smaller alkane molecules have lower molecular masses, weaker intermolecular forces, and lower boiling points. Thus, Refinery gas (

C_1-C_4

) has the lowest boiling point, followed by Petrol (

C_5-C_{10}

), Kerosene (

C_{10}-C_{16}

), and Diesel oil (

C_{14}-C_{20}

) with the highest boiling point among the listed options.

Step-by-Step Solution

1

Determine the relationship between carbon chain length and boiling point in alkane fractions.

Smaller hydrocarbon molecules have weaker intermolecular van der Waals forces and therefore lower boiling points.

Boiling point increases as the number of carbon atoms per molecule increases.

2

Identify the carbon chain lengths for each given fraction.

Refinery gas (

C_1-C_4

), Petrol (

C_5-C_{10}

), Kerosene (

C_{10}-C_{16}

), and Diesel oil (

C_{14}-C_{20}

).

Fractional distillation separates crude oil based on boiling point ranges governed by molecular sizes.

3

Arrange the fractions from smallest carbon number to largest carbon number.

Refinery gas

\rightarrow

Petrol

\rightarrow

Kerosene

\rightarrow

Diesel oil.

This sequence directly corresponds to increasing boiling point range.

Key Concept

Fractional Distillation and Boiling Point Trends of Petroleum Fractions

Estimated Time:45s

Question 5Question →

In petroleum refining, catalytic cracking breaks down long-chain hydrocarbons into smaller, economically valuable fractions. If one mole of dodecane ( $C_{12}H_{26}$ ) undergoes thermal-catalytic cracking to yield one mole of hexane ( $C_6H_{14}$ ) and two moles of an alkene product $X$ , what is the molecular formula of $X$ ?

C_3H_6

C_3H_8

C_6H_{12}

C_6H_{14}

Show answer & explanation

Answer: $C_3H_6$

Answer

The molecular formula of alkene

X

is

C_3H_6

(propene).

In catalytic cracking, atomic mass is conserved. Subtracting the atoms of hexane (

C_6H_{14}

) from dodecane (

C_{12}H_{26}

) leaves 6 carbon atoms and 12 hydrogen atoms. Since 2 moles of product

X

are formed (

2X = C_6H_{12}

), dividing by 2 yields

C_3H_6

, which is propene, a valid alkene.

Step-by-Step Solution

1

Write the balanced stoichiometric equation for the cracking reaction.

C_{12}H_{26} \rightarrow C_6H_{14} + 2X

Cracking conserves the total number of carbon and hydrogen atoms between reactants and products.

2

Determine the remaining number of carbon and hydrogen atoms allocated to

2X

.

Carbons:

12 - 6 = 6

, Hydrogens:

26 - 14 = 12

. Total fragment formula is

C_6H_{12}

.

Subtract the atoms present in one mole of hexane from dodecane.

3

Divide the carbon and hydrogen count by 2 to find the formula of 1 mole of product

X

.

Carbon count for

X = \frac{6}{2} = 3

, Hydrogen count for

X = \frac{12}{2} = 6

. Formula of

X = C_3H_6

.

Since two moles of alkene

X

are produced, each mole contains half the remaining atoms, fitting the general formula for alkenes (

C_nH_{2n}

).

Key Concept

Catalytic Cracking Stoichiometry of Alkanes

Question 6Question →

Match each chemical process or reaction involving alkanes and petroleum refining in Column A with its corresponding chemical description or primary purpose in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Catalytic Cracking matches with the thermal breakdown of long-chain hydrocarbons into shorter alkanes and alkenes. Reforming matches with converting straight-chain alkanes into branched or aromatic hydrocarbons to boost octane rating. Complete Combustion matches with reacting alkanes in excess oxygen to produce

CO_2

and

H_2O

. Free-Radical Substitution matches with replacing hydrogen atoms with halogens under UV light.

Catalytic Cracking breaks larger hydrocarbon molecules into smaller, more useful molecules (alkanes and alkenes). Reforming increases fuel quality (octane rating) by isomerizing straight chains to branched chains or aromatics. Complete Combustion converts alkanes in excess oxygen to carbon dioxide and water. Free-Radical Substitution halogenates alkanes in the presence of UV light.

Step-by-Step Solution

1

Identify the primary function of Catalytic Cracking.

Cracking involves breaking heavy petroleum fractions into smaller alkanes and alkenes.

Heavy oils have low demand, whereas lighter fractions like petrol and gases have high industrial demand.

2

Identify the structural transformation involved in Reforming.

Reforming converts straight-chain alkanes into branched-chain alkanes and aromatic compounds.

Straight-chain alkanes cause engine knocking; branched and aromatic structures improve fuel efficiency by increasing the octane rating.

3

Determine the products of Complete Combustion of alkanes.

Alkanes react completely with excess oxygen to yield

CO_2(g)

and

H_2O(g)

.

Hydrocarbon oxidation in excess

O_2

yields fully oxidized carbon dioxide and water.

4

Determine the mechanism for alkane halogenation.

Halogenation of alkanes requires ultraviolet light to generate free radicals for substitution.

Alkanes are unreactive saturated hydrocarbons (paraffins) and require UV light to initiate homeolytic fission of chlorine or bromine molecules.

Key Concept

Chemical reactions of alkanes and industrial petroleum refining processes

Alkanes: Properties, Reactions, and Petroleum Refining

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