Non-Metals and Their Compounds Practice Questions — JAMB UTME — Page 2

Question 21Question →

A sample of well water contains dissolved calcium hydrogentricarbonate(IV), $\text{Ca(HCO}_3)_2$ , and magnesium tetraoxosulfate(VI), $\text{MgSO}_4$ . After boiling the water sample and filtering off the precipitate formed, which type of hardness is removed, and which salt remains in the filtrate causing persistent hardness?

Temporary hardness is removed; magnesium tetraoxosulfate(VI) remains in the filtrate.

Permanent hardness is removed; calcium hydrogentricarbonate(IV) remains in the filtrate.

Temporary hardness is removed; calcium hydrogentricarbonate(IV) remains in the filtrate.

Permanent hardness is removed; magnesium tetraoxosulfate(VI) remains in the filtrate.

Show answer & explanation

Answer: Temporary hardness is removed; magnesium tetraoxosulfate(VI) remains in the filtrate.

Answer

Temporary hardness is removed, and magnesium tetraoxosulfate(VI) remains in the filtrate.

Boiling thermal decomposition converts soluble calcium hydrogentricarbonate(IV) into insoluble calcium trioxocarbonate(IV) precipitate, thereby removing temporary hardness. Soluble magnesium tetraoxosulfate(VI) remains dissolved in the filtrate, responsible for the remaining permanent hardness.

Step-by-Step Solution

1

Identify the nature of the dissolved salts present in the water sample.

\text{Ca(HCO}_3)_2

causes temporary hardness, while

\text{MgSO}_4

causes permanent hardness.

Hydrogentricarbonate salts of calcium and magnesium cause temporary hardness, whereas soluble sulfates and chlorides cause permanent hardness.

2

Analyze the chemical effect of boiling on the water sample.

\text{Ca(HCO}_3)_2 \xrightarrow{\Delta} \text{CaCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

Heat decomposes soluble calcium hydrogentricarbonate(IV) into insoluble calcium trioxocarbonate(IV), which precipitates out, removing temporary hardness.

3

Determine the species remaining in solution after filtration.

Magnesium tetraoxosulfate(VI),

\text{MgSO}_4

, remains unchanged in solution as the filtrate.

Magnesium tetraoxosulfate(VI) is thermally stable under boiling conditions and cannot be precipitated by heating alone, retaining permanent hardness.

Key Concept

Temporary vs. Permanent Water Hardness Removal

Question 22Question →

Oxides and oxygen allotropes exhibit distinct chemical behaviors depending on their bonding and acid-base character. Match each specified oxide or oxygen compound on the left with its corresponding chemical property or classification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium superoxide (

\text{KO}_2

) matches with the superoxide compound releasing oxygen gas upon reaction with water; Dinitrogen pentoxide (

\text{N}_2\text{O}_5

) matches with the acidic oxide yielding trioxonitrate(V) acid; Carbon monoxide (

\text{CO}

) matches with the neutral oxide showing no reaction with acids or bases; Beryllium oxide (

\text{BeO}

) matches with the amphoteric oxide dissolving in both acids and alkalis.

Oxides are categorized based on their chemical behavior toward water, acids, and bases. Potassium superoxide is a superoxide species that liberates oxygen gas when mixed with water. Dinitrogen pentoxide acts as an acidic oxide, reacting with water to form nitric acid. Carbon monoxide is neutral and does not form salts with acids or bases. Beryllium oxide shows dual acid-base character, making it an amphoteric oxide.

Step-by-Step Solution

1

Analyze potassium superoxide (

\text{KO}_2

)

Identified as a superoxide species containing

\text{O}_2^-

, which yields

\text{KOH}

,

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

, and liberates

\text{O}_2

gas when hydrolysed by water.

Superoxides contain oxygen in an oxidation state of

-\frac{1}{2}

and readily liberate oxygen gas upon contact with aqueous media.

2

Analyze dinitrogen pentoxide (

\text{N}_2\text{O}_5

)

Identified as an acidic oxide (acid anhydride) which dissolves in water according to

\text{N}_2\text{O}_5 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3

.

Non-metal oxides in high oxidation states typically dissolve in water to produce oxoacids.

3

Analyze carbon monoxide (

\text{CO}

)

Identified as a neutral oxide.

Neutral oxides like

\text{CO}

,

\text{N}_2\text{O}

, and

\text{NO}

do not react with water, acids, or alkalis to form salts.

4

Analyze beryllium oxide (

\text{BeO}

)

Identified as an amphoteric oxide.

Beryllium, like aluminium, forms an oxide that reacts with both acids to give beryllium salts and strong bases to form beryllate ions (

\text{[Be(OH)_4]^{2-}}

).

Key Concept

Classification of Oxides and Oxygen Species

Question 23Question →

During a laboratory demonstration, a sample containing $100\text{ cm}^3$ of ozone gas ( $\text{O}_3$ ) undergoes complete thermal decomposition into oxygen gas ( $\text{O}_2$ ) under constant temperature and pressure. What is the total volume of oxygen gas produced?

150\text{ cm}^3

100\text{ cm}^3

66.7\text{ cm}^3

200\text{ cm}^3

Show answer & explanation

Answer: $150\text{ cm}^3$

Answer

The total volume of oxygen gas produced is

150\text{ cm}^3

.

The thermal decomposition of ozone to oxygen is represented by the balanced equation

2\text{O}_3(\text{g}) \rightarrow 3\text{O}_2(\text{g})

. According to Gay-Lussac's law, gas volumes at constant temperature and pressure react in simple stoichiometric ratios. Here, 2 units of volume of ozone yield 3 units of volume of oxygen gas. Therefore,

100\text{ cm}^3

of ozone gas yields

\frac{3}{2} \times 100\text{ cm}^3 = 150\text{ cm}^3

of oxygen gas.

Step-by-Step Solution

1

Write the balanced chemical equation for the decomposition of ozone to oxygen.

2\text{O}_3(\text{g}) \rightarrow 3\text{O}_2(\text{g})

Establishing the balanced chemical equation determines the mole and volume ratios of the gaseous species involved.

2

Apply Gay-Lussac's Law of Combining Volumes to determine the volume ratio.

2\text{ volumes of }\text{O}_3 \text{ produce } 3\text{ volumes of }\text{O}_2

At constant temperature and pressure, the volumes of reacting gases and their products are in simple whole-number ratios given by their stoichiometric coefficients.

3

Calculate the volume of oxygen gas obtained from

100\text{ cm}^3

of ozone.

\text{Volume of }\text{O}_2 = 100\text{ cm}^3 \times \frac{3}{2} = 150\text{ cm}^3

Multiplying the initial volume of ozone by the stoichiometric ratio

\frac{3}{2}

yields the final volume of oxygen produced.

Key Concept

Stoichiometry of Gas Reactions and Allotropic Conversion of Ozone

Estimated Time:1m 0s

Question 24Question →

Match each industrial fuel gas or coal distillation product in Column A with its corresponding chemical composition or production method in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Water gas matches with the mixture of carbon monoxide and hydrogen (

CO + H_2

) formed by passing steam over incandescent coke; Producer gas matches with the mixture of carbon monoxide and nitrogen (

CO + N_2

) formed by passing air over red-hot coke; Coal gas matches with the gaseous fuel mixture containing hydrogen, methane, and carbon monoxide obtained during coal distillation; Coal tar matches with the dark viscous liquid byproduct rich in aromatic hydrocarbons.

Water gas is defined as a mixture of carbon monoxide and hydrogen gas made from steam and incandescent coke. Producer gas is defined as a mixture of carbon monoxide and nitrogen gas made from air and red-hot coke. Coal gas is the fuel mixture (

H_2, CH_4, CO

) obtained from coal carbonization. Coal tar is the thick, dark liquid distillate containing aromatic hydrocarbons produced alongside coke and coal gas.

Step-by-Step Solution

1

Differentiate between the synthesis and composition of the industrial fuel gases derived from coke.

Water gas is formed using steam (

H_2O

), yielding

CO + H_2

, whereas producer gas is formed using air (

O_2 + N_2

), yielding

CO + N_2

.

Reaction of hot carbon with water vapor produces hydrogen gas, while reaction with air introduces atmospheric nitrogen gas into the gaseous product.

2

Identify the physical state and chemical nature of products formed from the destructive distillation of coal.

Coal gas is the volatile gaseous product (

H_2, CH_4, CO

), whereas coal tar is the viscous liquid distillate containing aromatic organic compounds.

Heating coal without air breaks down complex organic matter into volatile fuel gases, liquid fractions (coal tar and ammoniacal liquor), and a solid residue (coke).

Key Concept

Industrial fuel gas synthesis and coal destructive distillation products

Question 25Question →

Both diamond and graphite consist entirely of elemental carbon in the solid state, yet they exhibit markedly different physical properties such as hardness and electrical conductivity. Which of the following statements accurately explains why they are classified as allotropes?

They represent different structural forms of the same element existing in the same physical state.

They consist of atoms of the same element with identical atomic numbers but different mass numbers.

They are compounds possessing the exact same molecular formula but different structural arrangements of atoms.

They are different physical states of carbon formed at different environmental temperatures.

Show answer & explanation

Answer: They represent different structural forms of the same element existing in the same physical state.

Answer

Diamond and graphite are classified as allotropes because they represent different structural forms of the same element in the same physical state.

The correct option accurately states that allotropes are different structural arrangements of the same element within the same physical state. In diamond, each carbon atom is tetrahedrally bonded (

sp^3

hybridized), whereas in graphite, carbon atoms form planar hexagonal layers (

sp^2

hybridized).

Step-by-Step Solution

1

Define allotropy in the context of main-group non-metals like carbon.

Allotropy is identified as the phenomenon where a single element exists in multiple structural forms within the same physical state (e.g., solid graphite, diamond, and fullerenes).

Understanding the precise chemical definition distinguishes allotropy from related chemical concepts.

2

Compare allotropy with alternative concepts such as isotopy, isomerism, and phase changes.

Isotopes differ by mass/neutrons, isomers are distinct compound structures with equal formulas, and physical states refer to solid/liquid/gas phases.

Eliminating definitions of isotopes, isomers, and state changes ensures accurate concept identification.

Key Concept

Allotropy of Carbon

Question 26Question →

When carbon(IV) oxide gas is continuously bubbled through an aqueous suspension of calcium trioxocarbonate(IV), $\text{CaCO}_3$ , the white suspension gradually turns into a clear solution. What is the chemical formula of the soluble compound formed?

Show answer & explanation

Answer: Ca(HCO3)2; Ca(HCO_3)_2; calcium hydrogen trioxocarbonate(IV); calcium hydrogencarbonate

Answer

\text{Ca(HCO}_3\text{)}_2

(Calcium hydrogen trioxocarbonate(IV))

When carbon(IV) oxide gas is bubbled into limewater or a suspension of

\text{CaCO}_3

, it initially forms a cloudy white precipitate of

\text{CaCO}_3

. Upon continued bubbling of excess

\text{CO}_2

, the

\text{CaCO}_3

reacts with water and

\text{CO}_2

to yield soluble calcium hydrogen trioxocarbonate(IV),

\text{Ca(HCO}_3\text{)}_2

, causing the cloudiness to disappear.

Step-by-Step Solution

1

Identify the reactants and the chemical transformation taking place when excess carbon(IV) oxide reacts with calcium trioxocarbonate(IV) in water.

Insoluble calcium trioxocarbonate(IV) (

\text{CaCO}_3

) reacts with dissolved carbon(IV) oxide (

\text{CO}_2

) and water (

\text{H}_2\text{O}

) according to the chemical equation:

\text{CaCO}_3\text{(s)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} \rightarrow \text{Ca(HCO}_3\text{)}_2\text{(aq)}

.

The reaction converts the insoluble trioxocarbonate(IV) salt into soluble calcium hydrogen trioxocarbonate(IV), which dissolves to clear the solution.

Key Concept

Formation of soluble hydrogen trioxocarbonate(IV) from carbon(IV) oxide and trioxocarbonate(IV) salts

Question 27Question →

What mass of anhydrous sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3$ , is obtained when $16.8\text{ g}$ of sodium hydrogentrioxocarbonate(IV), \text{NaHCO}_3, is completely decomposed by heating?
[Relative atomic masses: $\text{Na} = 23$ , $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ ]

10.6\text{ g}

21.2\text{ g}

16.8\text{ g}

5.3\text{ g}

Show answer & explanation

Answer: $10.6\text{ g}$

Answer

10.6\text{ g}

of sodium trioxocarbonate(IV) is produced.

Thermal decomposition of

2\text{ moles}

(

168\text{ g}

) of

\text{NaHCO}_3

yields

1\text{ mole}

(

106\text{ g}

) of

\text{Na}_2\text{CO}_3

, along with gaseous

\text{CO}_2

and water vapor. Heating

16.8\text{ g}

(

0.20\text{ mol}

) of

\text{NaHCO}_3

produces

0.10\text{ mol}

of

\text{Na}_2\text{CO}_3

, which evaluates to exactly

10.6\text{ g}

.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of sodium hydrogentrioxocarbonate(IV).

2\text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)

Establishing the stoichiometric mole ratio between the reactant and solid product.

2

Calculate the molar masses of

\text{NaHCO}_3

and

\text{Na}_2\text{CO}_3

.

Molar mass of

\text{NaHCO}_3 = 23 + 1 + 12 + (3 \times 16) = 84\text{ g/mol}

; Molar mass of

\text{Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106\text{ g/mol}

.

Converting given masses to moles and back.

3

Determine moles of

\text{NaHCO}_3

reacted and corresponding moles of

\text{Na}_2\text{CO}_3

formed.

Moles of

\text{NaHCO}_3 = \frac{16.8}{84} = 0.20\text{ mol}

. Moles of

\text{Na}_2\text{CO}_3 = \frac{0.20}{2} = 0.10\text{ mol}

.

The balanced equation dictates a

2:1

mole ratio between

\text{NaHCO}_3

and

\text{Na}_2\text{CO}_3

.

4

Calculate the mass of

\text{Na}_2\text{CO}_3

produced.

Mass

= 0.10\text{ mol} \times 106\text{ g/mol} = 10.6\text{ g}

.

Multiplying the moles of product by its molar mass yields the final mass.

Key Concept

Thermal decomposition of hydrogen trioxocarbonates and stoichiometric mass-mass calculations

Question 28Question →

Match each nitrogen compound or microorganism in the left column with its corresponding characteristic property or role in the nitrogen cycle in the right column.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Nitrosomonas species matches oxidation of ammonium ions to nitrite ions; Nitrogen dioxide matches reddish-brown poisonous gas forming trioxonitrate(V) and dioxonitrate(III) acids; Dinitrogen monoxide matches neutral oxide that relights a glowing splint; Rhizobium species matches symbiotic nitrogen fixation in legume root nodules.

Each nitrogen species or organism is accurately paired with its specific chemical property or nitrogen cycle function: Nitrosomonas oxidizes ammonium to nitrite, Nitrogen dioxide is a reddish-brown acidic mixed anhydride, Dinitrogen monoxide is a neutral oxide supporting combustion, and Rhizobium fixes nitrogen in legume root nodules.

Step-by-Step Solution

1

Classify the biological roles of the given microorganisms in the nitrogen cycle

Nitrosomonas species catalyze the conversion of ammonium ions (

NH_4^+

) to nitrites (

NO_2^-

), whereas Rhizobium species fix free atmospheric nitrogen (

N_2

) inside the root nodules of leguminous plants.

Different soil bacteria perform distinct biochemical transformations within the nitrogen cycle.

2

Identify the physical and chemical properties of the specified oxides of nitrogen

Nitrogen dioxide (

NO_2

) is a brown acidic gas forming

HNO_3

and

HNO_2

with water, whereas dinitrogen monoxide (

N_2O

) is a neutral gas supporting combustion.

Color, acid-base nature, and thermal decomposition characteristics uniquely differentiate oxides of nitrogen.

Key Concept

Chemical properties of nitrogen oxides and biological processes of the nitrogen cycle

Question 29Question →

During the industrial preparation of nitrogen gas by fractional distillation of liquid air, nitrogen, argon, and oxygen boil off at different temperatures: $-196\ ^\circ\text{C}$ , $-186\ ^\circ\text{C}$ , and $-183\ ^\circ\text{C}$ , respectively. Which gas distills over first at the top of the fractionating column, and why?

Nitrogen gas, because its lower boiling point of

-196\ ^\circ\text{C}

makes it the most volatile component.

Oxygen gas, because its boiling point of

-183\ ^\circ\text{C}

is higher, allowing it to vaporize first upon heating.

Argon gas, because monoatomic noble gases have no intermolecular forces and evaporate before diatomic molecules.

Oxygen gas, because its higher molar mass of

32\ \text{g/mol}

causes it to diffuse to the top faster than nitrogen.

Show answer & explanation

Answer: Nitrogen gas, because its lower boiling point of $-196\ ^\circ\text{C}$ makes it the most volatile component.

Answer

Nitrogen gas distills over first because its lower boiling point of

-196\ ^\circ\text{C}

makes it the most volatile component in liquid air.

In the industrial liquefaction and fractional distillation of air, nitrogen has a boiling point of

-196\ ^\circ\text{C}

(

77\ \text{K}

), which is lower than that of argon (

-186\ ^\circ\text{C}

) and oxygen (

-183\ ^\circ\text{C}

). Because nitrogen is the most volatile component (has the lowest boiling point), it vaporizes first when the liquefied air is slowly warmed, allowing it to collect at the top of the fractionating column as a gaseous fraction.

Step-by-Step Solution

1

Compare the boiling points of the components of liquid air

Nitrogen boils at

-196\ ^\circ\text{C}

(

77\ \text{K}

), argon at

-186\ ^\circ\text{C}

(

87\ \text{K}

), and oxygen at

-183\ ^\circ\text{C}

(

90\ \text{K}

).

The component with the lowest boiling point requires the least thermal energy to change from liquid to gas state.

2

Determine which component vaporizes first as temperature rises during fractional distillation

As liquid air warms from

-200\ ^\circ\text{C}

, nitrogen reaches its boiling point first at

-196\ ^\circ\text{C}

.

Lower boiling point corresponds to higher vapor pressure/volatility at a given temperature.

3

Identify the gas collected at the top of the fractionating column

Nitrogen gas vaporizes first and is collected as distillate from the top of the column.

More volatile vapors ascend to the top of fractional distillation columns while less volatile components remain liquid longer at the base.

Key Concept

Industrial Isolation of Nitrogen Gas via Fractional Distillation of Liquid Air

Estimated Time:1m 0s

Question 30Question →

In laboratory chemistry, specific procedures and chemical tests are used to identify and handle ammonia ( $NH_3$ ) and trioxonitrate(V) acid ( $HNO_3$ ). Match each laboratory procedure on the left with its corresponding chemical observation or outcome on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Matching pairs: (1) Exposing ammonia to HCl vapor matches Observation of dense white fumes of ammonium chloride. (2) Performing the brown ring test matches Formation of a brown ring of iron(II) nitrosyl complex. (3) Passing moist ammonia through quicklime matches Efficient drying of the alkaline gas without chemical reaction. (4) Reacting concentrated trioxonitrate(V) acid with copper matches Evolution of reddish-brown gas, nitrogen(IV) oxide.

Each procedure pairs directly with its established chemical outcome: ammonia gas reacts with hydrogen chloride vapor to yield dense white ammonium chloride fumes; the brown ring test relies on forming a brown iron(II) nitrosyl complex; quicklime (

CaO

) safely dries basic ammonia without chemical reaction; and concentrated trioxonitrate(V) acid oxidizes copper to release reddish-brown nitrogen(IV) oxide gas.

Step-by-Step Solution

1

Analyze the gas-phase reaction of ammonia with hydrogen chloride vapor.

NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)

, producing dense white fumes.

Ammonia gas is alkaline and reacts with acidic hydrogen chloride gas to produce fine solid particles of ammonium chloride.

2

Identify the chemical process involved in the brown ring test for trioxonitrate(V) ions.

Iron(II) ions reduce trioxonitrate(V) to nitrogen(II) oxide (

NO

), which then combines with excess

Fe^{2+}

ions to yield

[Fe(H_2O)_5NO]^{2+}

.

Concentrated

H_2SO_4

forms a dense bottom layer, creating a distinct interface where the brown nitrosyl complex accumulates.

3

Select the appropriate desiccant for ammonia gas.

Unslaked lime / quicklime (

CaO

) dries ammonia effectively.

Acidic drying agents like

H_2SO_4

or

P_4O_{10}

react with

NH_3

, while anhydrous

CaCl_2

forms an addition complex (

CaCl_2 \cdot 8NH_3

); therefore, basic

CaO

must be used.

4

Examine the reaction between concentrated trioxonitrate(V) acid and copper metal.

Cu(s) + 4HNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g) + 2H_2O(l)

, evolving nitrogen(IV) oxide gas.

Concentrated

HNO_3

is a powerful oxidizing agent, yielding reddish-brown

NO_2

gas upon reaction with copper.

Key Concept

Laboratory Preparation, Identification, and Reactions of Ammonia and Trioxonitrate(V) Acid

Question 31Question →

During the catalytic oxidation step in the Ostwald process for the industrial preparation of trioxonitrate(V) acid, ammonia gas reacts with excess oxygen according to the reaction equation:

4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)

What is the volume of nitrogen(II) oxide gas (

\text{NO}

) produced at STP when

17.0\text{ g}

of ammonia gas is completely oxidized?
(Molar mass of

\text{NH}_3 = 17.0\text{ g mol}^{-1}

; Molar volume of gas at STP

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

)

24.0\text{ dm}^3

22.4\text{ dm}^3

5.6\text{ dm}^3

89.6\text{ dm}^3

Show answer & explanation

Answer: $22.4\text{ dm}^3$

Answer

22.4\text{ dm}^3

One mole of ammonia (

17.0\text{ g}

) yields exactly one mole of nitrogen(II) oxide gas according to the 1:1 mole ratio in the balanced chemical equation. At standard temperature and pressure (STP), one mole of any ideal gas occupies

22.4\text{ dm}^3

. Therefore, the volume of nitrogen(II) oxide gas produced is

22.4\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the amount of ammonia gas in moles.

\text{Moles of NH}_3 = \frac{17.0\text{ g}}{17.0\text{ g mol}^{-1}} = 1.0\text{ mol}

Dividing given mass by molar mass determines the number of moles of reactant present.

2

Determine the mole ratio between ammonia (

\text{NH}_3

) and nitrogen(II) oxide (

\text{NO}

).

\text{Mole ratio of NH}_3 : \text{NO} = 4 : 4 = 1 : 1

. Therefore,

1.0\text{ mol}

of

\text{NH}_3

produces

1.0\text{ mol}

of

\text{NO}

.

The balanced chemical equation shows that 4 moles of

\text{NH}_3

produce 4 moles of

\text{NO}

.

3

Calculate the volume of nitrogen(II) oxide gas produced at STP.

\text{Volume of NO} = 1.0\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 22.4\text{ dm}^3

Multiplying moles of product by the molar gas volume at STP (

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

) yields the total gas volume.

Key Concept

Gas Stoichiometry and Ostwald Process Oxidation

Question 32Question →

Match each sulfur-related substance or allotrope in Column A with its corresponding physical or chemical property in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Rhombic sulfur matches the stable octahedral crystalline allotrope below

95.6^\circ\text{C}

; Hydrogen sulfide matches the poisonous gas with a rotten-egg smell that turns lead(II) ethanoate paper black; Sulfur(IV) oxide matches the gaseous reducing agent that turns acidified potassium dichromate(VI) solution from orange to green; Plastic sulfur matches the amorphous, rubber-like allotrope formed by rapidly cooling boiling sulfur in cold water.

Rhombic sulfur is the stable octahedral crystalline allotrope at room temperature (

< 95.6^\circ\text{C}

). Hydrogen sulfide is a toxic gas characterized by a rotten-egg odor and the ability to turn moist lead(II) ethanoate paper black due to

PbS

formation. Sulfur(IV) oxide is a gaseous reducing agent that changes orange acidified potassium dichromate(VI) to green chromium(III). Plastic sulfur is an amorphous, elastic allotrope made by quenching boiling sulfur in cold water.

Step-by-Step Solution

1

Identify the physical properties and structures of sulfur allotropes.

Rhombic sulfur (

\alpha

-sulfur) consists of octahedral crystals stable below

95.6^\circ\text{C}

, while plastic sulfur is formed by pouring boiling sulfur into cold water, producing an amorphous, rubbery solid.

Allotropes exhibit different structural arrangements and stability ranges based on temperature and cooling rate.

2

Analyze qualitative test reactions for hydrogen sulfide (

H_2S

) and sulfur(IV) oxide (

SO_2

).

Hydrogen sulfide gas has a distinctive rotten-egg odor and reacts with lead(II) ethanoate to precipitate black lead(II) sulfide (

PbS

). Sulfur(IV) oxide is a reducing agent that reduces orange

Cr_2O_7^{2-}

to green

Cr^{3+}

.

Each gas has unique chemical properties and reagent tests suitable for laboratory identification.

Key Concept

Physical properties of sulfur allotropes and chemical identification tests for hydrogen sulfide and sulfur(IV) oxide.

Question 33Question →

What is the volume of chlorine gas produced at s.t.p., in $\text{dm}^3$ , when $8.7\text{ g}$ of manganese(IV) oxide ( $MnO_2$ ) reacts completely with excess concentrated hydrochloric acid ( $HCl$ )? [Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{mol}^{-1}$ , $Mn = 55$ , $O = 16$ ]

Show answer & explanation

Answer: 2.24

Answer

The volume of chlorine gas produced at s.t.p. is

2.24\text{ dm}^3

.

In the laboratory preparation of chlorine, manganese(IV) oxide acts as an oxidizing agent according to

MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O

. Since

8.7\text{ g}

of

MnO_2

represents

0.1\text{ mol}

, exactly

0.1\text{ mol}

of

Cl_2

gas is produced. At s.t.p.,

0.1\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 2.24\text{ dm}^3

.

Step-by-Step Solution

1

Write the balanced equation for the preparation of chlorine gas from manganese(IV) oxide and concentrated hydrochloric acid.

MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O

Establishes the stoichiometric relationship between

MnO_2

and

Cl_2

.

2

Calculate the molar mass of manganese(IV) oxide (

MnO_2

).

87\text{ g mol}^{-1}

Required to convert the given mass of

MnO_2

into moles.

3

Calculate the moles of

MnO_2

supplied.

\frac{8.7\text{ g}}{87\text{ g mol}^{-1}} = 0.1\text{ mol}

Determines the exact amount of reactant taking part in the reaction.

4

Multiply moles of

Cl_2

by the molar volume of a gas at s.t.p.

0.1\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

One mole of any ideal gas occupies

22.4\text{ dm}^3

at s.t.p.

Key Concept

Laboratory preparation and stoichiometry of chlorine gas evolution.

Estimated Time:1m 30s

Question 34Question →

Under the same conditions of temperature and pressure, $50\text{ cm}^3$ of sulfur(IV) oxide diffuses through a porous plug in $20\text{ seconds}$ . How long will it take for an equal volume of methane gas ( $CH_4$ ) to diffuse through the same plug under identical conditions? $[H = 1, C = 12, O = 16, S = 32]$

10\text{ seconds}

5\text{ seconds}

40\text{ seconds}

80\text{ seconds}

Show answer & explanation

Answer: $10\text{ seconds}$

Answer

The time required for an equal volume of methane gas to diffuse is

10\text{ seconds}

.

According to Graham's Law of diffusion, the time required for a fixed volume of gas to diffuse is directly proportional to the square root of its molar mass (

t \propto \sqrt{M}

). Since the molar mass of sulfur(IV) oxide (

SO_2

) is

64\text{ g mol}^{-1}

and that of methane (

CH_4

) is

16\text{ g mol}^{-1}

, the ratio of their diffusion times is

\sqrt{\frac{16}{64}} = \frac{1}{2}

. Therefore, methane takes half as long as sulfur(IV) oxide (

20 \times 0.5 = 10\text{ seconds}

).

Step-by-Step Solution

1

Calculate the molar masses of sulfur(IV) oxide (

SO_2

) and methane (

CH_4

).

M(SO_2) = 32 + (2 \times 16) = 64\text{ g mol}^{-1}

and

M(CH_4) = 12 + (4 \times 1) = 16\text{ g mol}^{-1}

.

Molar masses are required to apply Graham's Law of diffusion.

2

Set up Graham's Law equation relating diffusion time (

t

) to molar mass (

M

) for equal volumes.

\frac{t(CH_4)}{t(SO_2)} = \sqrt{\frac{M(CH_4)}{M(SO_2)}}

.

The rate of diffusion is inversely proportional to the square root of density or molar mass (

R = \frac{V}{t} \propto \frac{1}{\sqrt{M}}

).

3

Substitute the known values into the equation and solve for

t(CH_4)

.

\frac{t(CH_4)}{20} = \sqrt{\frac{16}{64}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

, so

t(CH_4) = 20 \times \frac{1}{2} = 10\text{ seconds}

.

Because methane has a smaller molar mass, it diffuses faster and takes less time to diffuse.

Key Concept

Graham's Law of Diffusion

Question 35Question →

In the industrial Contact Process, sulfur(VI) oxide ( $SO_3$ ) is hydrated to form tetraoxosulfate(VI) acid ( $H_2SO_4$ ). What volume of $SO_3$ gas, measured in $\text{dm}^3$ at s.t.p., is theoretically required to produce $196\text{ g}$ of pure $H_2SO_4$ ? [Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{mol}^{-1}$ , $H = 1.0$ , $S = 32.0$ , $O = 16.0$ ]

Show answer & explanation

Answer: 44.8

Answer

44.8 dm³

The molar mass of

H_2SO_4

is

98.0\text{ g mol}^{-1}

, so

196\text{ g}

corresponds to

2.0\text{ moles}

. According to the equation

SO_3 + H_2O \rightarrow H_2SO_4

,

1\text{ mole}

of

SO_3

yields

1\text{ mole}

of

H_2SO_4

. Therefore,

2.0\text{ moles}

of

SO_3

gas is needed, which occupies

2.0 \times 22.4\text{ dm}^3 = 44.8\text{ dm}^3

at s.t.p.

Step-by-Step Solution

1

Calculate the molar mass of tetraoxosulfate(VI) acid (

H_2SO_4

).

\text{Molar mass of } H_2SO_4 = 2(1.0) + 32.0 + 4(16.0) = 98.0\text{ g mol}^{-1}

.

Essential to convert mass of acid to chemical amount in moles.

2

Calculate the number of moles of

H_2SO_4

in

196\text{ g}

.

\text{Moles of } H_2SO_4 = \frac{196\text{ g}}{98.0\text{ g mol}^{-1}} = 2.0\text{ mol}

.

Finds the quantitative molar requirement.

3

Calculate the required volume of

SO_3

gas at s.t.p.

\text{Volume of } SO_3 = 2.0\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 44.8\text{ dm}^3

.

Based on the 1:1 mole ratio of

SO_3

to

H_2SO_4

and standard molar gas volume.

Key Concept

Molar gas volume and stoichiometric relationships in the Contact Process

Question 36Question →

During the laboratory preparation of dry hydrogen chloride gas, concentrated tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) is used as the drying agent rather than quicklime ( $\text{CaO}$ ). Which of the following explains why quicklime cannot be used to dry hydrogen chloride gas?

Quicklime is a basic oxide that chemically reacts with acidic hydrogen chloride gas to form calcium chloride and water.

Quicklime acts as a strong oxidizing agent that oxidizes hydrogen chloride gas into chlorine gas.

Quicklime is an acidic oxide that increases the pH of the solution by releasing hydrogen ions.

Quicklime physically absorbs hydrogen chloride gas without chemical change, altering the molar volume of the gas at STP.

Show answer & explanation

Answer: Quicklime is a basic oxide that chemically reacts with acidic hydrogen chloride gas to form calcium chloride and water.

Answer

Quicklime is a basic oxide that chemically reacts with acidic hydrogen chloride gas to form calcium chloride and water.

Drying agents must not react chemically with the gas being dried. Because hydrogen chloride (

\text{HCl}

) is an acidic gas and quicklime (

\text{CaO}

) is a basic oxide, they undergo a neutralization reaction forming calcium chloride and water. Concentrated tetraoxosulfate(VI) acid (

\text{H}_2\text{SO}_4

) is acidic and does not react with

\text{HCl}

, making it the correct drying agent.

Step-by-Step Solution

1

Identify the chemical nature of hydrogen chloride gas.

Hydrogen chloride (

\text{HCl}

) is an acidic gas.

It dissolves in water to form hydrochloric acid and reacts readily with bases.

2

Identify the chemical nature of quicklime (

\text{CaO}

).

Calcium oxide (

\text{CaO}

) is a basic oxide.

Group 2 metal oxides are basic oxides.

3

Determine the interaction between an acidic gas and a basic drying agent.

They undergo an acid-base neutralization reaction:

\text{CaO(s)} + 2\text{HCl(g)} \rightarrow \text{CaCl}_2\text{(s)} + \text{H}_2\text{O(l)}

.

A drying agent must be chemically unreactive towards the gas being dried.

Key Concept

Chemical compatibility of drying agents with gases

Estimated Time:1m 0s

Question 37Question →

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 corresponds to filling weather balloons; Neon corresponds to reddish-orange advertising discharge tubes; Argon corresponds to an inert gas shield during electric arc welding; Krypton corresponds to high-intensity airport runway lamps.

Helium is chosen for balloons because it is light and non-flammable. Neon produces a characteristic reddish-orange light when ionized in sign tubes. Argon is abundant and unreactive, ideal for protecting molten metals during arc welding. Krypton increases filament efficiency and brightness in specialized high-intensity beacons.

Step-by-Step Solution

1

Evaluate the unique physical characteristics and electronic stability of each Group 18 element.

Helium has low density and

1s^2

duplet stability; Neon, Argon, and Krypton have complete

ns^2 np^6

octets with specific emission spectra and densities.

Specific industrial uses rely on unreactivity, density, optical emission, or thermal conductivity.

2

Correlate each gas to its technological application.

Helium

\rightarrow

weather balloons; Neon

\rightarrow

orange signboards; Argon

\rightarrow

arc welding shield; Krypton

\rightarrow

runway beacons.

Matching properties ensures optimal performance and safety in industrial contexts.

Key Concept

Noble gases possess full valence energy levels (

1s^2

for

He

,

ns^2 np^6

for others), making them chemically unreactive and suitable for specialized optical, atmospheric, and cryogenic applications.

Question 38Question →

During the industrial processing of coal by destructive distillation in the absence of air, several products are obtained. Which of the following substances is recovered as the solid residue remaining in the retort?

Coke

Coal tar

Ammoniacal liquor

Coal gas

Show answer & explanation

Answer: Coke

Answer

Coke is the solid residue obtained from the destructive distillation of coal.

Destructive distillation involves heating coal in an airtight retort at high temperatures. Volatile compounds escape as vapors and are condensed into liquid coal tar, ammoniacal liquor, and uncondensed coal gas. The non-volatile, porous carbonaceous residue that remains inside the retort is coke.

Step-by-Step Solution

1

Analyze the conditions of destructive distillation of coal.

Coal is heated strongly in the absence of air to prevent combustion and break down complex organic compounds into volatile and non-volatile fractions.

Thermal decomposition drives off gaseous and liquid vapors while leaving fixed carbon behind.

2

Classify the resulting products by their physical states.

Coal gas is gaseous, coal tar and ammoniacal liquor condense into liquids, and coke remains as a solid in the retort.

Non-volatile elemental carbon stays behind as coke, whereas volatile components vaporize.

Key Concept

Products of coal destructive distillation

Question 39Question →

Calcium hydride ( $\text{CaH}_2$ ) reacts vigorously with water to produce calcium hydroxide and hydrogen gas. What volume of dry hydrogen gas, in $\text{dm}^3$ , measured at standard temperature and pressure (s.t.p.), is liberated when $10.5\text{ g}$ of pure calcium hydride reacts completely with excess water?

[Relative atomic masses: $\text{Ca} = 40$ , $\text{H} = 1$ ; Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{ mol}^{-1}$ ]

Show answer & explanation

Answer: 11.2

Answer

The volume of dry hydrogen gas liberated at s.t.p. is 11.2 dm³.

Calcium hydride reacts with water according to the reaction CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂. Given 10.5 g of CaH₂ (molar mass 42 g/mol), there are 0.25 moles of CaH₂. Based on the 1:2 stoichiometric ratio, 0.50 moles of H₂ gas are generated. Multiplying by the molar volume at s.t.p. (22.4 dm³/mol) yields 11.2 dm³.

Step-by-Step Solution

1

Write the balanced chemical equation for the reaction of calcium hydride with water

CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂

Establishing the stoichiometric mole ratio between the reactant CaH₂ and the product H₂ gas.

2

Calculate the molar mass of CaH₂

42 g/mol

Molar mass is required to convert the given mass of CaH₂ into moles.

3

Determine the amount of CaH₂ in moles

0.25 mol

Moles = Mass / Molar mass = 10.5 g / 42 g/mol.

4

Calculate the moles of H₂ gas liberated using the 1:2 stoichiometric ratio

0.50 mol

1 mole of CaH₂ produces 2 moles of H₂ gas.

5

Calculate the volume of H₂ gas produced at standard temperature and pressure (s.t.p.)

11.2 dm³

Volume at s.t.p. = Moles × Molar volume at s.t.p. = 0.50 mol × 22.4 dm³/mol.

Key Concept

Laboratory and industrial preparation of hydrogen using metal hydrides and mole-volume stoichiometric calculations at s.t.p.

Estimated Time:2m 0s

Question 40Question →

Liquid air undergoes fractional distillation to isolate noble gases and other atmospheric components. Based on their boiling points, in which sequence are nitrogen, argon, and oxygen collected as vapors from first to last?

Nitrogen, Argon, Oxygen

Oxygen, Argon, Nitrogen

Argon, Nitrogen, Oxygen

Nitrogen, Oxygen, Argon

Show answer & explanation

Answer: Nitrogen, Argon, Oxygen

Answer

Nitrogen, Argon, Oxygen

During the fractional distillation of liquid air, components boil off in order of increasing boiling points (most volatile to least volatile). Nitrogen has the lowest boiling point (

-196^\circ\text{C}

), followed by Argon (

-186^\circ\text{C}

), and finally Oxygen (

-183^\circ\text{C}

). Thus, the collection sequence is Nitrogen, Argon, Oxygen.

Step-by-Step Solution

1

Identify the boiling points of the three components in liquid air

Nitrogen (

N_2

) =

-196^\circ\text{C}

, Argon (

Ar

) =

-186^\circ\text{C}

, Oxygen (

O_2

) =

-183^\circ\text{C}

Fractional distillation separates liquefied gases based on differences in their boiling points.

2

Arrange the gases in increasing order of boiling points (from most volatile to least volatile)

-196^\circ\text{C} < -186^\circ\text{C} < -183^\circ\text{C}

, corresponding to Nitrogen

\rightarrow

Argon

\rightarrow

Oxygen

The substance with the lowest boiling point vaporizes and boils off first during distillation.

Key Concept

Isolation of noble gases from liquid air by fractional distillation

Non-Metals and Their Compounds

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept