Non-Metals and Their Compounds Practice Questions — JAMB UTME — Page 6

Question 101Question →

In the industrial manufacture of tetraoxosulfate(VI) acid via the Contact Process, sulfur(VI) oxide ( $SO_3$ ) gas is absorbed into concentrated tetraoxosulfate(VI) acid to form oleum ( $H_2S_2O_7$ ) rather than being dissolved directly in water. What is the primary chemical reason for avoiding direct dissolution in water?

Direct reaction with water is violently exothermic and creates a dense mist of acid droplets that is extremely difficult to condense.

Water oxidizes sulfur(VI) oxide into trioxosulfate(IV) acid instead of tetraoxosulfate(VI) acid.

Direct addition of water shifts the dynamic equilibrium position backward, significantly reducing the yield of product.

Water fails to react stoichiometrically with sulfur(VI) oxide gas at standard molar volume conditions.

Show answer & explanation

Answer: Direct reaction with water is violently exothermic and creates a dense mist of acid droplets that is extremely difficult to condense.

Answer

Direct reaction with water is violently exothermic and creates a dense mist of acid droplets that is extremely difficult to condense.

Direct addition of sulfur(VI) oxide gas to water liberates excessive thermal energy, rapidly boiling the water and creating a fog or mist of fine tetraoxosulfate(VI) acid droplets that cannot be collected easily in industrial towers. Absorbing

SO_3

in 98%

H_2SO_4

forms oleum (

H_2S_2O_7

) controlledly without mist generation.

Step-by-Step Solution

1

Analyze the chemical interaction between

SO_3

and

H_2O

The reaction

SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)

is extremely exothermic.

Large enthalpy of hydration releases intense localized thermal energy.

2

Identify the physical consequence of direct hydration

The heat vaporizes surrounding water and acid, forming a micro-droplet acid aerosol/mist.

Fine mist droplets cannot easily settle or condense using conventional absorption towers.

3

Evaluate the industrial solution in the Contact Process

SO_3

is absorbed smoothly into 98%

H_2SO_4

to form oleum (

H_2S_2O_7

), which is subsequently diluted safely with water.

Absorption in concentrated acid avoids mist formation while maintaining high efficiency.

Key Concept

Contact Process

SO_3

Absorption Mechanics

Question 102Question →

During the first stage of the Contact Process for the industrial manufacture of tetraoxosulfate(VI) acid, pure sulfur is burned in dry air to produce sulfur(IV) oxide gas according to the equation:

S(s) + O_2(g) \rightarrow SO_2(g)

What volume of sulfur(IV) oxide gas, in

\text{dm}^3

measured at standard temperature and pressure (STP), is produced by the complete combustion of

16.0\text{ g}

of sulfur?
[Relative atomic mass:

S = 32

; Molar volume of gas at STP =

22.4\text{ dm}^3\text{ mol}^{-1}

]

Show answer & explanation

Answer: 11.2

Answer

The volume of sulfur(IV) oxide gas produced at STP is

11.2\text{ dm}^3

.

According to the balanced equation

S(s) + O_2(g) \rightarrow SO_2(g)

,

1\text{ mol}

(

32\text{ g}

) of sulfur yields

1\text{ mol}

(

22.4\text{ dm}^3

at STP) of sulfur(IV) oxide gas. Therefore,

16.0\text{ g}

of sulfur corresponds to

\frac{16.0}{32} = 0.50\text{ mol}

, which produces

0.50 \times 22.4\text{ dm}^3 = 11.2\text{ dm}^3

of

SO_2

gas at STP.

Step-by-Step Solution

1

Calculate the amount in moles of sulfur reacted.

0.50\text{ mol}

of sulfur.

Using the formula

\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{16.0\text{ g}}{32.0\text{ g mol}^{-1}} = 0.50\text{ mol}

.

2

Use the mole ratio from the balanced chemical equation to find moles of sulfur(IV) oxide gas formed.

0.50\text{ mol}

of

SO_2(g)

.

The equation shows a 1:1 stoichiometric ratio between

S(s)

and

SO_2(g)

.

3

Calculate the gas volume at standard temperature and pressure (STP).

11.2\text{ dm}^3

.

Multiply the moles of gas by the molar volume at STP:

V = 0.50\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3

.

Key Concept

Stoichiometric Volume Calculations for Gas Generation in the Contact Process

Question 103Question →

Match each noble gas listed on the left with its corresponding primary industrial application or physical property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Helium matches deep-sea diving breathing mixtures (heliox); Neon matches orange-red advertising discharge tubes; Argon matches inert shielding atmosphere in arc welding; Krypton matches high-efficiency airport runway lighting.

Each noble gas possesses distinct physical properties leading to specific applications: Helium's low blood solubility makes it essential for diving mixtures; Neon's electrical excitation spectrum yields orange-red sign lighting; Argon's abundance and chemical inertness provide a protective shield during welding; Krypton's high atomic mass improves filament life in specialized high-intensity lighting.

Step-by-Step Solution

1

Identify the low solubility property of Helium

Helium replaces nitrogen in deep-sea breathing gas (heliox).

Prevents painful decompression sickness because helium is significantly less soluble in human blood under high pressure.

2

Determine the atomic spectrum property of Neon

Neon produces a characteristic orange-red glow in gas discharge lamps.

Electron transitions in excited neon gas release photons with wavelengths corresponding to reddish-orange light.

3

Analyze the industrial application of Argon in metallurgy

Argon serves as an inert protective blanket in electric arc welding.

Being non-reactive and atmospheric abundant, it displaces atmospheric oxygen and nitrogen during metal joining.

4

Relate Krypton's atomic mass to incandescent lighting efficiency

Krypton is used in high-intensity airport runway bulbs.

Heavy noble gas atoms retard the thermal evaporation of tungsten filaments.

Key Concept

Specific industrial applications and unique physical/chemical characteristics of noble gases (Group 18).

Estimated Time:1m 30s

Question 104Question →

Helium is used in deep-sea diving gas mixtures (heliox) to prevent decompression sickness and in meteorological balloons to provide lift. Which set of properties makes helium suitable for both of these applications?

Low solubility in blood under pressure, low density, and non-flammability

High solubility in blood under pressure, low boiling point, and strong reducing ability

High density, rapid reaction with oxygen, and high thermal conductivity

Ease of liquefaction at room temperature, high solubility, and low ionization energy

Show answer & explanation

Answer: Low solubility in blood under pressure, low density, and non-flammability

Answer

Low solubility in blood under pressure, low density, and non-flammability

Helium's

1s^2

stable electronic duplet makes it non-flammable and chemically inert. Its small molar mass (

4\text{ g/mol}

) makes it less dense than air, providing lift for weather balloons. Furthermore, its exceptionally low solubility in blood under pressure prevents decompression sickness ('the bends') in deep-sea divers.

Step-by-Step Solution

1

Analyze the requirements for deep-sea diving gas mixtures (heliox).

The gas mixed with oxygen must have low solubility in blood at high underwater pressures to prevent nitrogen narcosis and painful bubble formation upon decompression.

Helium replaces nitrogen in heliox because of its minimal blood solubility under pressure.

2

Analyze the requirements for meteorological (weather) balloons.

The gas must be less dense than air to provide upward buoyant force and non-flammable to prevent explosion hazards.

Helium has a molar mass of 4 g/mol (much lower than air's average of 29 g/mol) and a complete

1s^2

valence shell, making it non-flammable and safe compared to hydrogen.

3

Synthesize the properties into a single matching choice.

The combination of low blood solubility under pressure, low density, and chemical non-flammability correctly describes helium's behavior.

This set of physical and chemical properties uniquely qualifies helium for both applications.

Key Concept

Properties and Applications of Helium

Estimated Time:1m 0s

Question 105Question →

Match each noble gas listed on the left with its correct industrial application or property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Helium pairs with deep-sea diving gas mixtures; Neon pairs with orange-red discharge advertising lamps; Argon pairs with inert shield in arc welding; Radon pairs with cancer radiotherapy.

Each noble gas has specific industrial applications: Helium is utilized in deep-sea diving mixtures due to low blood solubility; Neon is used in glowing discharge signs; Argon serves as an inert protective atmosphere in high-temperature welding; Radon is radioactive and used in cancer radiotherapy.

Step-by-Step Solution

1

Analyze the unique physical and chemical properties of each noble gas listed.

Helium is non-flammable with low blood solubility; Neon exhibits characteristic light emission under electrical discharge; Argon is chemically inert and relatively cheap; Radon is radioactive.

Matching each noble gas to its primary industrial use depends on these distinct physical and chemical properties.

2

Correlate each gas to its specific practical application.

Helium matches diving gas mixture dilution (Heliox); Neon matches advertising discharge signs; Argon matches metal arc welding inert environment; Radon matches cancer treatment radiotherapy.

This establishes the precise pairs based on standard JAMB Chemistry syllabus requirements for noble gases.

Key Concept

Industrial applications and properties of Group 0 elements

Question 106Question →

Chlorine gas dissolves in water to produce a pale greenish-yellow mixture known as chlorine water. When this solution is left exposed to bright sunlight for a prolonged period, the solution loses its color and gas bubbles are observed escaping from the container. Which gas is evolved, and which specific component of chlorine water decomposes to produce it?

Oxygen gas (

\text{O}_2

), produced by the light-catalyzed decomposition of hypochlorous acid (

\text{HClO}

)

Hydrogen gas (

\text{H}_2

), produced by the thermal decomposition of hydrochloric acid (

\text{HCl}

)

Chlorine gas (

\text{Cl}_2

), produced by the volatilization of dissolved chloride ions

Oxygen gas (

\text{O}_2

), produced by the direct decomposition of hydrochloric acid (

\text{HCl}

)

Show answer & explanation

Answer: Oxygen gas ( $\text{O}_2$ ), produced by the light-catalyzed decomposition of hypochlorous acid ( $\text{HClO}$ )

Answer

Oxygen gas (

\text{O}_2

), produced by the light-catalyzed decomposition of hypochlorous acid (

\text{HClO}

)

When chlorine dissolves in water, it undergoes a reversible reaction to form hydrochloric acid (

\text{HCl}

) and hypochlorous acid (

\text{HClO}

). Hypochlorous acid is a weak, unstable oxoacid that decomposes in the presence of sunlight to yield additional hydrochloric acid and oxygen gas (

2\text{HClO} \rightarrow 2\text{HCl} + \text{O}_2

). The release of oxygen gas bubbles accounts for the observed gas evolution, while the removal of green

\text{Cl}_2

shifts the initial equilibrium until the pale color disappears completely.

Step-by-Step Solution

1

Identify the equilibrium products of chlorine gas dissolved in water.

Chlorine reacts reversibly with water to form hydrochloric acid (

\text{HCl}

) and hypochlorous acid (

\text{HClO}

):

\text{Cl}_2\text{(g)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HCl(aq)} + \text{HClO(aq)}

.

Chlorine water is an equilibrium mixture containing both acids.

2

Determine the photochemical behavior of the oxoacid component in sunlight.

Hypochlorous acid (

\text{HClO}

) is photochemically unstable and decomposes under bright sunlight:

2\text{HClO(aq)} \xrightarrow{\text{sunlight}} 2\text{HCl(aq)} + \text{O}_2\text{(g)}

.

The weak

\text{O-Cl}

bond in hypochlorous acid readily breaks upon absorption of ultraviolet/visible light energy.

3

Conclude the identity of the evolved gas and the decomposing species.

The evolved gas is oxygen (

\text{O}_2

), and the decomposing species is hypochlorous acid (

\text{HClO}

).

Hydrochloric acid remains in solution, causing the overall solution to become gradually more acidic as hypochlorous acid decomposes.

Key Concept

Photochemical decomposition of hypochlorous acid in chlorine water

Estimated Time:1m 0s

Question 107Question →

During the industrial isolation of atmospheric gases from liquid air, argon is obtained in a significantly higher yield than any other noble gas. Which chemical property of argon makes it superior to nitrogen for filling high-temperature electric filament bulbs?

Argon is a monatomic gas with a stable octet configuration, rendering it completely chemically inert even at white-hot temperatures where nitrogen would react with the tungsten filament.

Argon has a higher boiling point than nitrogen, causing it to condense inside the bulb and cool the filament during operation.

Argon acts as an efficient reducing agent that chemically converts any evaporated tungsten oxide back into metallic tungsten.

Argon forms a stable diatomic gas that readily reacts with trace oxygen to produce a protective nitride coating on the filament.

Show answer & explanation

Answer: Argon is a monatomic gas with a stable octet configuration, rendering it completely chemically inert even at white-hot temperatures where nitrogen would react with the tungsten filament.

Answer

Argon is a monatomic gas with a stable octet configuration, rendering it completely chemically inert even at white-hot temperatures where nitrogen would react with the tungsten filament.

The correct answer highlights that argon has a stable octet valence shell (

3s^2 3p^6

). At white-hot temperatures inside an incandescent light bulb, nitrogen gas can react with tungsten to form nitrides, shortening the filament life. Argon is completely inert and monatomic, preventing oxidation or chemical degradation of the filament.

Step-by-Step Solution

1

Analyze the electronic structure of argon.

Argon (atomic number 18) has an electronic configuration of 2,8,8 (or

1s^2 2s^2 2p^6 3s^2 3p^6

), possessing a full octet in its outermost shell.

A full octet gives argon extreme chemical inertness.

2

Compare argon's chemical reactivity at high temperatures with nitrogen.

Nitrogen (

N_2

) can react with tungsten at white-hot temperatures (above

2000^\circ\text{C}

) to form tungsten nitride, whereas argon remains completely unreactive.

High heat in light bulbs can break nitrogen's triple bond and cause chemical reaction with the metal filament, whereas argon cannot react.

Key Concept

Chemical inertness of noble gases due to stable octet electronic structure and their application as protective atmospheres.

Question 108Question →

During the laboratory preparation of oxygen gas, a sample of

17.0\text{ g}

of hydrogen peroxide (

\text{H}_2\text{O}_2

) decomposes completely in the presence of a manganese(IV) oxide catalyst according to the equation:

2\text{H}_2\text{O}_2(\text{aq}) \xrightarrow{\text{MnO}_2} 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})

What volume of oxygen gas, measured at standard temperature and pressure (STP), is released in this process?
[Molar mass of

\text{H}_2\text{O}_2 = 34.0\text{ g mol}^{-1}

; Molar volume of gas at STP

= 22.4\text{ dm}^3\text{ mol}^{-1}

]

2.80\text{ dm}^3

5.60\text{ dm}^3

6.00\text{ dm}^3

11.2\text{ dm}^3

Show answer & explanation

Answer: $5.60\text{ dm}^3$

Answer

5.60\text{ dm}^3

The decomposition of

17.0\text{ g}

of

\text{H}_2\text{O}_2

yields

0.50\text{ mol}

of reactant. According to the balanced chemical equation,

2\text{ moles}

of

\text{H}_2\text{O}_2

yield

1\text{ mole}

of

\text{O}_2

, producing

0.25\text{ mol}

of oxygen gas. At standard temperature and pressure (STP),

0.25\text{ mol}

occupies

0.25 \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.60\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the amount in moles of hydrogen peroxide (

\text{H}_2\text{O}_2

) reactant.

n(\text{H}_2\text{O}_2) = \frac{17.0\text{ g}}{34.0\text{ g mol}^{-1}} = 0.50\text{ mol}

Converting mass to moles using the molar mass provides the quantity of reactant available.

2

Determine the moles of oxygen gas (

\text{O}_2

) formed using equation stoichiometry.

Since

2\text{ mol } \text{H}_2\text{O}_2 \rightarrow 1\text{ mol } \text{O}_2

,

n(\text{O}_2) = \frac{0.50\text{ mol}}{2} = 0.25\text{ mol}

The balanced chemical equation shows a 2:1 molar ratio between reactant and gaseous product.

3

Calculate the volume of oxygen gas produced at STP.

V(\text{O}_2) = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.60\text{ dm}^3

Multiplying the calculated moles of gas by the standard molar gas volume gives the volume at STP.

Key Concept

Stoichiometric calculations and gas molar volume at STP for oxygen preparation.

Estimated Time:1m 30s

Question 109Question →

Under the same conditions of temperature and pressure, how many times faster does protium gas ( $\text{H}_2$ ) diffuse through a porous plug compared to tritium gas ( $\text{T}_2$ )? [Relative atomic masses: protium, $^{1}\text{H} = 1.0$ ; tritium, $^{3}\text{H} = 3.0$ ]

1.73

3.00

0.58

9.00

Show answer & explanation

Answer: $1.73$

Answer

Protium gas diffuses approximately

1.73

times faster than tritium gas.

The relative rate of diffusion of protium gas (

\text{H}_2

) relative to tritium gas (

\text{T}_2

) is governed by Graham's Law, which states that

R_1 / R_2 = \sqrt{M_2 / M_1}

. Given

M(\text{H}_2) = 2\text{ g mol}^{-1}

and

M(\text{T}_2) = 6\text{ g mol}^{-1}

, the ratio is

\sqrt{6 / 2} = \sqrt{3} \approx 1.73

. Thus, protium gas diffuses

1.73

times faster.

Step-by-Step Solution

1

Calculate the molar mass of diatomic protium gas (

\text{H}_2

) and tritium gas (

\text{T}_2

).

M(\text{H}_2) = 2 \times 1.0 = 2.0\text{ g mol}^{-1}

and

M(\text{T}_2) = 2 \times 3.0 = 6.0\text{ g mol}^{-1}

.

Graham's law of diffusion requires the molar masses of the gaseous species.

2

Apply Graham's Law of diffusion:

\frac{R_{\text{H}_2}}{R_{\text{T}_2}} = \sqrt{\frac{M_{\text{T}_2}}{M_{\text{H}_2}}}

.

\frac{R_{\text{H}_2}}{R_{\text{T}_2}} = \sqrt{\frac{6.0}{2.0}} = \sqrt{3.0}

.

The rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass.

3

Evaluate the square root to determine the ratio.

\sqrt{3.0} \approx 1.73

.

This yields the relative diffusion rate of protium gas compared to tritium gas.

Key Concept

Graham's Law of Diffusion applied to Hydrogen Isotopes

Estimated Time:1m 15s

Non-Metals and Their Compounds

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