Non-Metals and Their Compounds Practice Questions — JAMB UTME

Question 1Question →

Match each chemical process, reaction requirement, or observation involving ammonia ( $NH_3$ ) and trioxonitrate(V) acid ( $HNO_3$ ) on the left with its corresponding reagent or product on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair: (1) Drying agent for ammonia with Calcium oxide; (2) Ostwald process catalyst with Platinum-rhodium gauze; (3) Laboratory prep reagents with Calcium hydroxide and ammonium chloride; (4) Brown gas evolved with copper with Nitrogen dioxide.

Each item on the left correctly corresponds to its specific chemical drying agent, catalyst, preparation mixture, or redox product. Quicklime (

CaO

) dries

NH_3

safely without reacting with it. Platinum-rhodium gauze catalyzes

NH_3

oxidation in the Ostwald process. Heating

Ca(OH)_2

with

NH_4Cl

produces

NH_3

gas. Concentrated

HNO_3

oxidizes copper metal to yield nitrogen dioxide (

NO_2

) gas.

Step-by-Step Solution

1

Determine the appropriate drying agent for basic ammonia gas.

Ammonia reacts with acidic drying agents (

H_2SO_4

,

P_4O_{10}

) and forms an addition complex with

CaCl_2

. Therefore, unslaked lime (Calcium oxide,

CaO

) must be used.

A gas can only be dried by a drying agent with which it does not chemically react.

2

Identify the industrial catalyst for the oxidation step in trioxonitrate(V) acid production.

Ammonia is oxidized to nitrogen(II) oxide (

NO

) over a Platinum-rhodium (

Pt/Rh

) gauze catalyst during the Ostwald process.

The catalyst allows oxidation to proceed selectively and rapidly at high temperatures.

3

Recall the standard reactants used for laboratory synthesis of ammonia.

A solid mixture of Calcium hydroxide (

Ca(OH)_2

) and Ammonium chloride (

NH_4Cl

) is heated.

Action of a strong base on an ammonium salt liberates ammonia gas, water, and calcium chloride.

4

Identify the gaseous reduction product of concentrated trioxonitrate(V) acid reacting with copper.

Reddish-brown fumes of Nitrogen dioxide (

NO_2

) are liberated.

Concentrated

HNO_3

is a strong oxidizing agent that is reduced to

NO_2

gas when it oxidizes metals like copper.

Key Concept

Preparation, industrial processes, and characteristic chemical reactions of ammonia and trioxonitrate(V) acid.

Question 2Question →

In the industrial manufacture of tetraoxosulfate(VI) acid by the Contact Process, sulfur(VI) oxide gas ( $SO_3$ ) is absorbed in concentrated $H_2SO_4$ to form oleum ( $H_2S_2O_7$ ) rather than being dissolved directly in water. Which of the following best explains the primary reason for this operational step?

Direct dissolution in water is an extremely exothermic reaction that produces a dense acid mist which is difficult to condense.

Direct dissolution in water decomposes tetraoxosulfate(VI) acid into sulfur(IV) oxide gas and oxygen.

Direct dissolution in water causes the vanadium(V) oxide catalyst to poison and deactivate rapidly.

Sulfur(VI) oxide is insoluble in pure water under atmospheric temperature and pressure.

Show answer & explanation

Answer: Direct dissolution in water is an extremely exothermic reaction that produces a dense acid mist which is difficult to condense.

Answer

Direct dissolution in water is an extremely exothermic reaction that produces a dense acid mist which is difficult to condense.

Dissolving sulfur(VI) oxide directly in water releases an extreme amount of heat, causing the liquid to boil instantly and forming a fog-like mist of liquid tetraoxosulfate(VI) acid droplets suspended in steam. This mist is difficult to condense and collect industrially. Absorbing the gas in 98% concentrated tetraoxosulfate(VI) acid forms oleum smoothly without misting.

Step-by-Step Solution

1

Examine the thermochemistry of the direct hydration reaction

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

.

The direct combination of sulfur(VI) oxide gas with liquid water is highly exothermic.

A massive amount of heat energy is liberated instantly during the hydration process.

2

Analyze the physical impact of the heat generated during direct absorption.

The intense heat vaporizes water rapidly, converting the acid into tiny aerosol droplets (dense mist).

This fine acid mist resists condensation and escapes into the atmosphere with the exhaust gases.

3

Identify the industrial workaround used in the Contact Process.

SO_3

gas is absorbed smoothly into 98% concentrated

H_2SO_4

to yield oleum (

H_2S_2O_7

), which is subsequently diluted with a controlled volume of water to yield concentrated acid.

Dissolving

SO_3

in concentrated acid avoids the violent mist-producing reaction while yielding high-purity tetraoxosulfate(VI) acid.

Key Concept

Absorption of sulfur(VI) oxide in the Contact Process

Estimated Time:1m 0s

Question 3Question →

Producer gas is manufactured industrially by passing air over red-hot coke, yielding a mixture consisting approximately of $1$ mole of $\text{CO}$ to $2$ moles of $\text{N}_2$ . Water gas is produced by passing steam over incandescent coke, yielding an equimolar mixture of $\text{CO}$ and $\text{H}_2$ . If equal volumes of producer gas and water gas are allowed to effuse through identical porous barriers under the same conditions of temperature and pressure, which of the following statements correctly compares their initial rates of effusion? $(\text{Atomic masses: } \text{H} = 1, \text{C} = 12, \text{N} = 14, \text{O} = 16)$

Water gas effuses faster than producer gas because its average molar mass (

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

) is less than that of producer gas (

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

).

Producer gas effuses faster than water gas because the rate of effusion is directly proportional to the average molar mass of the gaseous mixture.

Both gas mixtures effuse at the exact same rate because carbon(II) oxide and nitrogen gas are gaseous allotropes of carbon with identical molecular masses.

Water gas effuses at twice the rate of producer gas because one mole of any gas occupies

22.4\text{ dm}^3

at STP while producer gas occupies

44.8\text{ dm}^3

per mole of carbon.

Show answer & explanation

Answer: Water gas effuses faster than producer gas because its average molar mass ( $15\text{ g mol}^{-1}$ ) is less than that of producer gas ( $28\text{ g mol}^{-1}$ ).

Answer

Water gas effuses faster than producer gas because its average molar mass (

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

) is less than that of producer gas (

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

).

Water gas consists of an equimolar mixture of carbon(II) oxide and hydrogen gas, giving an average molar mass of

\frac{28 + 2}{2} = 15\text{ g mol}^{-1}

. Producer gas consists of carbon(II) oxide and nitrogen gas in a

1:2

molar ratio, giving an average molar mass of

\frac{28 + 2(28)}{3} = 28\text{ g mol}^{-1}

. According to Graham's Law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass (

R \propto \frac{1}{\sqrt{M}}

). Because water gas has a significantly lower average molar mass (

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

) than producer gas (

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

), water gas effuses faster.

Step-by-Step Solution

1

Calculate the average molar mass of producer gas

Molar mass of

\text{CO} = 12 + 16 = 28\text{ g mol}^{-1}

, Molar mass of

\text{N}_2 = 2 \times 14 = 28\text{ g mol}^{-1}

. For a

1:2

mole ratio of

\text{CO} : \text{N}_2

,

\bar{M}_{\text{producer}} = \frac{1(28) + 2(28)}{1 + 2} = 28\text{ g mol}^{-1}

.

To compare effusion rates, the average molar mass of the gas mixture must first be determined.

2

Calculate the average molar mass of water gas

Molar mass of

\text{CO} = 28\text{ g mol}^{-1}

, Molar mass of

\text{H}_2 = 2 \times 1 = 2\text{ g mol}^{-1}

. For an equimolar (

1:1

) mixture of

\text{CO} : \text{H}_2

,

\bar{M}_{\text{water gas}} = \frac{1(28) + 1(2)}{1 + 1} = 15\text{ g mol}^{-1}

.

The average molar mass of water gas is required to apply Graham's Law.

3

Apply Graham's Law of Effusion to compare rates

\frac{R_{\text{water gas}}}{R_{\text{producer gas}}} = \sqrt{\frac{\bar{M}_{\text{producer}}}{\bar{M}_{\text{water gas}}}} = \sqrt{\frac{28}{15}} \approx 1.37. Since

\bar{M}_{\text{water gas}} < \bar{M}_{\text{producer}}

, water gas effuses faster.

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (

R \propto \frac{1}{\sqrt{M}}

).

Key Concept

Composition of industrial fuel gases (producer gas vs water gas) and application of Graham's Law of effusion to gas mixtures

Question 4Question →

Complete the following statement regarding the allotropic transition of sulfur by filling in the blank with the correct term.

Fill in the blanks below

At temperatures above

96^\circ\text{C}

, rhombic sulfur changes reversibly into sulfur.

Show answer & explanation

Answer

Monoclinic (or prismatic) sulfur is the stable crystalline allotrope of sulfur above the transition temperature of

96^\circ\text{C}

.

Rhombic sulfur (

\alpha

-sulfur) is stable at room temperature up to

96^\circ\text{C}

. When heated above this transition temperature (

96^\circ\text{C}

), it slowly transforms into monoclinic sulfur (

\beta

-sulfur), which remains stable up to its melting point.

Step-by-Step Solution

1

Identify the two main crystalline allotropes of sulfur and their transition temperature.

Rhombic sulfur (alpha-sulfur) and monoclinic sulfur (beta-sulfur) exist in dynamic equilibrium at the transition temperature of

96^\circ\text{C}

.

Sulfur exhibits enantiotropic allotropy where the stability of each crystalline form depends on temperature.

2

Determine which allotrope is stable above

96^\circ\text{C}

.

Monoclinic (prismatic) sulfur is stable between

96^\circ\text{C}

and its melting point of

119^\circ\text{C}

.

Below

96^\circ\text{C}

, rhombic sulfur is the stable form; heating it above

96^\circ\text{C}

transforms it into monoclinic sulfur.

Key Concept

Transition temperature of sulfur allotropes

Estimated Time:45s

Question 5Question →

Match each noble gas to its specific industrial application based on its unique physical properties, electronic configuration, or behavior during fractional distillation of liquid air.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Helium matches with oxygen mixtures for deep-sea diving due to low blood solubility; Argon matches with inert shielding in arc welding and electric bulbs; Neon matches with orange-red high-voltage advertising signage glow; Krypton matches with high-speed flash lamps and airport runway lights.

Each noble gas possesses distinct physical and chemical attributes dictated by its electronic structure (

ns^2 np^6

or

1s^2

) and position in liquefaction/fractional distillation order. Helium's low solubility under pressure makes it vital for diving gas blends. Argon provides an economical inert environment for welding and lighting. Neon produces the signature orange-red discharge for neon signs, while Krypton provides high-luminance white emission for photographic flashes and runway lights.

Step-by-Step Solution

1

Analyze the physical properties and biological solubility of Helium.

Helium has a non-polar

1s^2

doublet configuration, extremely weak dispersion forces, and negligible solubility in blood, identifying it as the gas mixed with oxygen for deep-sea diving.

Preventing nitrogen narcosis and decompression sickness requires a non-toxic gas with minimal blood solubility.

2

Analyze the industrial abundance and thermal stability applications of Argon.

Argon (

[Ne]3s^2 3p^6

) is chemically inert and abundant in atmospheric air. It prevents oxidation during metallurgy/welding and retards tungsten filament sublimation.

High-temperature arc welding requires an inert shroud gas to displace atmospheric oxygen and nitrogen.

3

Evaluate the discharge emission spectrum of Neon.

Low-pressure electric discharge through Neon produces electronic transitions yielding a bright orange-red light, characteristic of neon advertising signs.

Excitation of valence electrons in Neon produces distinctive spectral emission in the red-orange wavelength region.

4

Evaluate the optical flash applications of Krypton.

Krypton's multi-line bright white emission under rapid electrical discharge makes it the correct choice for high-speed photographic flash bulbs and airport runway signals.

Heavy noble gases produce brilliant white light discharge suitable for specialized optical equipment.

Key Concept

Physical properties, isolation, electronic stability, and industrial applications of noble gases

Question 6Question →

Which of the following oxides reacts with both hydrochloric acid and sodium hydroxide solution to form a salt and water?

Zinc oxide (

\text{ZnO}

)

Calcium oxide (

\text{CaO}

)

Carbon(IV) oxide (

\text{CO}_2

)

Carbon(II) oxide (

\text{CO}

)

Show answer & explanation

Answer: Zinc oxide ( $\text{ZnO}$ )

Answer

Zinc oxide (

\text{ZnO}

)

Zinc oxide (

\text{ZnO}

) is amphoteric, allowing it to act as a base in the presence of an acid and as an acid in the presence of a base, forming salt and water in both cases.

Step-by-Step Solution

1

Identify the property of reacting with both acids and bases.

Oxides that react with both acids and alkalis to form salt and water are classified as amphoteric oxides.

Amphoteric oxides exhibit both basic and acidic chemical behavior.

2

Select the amphoteric oxide from the options.

Zinc oxide (

\text{ZnO}

) reacts with

\text{HCl}

to form

\text{ZnCl}_2

and

\text{H}_2\text{O}

, and with

\text{NaOH}

to form

\text{Na}_2\text{ZnO}_2

and

\text{H}_2\text{O}

.

Zinc, aluminium, and lead oxides are key examples of amphoteric oxides.

Key Concept

Classification of Oxides: Amphoteric Oxides

Estimated Time:45s

Question 7Question →

An industrial chemist passes steam over incandescent coke at $1000^\circ\text{C}$ to synthesize water gas. If $120\text{ g}$ of pure carbon is completely consumed in this reaction, what is the total volume of the resulting fuel gas mixture collected at s.t.p.? $[\text{Relative atomic mass: } C = 12; \text{Molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]$

224\text{ dm}^3

448\text{ dm}^3

672\text{ dm}^3

112\text{ dm}^3

Show answer & explanation

Answer: $448\text{ dm}^3$

Answer

The total volume of the resulting water gas mixture collected at s.t.p. is

448\text{ dm}^3

.

Water gas is produced by passing steam over incandescent carbon at high temperatures according to the equation

C_{(s)} + H_2O_{(g)} \rightarrow CO_{(g)} + H_{2(g)}

. One mole of carbon yields two moles of gaseous products (

1\text{ mole } CO

and

1\text{ mole } H_2

). Given

120\text{ g}

of carbon (

10\text{ moles}

), the total moles of gas produced is

20\text{ moles}

. Multiplying by the molar volume at s.t.p. (

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

) gives

448\text{ dm}^3

.

Step-by-Step Solution

1

Write the balanced thermochemical equation for water gas formation

C_{(s)} + H_2O_{(g)} \rightarrow CO_{(g)} + H_{2(g)}

Passing steam over red-hot coke produces water gas, which is an equimolar mixture of carbon(II) oxide and hydrogen gas.

2

Calculate the moles of carbon reacted

\text{Moles of } C = \frac{120\text{ g}}{12\text{ g mol}^{-1}} = 10\text{ mol}

Moles equal mass divided by molar mass.

3

Determine the total moles of gaseous products generated

From stoichiometry,

1\text{ mol } C \rightarrow 1\text{ mol } CO + 1\text{ mol } H_2 = 2\text{ mol of gas mixture}

. Therefore,

10\text{ mol } C \rightarrow 20\text{ mol of gas mixture}

.

Water gas comprises both

CO

and

H_2

gases.

4

Calculate total volume at s.t.p.

\text{Volume} = 20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 448\text{ dm}^3

At s.t.p.,

1\text{ mole}

of any gas or gas mixture occupies

22.4\text{ dm}^3

.

Key Concept

Manufacture and Stoichiometry of Industrial Fuel Gases (Water Gas)

Estimated Time:2m 0s

Question 8Question →

A solid mixture containing $16.8\text{ g}$ of $\text{NaHCO}_3$ and $10.6\text{ g}$ of $\text{Na}_2\text{CO}_3$ is heated strongly in an open crucible until no further change in mass occurs. Given the molar volume of any gas at s.t.p. is $22.4\text{ dm}^3\text{mol}^{-1}$ and relative atomic masses $(\text{Na}=23, \text{H}=1, \text{C}=12, \text{O}=16)$ , what is the volume of $\text{CO}_2$ gas evolved at s.t.p. and the total mass of the solid residue remaining?

2.24\text{ dm}^3

of

\text{CO}_2

and

21.2\text{ g}

of solid residue

4.48\text{ dm}^3

of

\text{CO}_2

and

10.6\text{ g}

of solid residue

2.40\text{ dm}^3

of

\text{CO}_2

and

21.2\text{ g}

of solid residue

4.48\text{ dm}^3

of

\text{CO}_2

and

12.4\text{ g}

of solid residue

Show answer & explanation

Answer: $2.24\text{ dm}^3$ of $\text{CO}_2$ and $21.2\text{ g}$ of solid residue

Answer

The volume of carbon(IV) oxide evolved at s.t.p. is

2.24\text{ dm}^3

and the total mass of the solid residue remaining is

21.2\text{ g}

.

Heating

16.8\text{ g}

(

0.20\text{ mol}

) of

\text{NaHCO}_3

yields

0.10\text{ mol}

of

\text{CO}_2

gas, which occupies

2.24\text{ dm}^3

at s.t.p. (

0.10 \times 22.4\text{ dm}^3

). The reaction produces

0.10\text{ mol}

(

10.6\text{ g}

) of solid

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

. Adding this to the unreacted

10.6\text{ g}

of original

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

yields a total solid residue mass of

21.2\text{ g}

.

Step-by-Step Solution

1

Calculate the molar masses of the relevant substances

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

;

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

.

Molar masses are needed to convert mass to chemical amounts (moles).

2

Determine thermal stability of components and identify the decomposition reaction

Sodium trioxocarbonate(IV) (

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

) is thermally stable and does not decompose. Sodium hydrogentrioxocarbonate(IV) decomposes:

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

.

Alkali metal trioxocarbonates(IV) (except lithium) do not decompose on heating, whereas hydrogentrioxocarbonates(IV) decompose to form trioxocarbonate(IV), water vapor, and carbon(IV) oxide.

3

Calculate the moles of

\text{NaHCO}_3

and the volume of

\text{CO}_2

evolved

\text{Moles of NaHCO}_3 = \frac{16.8\text{ g}}{84\text{ g/mol}} = 0.20\text{ mol}

. From stoichiometry,

2\text{ mol NaHCO}_3 \rightarrow 1\text{ mol CO}_2

. Thus,

\text{moles of CO}_2 = \frac{0.20}{2} = 0.10\text{ mol}

. Volume of

\text{CO}_2 \text{ at s.t.p.} = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 2.24\text{ dm}^3

.

Stoichiometric mole ratio determines the yield of gaseous product at standard temperature and pressure.

4

Calculate the total mass of the solid residue remaining

From decomposition:

\text{moles of new Na}_2\text{CO}_3 = 0.10\text{ mol}

. Mass of new

\text{Na}_2\text{CO}_3 = 0.10\text{ mol} \times 106\text{ g/mol} = 10.6\text{ g}

. Total solid residue = original

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

+ produced

\text{Na}_2\text{CO}_3 = 10.6\text{ g} + 10.6\text{ g} = 21.2\text{ g}

.

The solid residue consists of both the thermally stable initial component and the newly formed trioxocarbonate(IV) salt.

Key Concept

Thermal decomposition of group 1 hydrogentrioxocarbonate(IV) salts vs trioxocarbonate(IV) salts and gas stoichiometry.

Question 9Question →

A $10.0\text{ g}$ sample of impure calcium trioxocarbonate(IV), $\text{CaCO}_3$ , was strongly heated until decomposition was complete. If the volume of carbon(IV) oxide gas evolved at STP was $1.792\text{ dm}^3$ , what is the percentage purity of the $\text{CaCO}_3$ sample? [Molar volume of gas at STP = $22.4\text{ dm}^3\text{mol}^{-1}$ ; relative atomic masses: $\text{Ca}=40, \text{C}=12, \text{O}=16$ ]

Show answer & explanation

Answer: 80

Answer

The percentage purity of the calcium trioxocarbonate(IV) sample is

80\%

.

Thermal decomposition of pure calcium trioxocarbonate(IV) releases carbon(IV) oxide gas according to

\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

. Dividing the gas volume (

1.792\text{ dm}^3

) by the molar gas volume at STP (

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

) yields

0.08\text{ mol}

of

\text{CO}_2

. Due to the 1:1 stoichiometry,

0.08\text{ mol}

of pure

\text{CaCO}_3

reacted. Multiplying by the molar mass of

\text{CaCO}_3

(

100\text{ g/mol}

) gives

8.0\text{ g}

of pure

\text{CaCO}_3

. The percentage purity is calculated as

\left(\frac{8.0\text{ g}}{10.0\text{ g}}\right) \times 100\% = 80\%

.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of calcium trioxocarbonate(IV).

\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

Establishes the 1:1 stoichiometric mole ratio between

\text{CaCO}_3

and

\text{CO}_2

.

2

Calculate the number of moles of

\text{CO}_2

gas produced at STP.

\text{Moles of CO}_2 = \frac{1.792\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.08\text{ mol}

Uses the molar gas volume relationship at standard temperature and pressure (

V / V_m

).

3

Calculate the mass of pure

\text{CaCO}_3

in the original sample.

\text{Mass of CaCO}_3 = 0.08\text{ mol} \times 100\text{ g/mol} = 8.0\text{ g}

Because

1\text{ mol}

of

\text{CaCO}_3

produces

1\text{ mol}

of

\text{CO}_2

,

0.08\text{ mol}

of pure

\text{CaCO}_3

reacted.

4

Calculate the percentage purity of the sample.

\text{Percentage purity} = \left(\frac{8.0\text{ g}}{10.0\text{ g}}\right) \times 100\% = 80\%

Compares the mass of active pure reactant to the total mass of the impure sample.

Key Concept

Stoichiometry of thermal decomposition of trioxocarbonate(IV) salts and gas molar volume calculations at STP.

Question 10Question →

A

10.0\text{ g}

sample of impure calcium carbonate (

\text{CaCO}_3

) is completely decomposed by strong heating according to the chemical equation:

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

The carbon(IV) oxide gas evolved is passed into an excess solution of sodium hydroxide, causing the mass of the solution to increase by

3.52\text{ g}

. Assuming the impurities present in the sample do not react or produce any gas, what is the percentage purity of the calcium carbonate sample?

[\text{Ca} = 40, \text{C} = 12, \text{O} = 16]

Show answer & explanation

Answer: 80

Answer

The percentage purity of the calcium carbonate sample is 80%.

Sodium hydroxide absorbs carbon(IV) oxide (

\text{CO}_2

) gas released during the thermal decomposition of calcium carbonate (

\text{CaCO}_3

). The

3.52\text{ g}

mass gain of the solution equals the mass of

\text{CO}_2

evolved. Dividing this mass by the molar mass of

\text{CO}_2

(

44\text{ g/mol}

) yields

0.08\text{ mol}

of

\text{CO}_2

. According to the 1:1 stoichiometric relationship,

0.08\text{ mol}

of pure

\text{CaCO}_3

decomposed. Multiplying by the molar mass of

\text{CaCO}_3

(

100\text{ g/mol}

) gives

8.00\text{ g}

of pure

\text{CaCO}_3

. The percentage purity is

(8.00\text{ g} / 10.0\text{ g}) \times 100\% = 80\%

.

Step-by-Step Solution

1

Calculate the molar masses of carbon(IV) oxide and calcium carbonate

Molar mass of CO2 = 44 g/mol, Molar mass of CaCO3 = 100 g/mol

Molar masses are required to convert between mass and moles.

2

Determine the moles of carbon(IV) oxide gas evolved

Moles of CO2 = 3.52 g / 44 g/mol = 0.08 mol

Sodium hydroxide reacts with and absorbs acidic carbon(IV) oxide, so mass increase equals the mass of CO2.

3

Determine the mass of pure calcium carbonate in the sample

Mass of pure CaCO3 = 0.08 mol * 100 g/mol = 8.00 g

The mole ratio of CaCO3 to CO2 in the thermal decomposition reaction is 1:1.

4

Calculate the percentage purity of the sample

(8.00 g / 10.0 g) * 100 = 80%

Percentage purity is the ratio of pure reactive substance mass to total sample mass expressed as a percentage.

Key Concept

Thermal decomposition of trioxocarbonates and percentage purity stoichiometry

Question 11Question →

A mixture of $50\text{ cm}^3$ of carbon(II) oxide and $30\text{ cm}^3$ of oxygen was ignited in an eudiometer tube to complete reaction. After cooling to room temperature and pressure, the resulting gas mixture was passed through concentrated potassium hydroxide solution. What is the volume of the residual gas remaining?

5\text{ cm}^3

20\text{ cm}^3

50\text{ cm}^3

55\text{ cm}^3

Show answer & explanation

Answer: $5\text{ cm}^3$

Answer

The volume of residual gas remaining is

5\text{ cm}^3

.

According to Gay-Lussac's Law of Combining Volumes,

2\text{ cm}^3

of

\text{CO}

combines with

1\text{ cm}^3

of

\text{O}_2

to produce

2\text{ cm}^3

of

\text{CO}_2

. For

50\text{ cm}^3

of

\text{CO}

, exactly

25\text{ cm}^3

of

\text{O}_2

is required, leaving

5\text{ cm}^3

of

\text{O}_2

unreacted. The reaction generates

50\text{ cm}^3

of

\text{CO}_2

. Passing the mixture through concentrated potassium hydroxide (

\text{KOH}

) removes all

50\text{ cm}^3

of

\text{CO}_2

via trioxocarbonate(IV) salt formation, leaving behind only the

5\text{ cm}^3

of unreacted oxygen gas.

Step-by-Step Solution

1

Write the balanced chemical equation for the combustion of carbon(II) oxide.

2\text{CO}_{(g)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)}

Establishing the combining volume ratio according to Gay-Lussac's Law.

2

Determine the reacting volumes and the limiting reactant.

2 volumes of

\text{CO}

react with 1 volume of

\text{O}_2

. Therefore,

50\text{ cm}^3

of

\text{CO}

reacts with

\frac{50}{2} = 25\text{ cm}^3

of

\text{O}_2

. Oxygen is in excess.

To find how much oxygen is consumed and how much remains unreacted.

3

Calculate the volume of products formed and unreacted gas remaining after combustion.

Volume of

\text{CO}_2

produced =

50\text{ cm}^3

. Volume of excess

\text{O}_2

remaining =

30\text{ cm}^3 - 25\text{ cm}^3 = 5\text{ cm}^3

.

Stoichiometric yield of

\text{CO}_2

equals the initial volume of

\text{CO}

burned.

4

Account for the absorption of carbon oxides by concentrated KOH.

\text{CO}_2

is an acidic oxide and reacts with potassium hydroxide:

\text{CO}_{2(g)} + 2\text{KOH}_{(aq)} \rightarrow \text{K}_2\text{CO}_{3(aq)} + \text{H}_2\text{O}_{(l)}

. All

50\text{ cm}^3

of

\text{CO}_2

is absorbed. Residual gas =

5\text{ cm}^3

of

\text{O}_2

.

Potassium hydroxide selectively absorbs carbon(IV) oxide gas.

Key Concept

Gay-Lussac's Law of Combining Volumes and chemical absorption properties of carbon oxides

Question 12Question →

When carbon(IV) oxide gas is passed over red-hot coke in a industrial furnace, it undergoes reduction to form a poisonous gas that serves as a vital reducing agent in blast furnace operations. What is the IUPAC name of the gas produced?

Show answer & explanation

Answer: Carbon(II) oxide; carbon(II) oxide; Carbon (II) oxide; carbon (II) oxide; Carbon monoxide; carbon monoxide

Answer

Carbon(II) oxide

Passing carbon(IV) oxide gas over red-hot carbon (coke) reduces

\text{CO}_2

to

\text{CO}

. The IUPAC name for

\text{CO}

is carbon(II) oxide, which is a major reducing agent in industrial metal extraction.

Step-by-Step Solution

1

Identify the chemical reaction between carbon(IV) oxide and carbon (coke).

The balanced chemical equation is

\text{CO}_2(g) + \text{C}(s) \rightarrow 2\text{CO}(g)

.

Red-hot coke acts as a reducing agent, reducing carbon(IV) oxide to carbon(II) oxide.

2

Determine the IUPAC name of the gaseous product

\text{CO}

.

The IUPAC systematic name for

\text{CO}

is carbon(II) oxide.

The Roman numeral (II) represents the +2 oxidation state of carbon in the oxide.

Key Concept

Reduction of Carbon(IV) Oxide to Carbon(II) Oxide by hot carbon

Question 13Question →

What volume of carbon(IV) oxide gas, in $\text{dm}^3$ , measured at STP, is produced by the complete thermal decomposition of $20.0\text{ g}$ of pure calcium trioxocarbonate(IV), $\text{CaCO}_3$ ? [ $\text{Ca} = 40$ , $\text{C} = 12$ , $\text{O} = 16$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{mol}^{-1}$ ]

Show answer & explanation

Answer: 4.48

Answer

4.48 dm³

Complete thermal decomposition of 20.0 g of CaCO₃ (molar mass 100 g/mol) generates 0.20 mol of CO₂ gas according to the equation CaCO₃(s) -> CaO(s) + CO₂(g). Since 1 mol of gas at STP occupies 22.4 dm³, 0.20 mol occupies 4.48 dm³.

Step-by-Step Solution

1

Determine the molar mass and number of moles of calcium trioxocarbonate(IV).

Molar mass of CaCO₃ = 100 g/mol; Moles of CaCO₃ = 20.0 g / 100 g/mol = 0.20 mol

Converting given mass to moles is required to apply stoichiometric ratios.

2

Apply the balanced reaction mole ratio to find moles of carbon(IV) oxide produced.

Moles of CO₂ = 0.20 mol

The equation CaCO₃(s) -> CaO(s) + CO₂(g) shows a 1:1 molar ratio between CaCO₃ and CO₂.

3

Multiply moles of CO₂ by molar gas volume at STP.

Volume of CO₂ = 0.20 mol × 22.4 dm³/mol = 4.48 dm³

At STP, 1 mole of any ideal gas occupies 22.4 dm³.

Key Concept

Thermal decomposition of trioxocarbonate(IV) salts and gas volume calculations at STP

Estimated Time:45s

Question 14Question →

Match each of the following oxides with its correct chemical classification and characteristic chemical behavior.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Dinitrogen monoxide matches as a neutral oxide; Barium peroxide matches as a true peroxide producing hydrogen peroxide with cold dilute acid; Lead(IV) oxide matches as a dioxide acting as an oxidizing agent to liberate chlorine gas from concentrated hydrochloric acid; Aluminium oxide matches as an amphoteric oxide reacting with both acids and bases.

The correct pairings accurately distinguish between oxide classes: dinitrogen monoxide is neutral, barium peroxide contains the peroxide anion yielding hydrogen peroxide with dilute acids, lead(IV) oxide is a dioxide acting as an oxidizing agent with concentrated hydrochloric acid, and aluminium oxide displays amphoteric properties by reacting with both acids and bases.

Step-by-Step Solution

1

Analyze Dinitrogen monoxide (

\text{N}_2\text{O}

)

Identify that oxides of non-metals like

\text{N}_2\text{O}

and

\text{CO}

are neutral and do not form salts with acids or bases.

Classification based on acid-base reactivity.

2

Differentiate between peroxides and dioxides using Barium peroxide (

\text{BaO}_2

) and Lead(IV) oxide (

\text{PbO}_2

)

True peroxides like

\text{BaO}_2

contain the

\text{O}_2^{2-}

ion and produce

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

with dilute acid. Dioxides like

\text{PbO}_2

contain metal in +4 oxidation state and act as oxidizing agents, yielding

\text{Cl}_2

gas with concentrated

\text{HCl}

.

Oxidation state analysis and chemical reaction product test.

3

Analyze Aluminium oxide (

\text{Al}_2\text{O}_3

)

Determine that metallic oxides of group 13 like

\text{Al}_2\text{O}_3

dissolve in both acids (forming

\text{Al}^{3+}

salts) and strong bases (forming aluminate complex salts), confirming amphoterism.

Amphoteric nature of specific metal oxides.

Key Concept

Classification of oxides (acidic, basic, amphoteric, neutral, peroxides, and dioxides)

Question 15Question →

In an industrial Contact Process plant, sulfur(IV) oxide (

SO_2

) gas is catalytically oxidized to sulfur(VI) oxide (

SO_3

) according to the equation:

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)

If

67.2\text{ dm}^3

of

SO_2

measured at STP is reacted with excess oxygen gas, and the reaction achieves a

90\%

conversion yield of

SO_3

, what is the mass in grams of tetraoxosulfate(VI) acid (

H_2SO_4

) produced when all the formed

SO_3

is absorbed in concentrated

H_2SO_4

and subsequently diluted with water? (Molar mass of

H_2SO_4 = 98\text{ g/mol}

, molar volume of gas at STP

= 22.4\text{ dm}^3\text{/mol}

)

Show answer & explanation

Answer: 264.6

Answer

264.6 g

The molar volume at STP (

22.4\text{ dm}^3\text{/mol}

) converts

67.2\text{ dm}^3

of

SO_2

into

3.0\text{ moles}

. Accounting for the

90\%

conversion efficiency yields

2.7\text{ moles}

of

SO_3

. Absorption of

SO_3

into concentrated

H_2SO_4

to form oleum (

H_2S_2O_7

) followed by dilution with water yields a

1:1

molar ratio of

H_2SO_4

relative to

SO_3

. Multiplying

2.7\text{ moles}

by the molar mass of

H_2SO_4

(

98\text{ g/mol}

) gives the correct mass of

264.6\text{ g}

.

Step-by-Step Solution

1

Calculate the moles of

SO_2

gas at STP.

n(SO_2) = \frac{67.2\text{ dm}^3}{22.4\text{ dm}^3\text{/mol}} = 3.0\text{ moles}

Molar volume of any ideal gas at STP is

22.4\text{ dm}^3\text{/mol}

.

2

Apply the

90\%

conversion efficiency to find the moles of

SO_3

produced.

n(SO_3) = 3.0\text{ mol} \times 0.90 = 2.7\text{ moles}

Only

90\%

of the reacted

SO_2

is converted to

SO_3

under operating conditions.

3

Relate the moles of

SO_3

to the moles of

H_2SO_4

produced.

n(H_2SO_4) = n(SO_3) = 2.7\text{ moles}

The absorption of

SO_3

into concentrated

H_2SO_4

forms oleum (

H_2S_2O_7

), which upon dilution with water yields

H_2SO_4

with an overall

1:1

stoichiometric molar equivalence to

SO_3

.

4

Calculate the total mass of

H_2SO_4

formed.

\text{Mass} = 2.7\text{ mol} \times 98\text{ g/mol} = 264.6\text{ g}

Mass equals number of moles multiplied by molar mass.

Key Concept

Stoichiometry of the Contact Process including STP gas conversion and oleum dilution stoichiometry

Question 16Question →

Arrange the following products obtained during the destructive distillation of coal in order of decreasing volatility (from the most volatile product to the solid residue left behind):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of products from most volatile to least volatile (solid residue) is Coal gas, Ammoniacal liquor, Coal tar, and Coke.

Destructive distillation of coal yields volatile gaseous products (coal gas), liquid condensate fractions (ammoniacal liquor and coal tar), and a non-volatile solid residue (coke). Arranging by decreasing volatility places the gaseous coal gas first, followed by ammoniacal liquor, coal tar, and finally coke.

Step-by-Step Solution

1

Identify the physical states and volatility of the products of destructive distillation of coal.

Coal gas is gaseous; ammoniacal liquor and coal tar are liquids of differing density/volatility; coke is a solid residue.

Destructive distillation involves heating coal in the absence of air to separate volatile compounds from non-volatile solids.

2

Rank the fractions based on volatility.

Gases evolve first without condensing (Coal gas), followed by light aqueous distillates (Ammoniacal liquor), heavy liquid fractions (Coal tar), and finally the solid non-volatile residue (Coke).

Volatility determines the sequence in which products escape and condense during industrial coal refining.

Key Concept

Destructive Distillation of Coal and By-product Volatility

Question 17Question →

Which of the following noble gases is extracted from air by fractional distillation and extensively used to provide an inert atmosphere during electric arc welding?

Argon

Helium

Neon

Krypton

Show answer & explanation

Answer: Argon

Answer

Argon is the noble gas obtained from liquid air by fractional distillation and used to create an inert atmosphere in electric arc welding.

Argon makes up roughly 0.93% of atmospheric air by volume and is isolated as a major byproduct during the fractional distillation of liquid air. Due to its completely filled valence electron shell, it is extremely unreactive, making it ideal for creating an inert atmosphere that protects molten metal from reacting with oxygen or nitrogen during electric arc welding.

Step-by-Step Solution

1

Identify the primary atmospheric noble gas isolated by fractional distillation.

Argon makes up about 0.93% of atmospheric air by volume and is commercially extracted during the fractional distillation of liquid air.

Air separation units liquefy air and separate its constituents based on boiling point differences.

2

Relate the chemical properties of Argon to its industrial application.

Because Argon is chemically inert, it acts as a protective shield during high-temperature arc welding to prevent hot metals from reacting with atmospheric oxygen and nitrogen.

A noble gas atmosphere prevents oxidation and corrosion of the weld joint.

Key Concept

Isolation and uses of noble gases

Estimated Time:45s

Question 18Question →

Match each oxygen species or oxide listed on the left with its characteristic chemical property or reaction behavior on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Dichlorine heptoxide matches with the acidic oxide yielding perchloric acid; Dinitrogen monoxide matches with the neutral oxide that decomposes to re-ignite a glowing splint; Sodium peroxide matches with the peroxide liberating hydrogen peroxide with cold dilute acid; Ozone matches with the triatomic allotrope turning moist KI-starch paper blue.

Each pair correctly aligns the oxide/allotrope with its fundamental structural class and laboratory reaction. Dichlorine heptoxide is an acid anhydride for perchloric acid; dinitrogen monoxide is a neutral oxide that thermally decomposes to support combustion; sodium peroxide yields hydrogen peroxide with cold dilute acid; and ozone is an allotrope of oxygen that oxidizes iodide to iodine, turning potassium iodide-starch paper blue.

Step-by-Step Solution

1

Analyze Dichlorine heptoxide (

\text{Cl}_2\text{O}_7

)

Highest oxide of chlorine with oxidation state +7. Dissolves in water according to

\text{Cl}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{HClO}_4

, making it the acid anhydride of perchloric acid.

Classification of non-metal higher oxides as acid anhydrides.

2

Analyze Dinitrogen monoxide (

\text{N}_2\text{O}

)

Neutral oxide that does not react with acids or bases. Upon thermal decomposition,

2\text{N}_2\text{O} \rightarrow 2\text{N}_2 + \text{O}_2

, supplying enough oxygen gas to support combustion and rekindle a glowing splint.

Properties of neutral oxides of nitrogen.

3

Analyze Sodium peroxide (

\text{Na}_2\text{O}_2

)

Contains the peroxide linkage

\text{O}_2^{2-}

. Reacts with cold dilute acids according to

\text{Na}_2\text{O}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}_2

.

Distinct chemical behavior of metallic peroxides versus normal oxides or dioxides.

4

Analyze Ozone (

\text{O}_3

)

Strong oxidizing allotrope of oxygen. Oxidizes

\text{I}^-

to

\text{I}_2

according to

\text{O}_3 + 2\text{KI} + \text{H}_2\text{O} \rightarrow \text{O}_2 + 2\text{KOH} + \text{I}_2

, turning starch paper blue.

Standard qualitative laboratory test for ozone gas.

Key Concept

Classification of oxides (acidic, neutral, peroxide) and chemical characterization of oxygen allotropes.

Question 19Question →

During industrial fuel gas manufacture, passing steam over incandescent coke yields water gas, whereas passing air over red-hot coke produces producer gas. Which of the following statements correctly explains why water gas possesses a significantly higher calorific value per unit volume than producer gas?

Water gas consists entirely of combustible gases (

\text{CO}

and

\text{H}_2

), whereas producer gas contains a high percentage of non-combustible nitrogen (

\text{N}_2

) from air.

The synthesis of water gas is an exothermic reaction that traps excess chemical enthalpy directly inside the resulting gas molecules.

Producer gas primarily consists of carbon(IV) oxide (

\text{CO}_2

) and steam (

\text{H}_2\text{O}

), which impede active combustion.

Water gas contains high concentrations of hydrocarbon allotropes of carbon that release double the combustion energy of carbon(II) oxide.

Show answer & explanation

Answer: Water gas consists entirely of combustible gases ( $\text{CO}$ and $\text{H}_2$ ), whereas producer gas contains a high percentage of non-combustible nitrogen ( $\text{N}_2$ ) from air.

Answer

Water gas possesses a higher calorific value because both of its major constituents, carbon(II) oxide (

\text{CO}

) and hydrogen (

\text{H}_2

), are flammable fuel gases. In contrast, producer gas contains approximately

60\%

atmospheric nitrogen (

\text{N}_2

), an inert, non-combustible gas that dilutes the energy density.

Water gas is synthesized by passing steam over incandescent coke at around

1000^\circ\text{C}

, producing an equimolar mixture of carbon(II) oxide and hydrogen gas (

\text{CO} + \text{H}_2

). Since both constituent gases burn in air releasing substantial heat,

100\%

of the gas volume is combustible. Producer gas, made by passing air over red-hot coke (

2\text{C} + \text{O}_2 + 3.76\text{N}_2 \rightarrow 2\text{CO} + 3.76\text{N}_2

), contains nearly

60\%

atmospheric nitrogen by volume. Nitrogen is non-combustible and absorbs heat during combustion, resulting in a substantially lower calorific value for producer gas.

Step-by-Step Solution

1

Analyze the chemical composition of water gas

Water gas is formed via

\text{C}(s) + \text{H}_2\text{O}(g) \rightarrow \text{CO}(g) + \text{H}_2(g)

. Both

\text{CO}

and

\text{H}_2

are active fuels.

Determining the active combustible components in water gas.

2

Analyze the chemical composition of producer gas

Producer gas is formed via

2\text{C}(s) + \text{O}_2(g) + 3.76\text{N}_2(g) \rightarrow 2\text{CO}(g) + 3.76\text{N}_2(g)

. Atmospheric

\text{N}_2

remains unreacted.

Identifying the non-combustible diluent present in producer gas.

3

Compare the heating values based on composition

Because nearly

100\%

of water gas contributes to combustion energy while over half of producer gas consists of inert

\text{N}_2

, water gas has a higher calorific value per unit volume.

Connecting molecular composition to volumetric calorific energy yield.

Key Concept

Industrial Fuel Gas Compositions and Calorific Values

Estimated Time:1m 30s

Question 20Question →

Fill in the missing chemical terms and oxidation numbers in the passage describing the reaction of hydrogen sulfide gas with an acidified oxidizing solution.

Fill in the blanks below

When hydrogen sulfide gas (

{\text{H}_2\text{S}}

) is bubbled into an acidified solution of potassium tetraoxomanganate(VII), the purple solution is decolorized because

{\text{H}_2\text{S}}

acts as a agent. During this redox process, sulfur is oxidized from an oxidation state of in

{\text{H}_2\text{S}}

to in the yellow elemental precipitate formed.

Show answer & explanation

Answer

The first blank is 'reducing', the second blank is '-2', and the third blank is '0'.

Hydrogen sulfide (

\text{H}_2\text{S}

) acts as a strong reducing agent in aqueous chemical reactions. When reacted with acidified

\text{KMnO}_4

, it reduces purple

\text{MnO}_4^-

ions to colorless

\text{Mn}^{2+}

ions, while sulfur itself is oxidized from an oxidation state of

-2

to

0

, forming a insoluble yellow precipitate of elemental sulfur.

Step-by-Step Solution

1

Determine the chemical role of

\text{H}_2\text{S}

when reacting with acidified

\text{KMnO}_4

.

Acidified

\text{KMnO}_4

is an oxidizing agent containing

\text{Mn}^{7+}

(purple), which is reduced to

\text{Mn}^{2+}

(colorless). Therefore,

\text{H}_2\text{S}

acts as a reducing agent.

A substance that causes reduction in another species while undergoing oxidation itself is a reducing agent.

2

Calculate the oxidation state of sulfur in hydrogen sulfide (

\text{H}_2\text{S}

).

With hydrogen in the

+1

state (

2 \times (+1) = +2

), the oxidation number of sulfur must be

-2

to make the neutral molecule sum to zero.

Sum of oxidation states in neutral compounds is zero.

3

Identify the oxidation state of sulfur in the elemental precipitate product.

Elemental sulfur (

S

) has an oxidation state of

0

.

The oxidation number of any uncombined element in its standard state is always zero.

Key Concept

Reducing properties of hydrogen sulfide gas and oxidation state changes in redox reactions

Non-Metals and Their Compounds

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