Non-Metals and Their Compounds Practice Questions — JAMB UTME — Page 3

Question 41Question →

During atmospheric lightning strikes, nitrogen gas reacts with oxygen gas to form nitrogen(II) oxide, which further oxidizes to nitrogen(IV) oxide gas (

NO_2

). What volume of

NO_2

gas, measured at

27^\circ\text{C}

and

1\text{ atm}

, is produced when

14.0\text{ g}

of nitrogen gas reacts completely according to the following equation?

N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)

[Molar gas volume at STP =

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

, atomic mass of

N = 14.0\text{ g mol}^{-1}

,

0^\circ\text{C} = 273\text{ K}

]

24.6\text{ dm}^3

22.4\text{ dm}^3

2.22\text{ dm}^3

15.0\text{ dm}^3

Show answer & explanation

Answer: $24.6\text{ dm}^3$

Answer

The correct volume of nitrogen(IV) oxide gas produced at

27^\circ\text{C}

and

1\text{ atm}

is

24.6\text{ dm}^3

.

The complete reaction converts

14.0\text{ g}

of

N_2

(

0.50\text{ mol}

) into

1.00\text{ mol}

of

NO_2

. At STP (

273\text{ K}

),

1.00\text{ mol}

occupies

22.4\text{ dm}^3

. Expanding to

27^\circ\text{C}

(

300\text{ K}

) using Charles's Law yields

22.4 \times (300 / 273) = 24.6\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the moles of nitrogen gas (

N_2

) reacted.

\text{Moles of } N_2 = \frac{14.0\text{ g}}{28.0\text{ g mol}^{-1}} = 0.50\text{ mol}

.

Molar mass of

N_2 = 2 \times 14.0 = 28.0\text{ g mol}^{-1}

.

2

Determine moles of

NO_2

gas produced using the mole ratio from the balanced equation.

\text{Moles of } NO_2 = 0.50\text{ mol } N_2 \times \frac{2\text{ mol } NO_2}{1\text{ mol } N_2} = 1.00\text{ mol } NO_2

.

The balanced chemical equation shows a

1:2

mole ratio between

N_2

and

NO_2

.

3

Calculate the volume of

1.00\text{ mol}

of

NO_2

at standard temperature and pressure (STP).

V_{\text{STP}} = 1.00\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 22.4\text{ dm}^3

.

One mole of any ideal gas occupies

22.4\text{ dm}^3

at STP (

273\text{ K}

and

1\text{ atm}

).

4

Convert the volume from STP (

273\text{ K}

) to the target temperature (

27^\circ\text{C} = 300\text{ K}

) at constant pressure.

V_2 = V_1 \times \frac{T_2}{T_1} = 22.4\text{ dm}^3 \times \frac{300\text{ K}}{273\text{ K}} = 24.615\text{ dm}^3 \approx 24.6\text{ dm}^3

.

According to Charles's Law, volume is directly proportional to absolute temperature when pressure is held constant.

Key Concept

Gas Stoichiometry and Temperature-Volume Relationship (Charles's Law)

Question 42Question →

A solid binary compound of oxygen and lead, $\text{PbO}_2$ , reacts with concentrated hydrochloric acid to yield lead(II) chloride, water, and chlorine gas, but fails to produce hydrogen peroxide when treated with cold dilute tetraoxosulfate(VI) acid. Based on this chemical behavior, which of the following statements correctly classifies $\text{PbO}_2$ and distinguishes it from a peroxide such as sodium peroxide ( $\text{Na}_2\text{O}_2$ )?

\text{PbO}_2

is a dioxide containing

\text{O}^{2-}

ions with lead in the

+4

oxidation state, whereas

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

is a peroxide containing

\text{O}_2^{2-}

ions.

\text{PbO}_2

is an amphoteric oxide that acts purely as a neutral compound, whereas

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

is an acidic anhydride.

\text{PbO}_2

is classified as a peroxide because its formula contains two oxygen atoms, whereas

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

is a basic monoxide.

\text{PbO}_2

contains the

\text{O}_2^{2-}

ion which oxidizes chloride ions to chlorine gas, whereas

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

contains

\text{O}^{2-}

ions.

Show answer & explanation

Answer: $\text{PbO}_2$ is a dioxide containing $\text{O}^{2-}$ ions with lead in the $+4$ oxidation state, whereas $\text{Na}_2\text{O}_2$ is a peroxide containing $\text{O}_2^{2-}$ ions.

Answer

Lead(IV) oxide (

\text{PbO}_2

) is classified as a dioxide containing

\text{O}^{2-}

ions with lead in the

+4

oxidation state, whereas sodium peroxide (

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

) is a peroxide containing the

\text{O}_2^{2-}

ion.

Lead(IV) oxide (

\text{PbO}_2

) is classified as a dioxide because it contains simple oxide ions (

\text{O}^{2-}

) with lead in the

+4

oxidation state. It acts as an oxidizing agent by reacting with concentrated

\text{HCl}

to liberate chlorine gas, but does not produce hydrogen peroxide when treated with cold dilute acids. In contrast, sodium peroxide (

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

) contains the peroxide ion (

\text{O}_2^{2-}

) and produces

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

upon reaction with cold dilute acids.

Step-by-Step Solution

1

Analyze the structural difference between peroxides and dioxides.

Peroxides contain the peroxide anion

(\text{O}_2^{2-})

where oxygen has an oxidation state of

-1

. Dioxides contain standard oxide anions

(\text{O}^{2-})

where oxygen has an oxidation state of

-2

and the metal is in the

+4

oxidation state.

Chemical classification depends on the specific ionic species present in the oxide crystal lattice.

2

Evaluate the reaction of

\text{PbO}_2

with dilute acid.

\text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow \text{Pb(SO}_4)_2 + 2\text{H}_2\text{O}

(no

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

formed).

True peroxides yield hydrogen peroxide

(\text{H}_2\text{O}_2)

upon treatment with cold dilute acids, whereas dioxides do not.

3

Evaluate the reaction of

\text{PbO}_2

with concentrated hydrochloric acid.

\text{PbO}_2 + 4\text{HCl} \rightarrow \text{PbCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2\uparrow

.

The

\text{Pb}^{4+}

ion acts as an oxidizing agent, oxidizing

\text{Cl}^-

to

\text{Cl}_2

gas, confirming its behavior as a higher dioxide.

Key Concept

Distinction between Peroxides and Dioxides

Estimated Time:1m 30s

Question 43Question →

Complete the following statement regarding the qualitative test for hydrogen sulfide gas by filling in the missing physical characteristic of the precipitate formed.

Fill in the blanks below

When hydrogen sulfide gas (

H_2S

) is bubbled into an aqueous solution of lead(II) ethanoate, a insoluble precipitate of lead(II) sulfide (

PbS

) is observed.

Show answer & explanation

Answer

black

When hydrogen sulfide gas comes into contact with lead(II) ethanoate, lead(II) sulfide (

PbS

) is formed. Lead(II) sulfide is insoluble in water and has a characteristic black color, making 'black' the correct term to complete the sentence.

Step-by-Step Solution

1

Identify the chemical reaction between hydrogen sulfide gas (

H_2S

) and aqueous lead(II) ethanoate solution (

(CH_3COO)_2Pb

).

The double displacement reaction produces lead(II) sulfide precipitate (

PbS

) and ethanoic acid (

CH_3COOH

).

The equation is:

(CH_3COO)_2Pb_{(aq)} + H_2S_{(g)} \rightarrow PbS_{(s)} + 2CH_3COOH_{(aq)}

.

2

Determine the physical property (color) of the resulting lead(II) sulfide precipitate.

Lead(II) sulfide (

PbS

) is a distinct black solid.

This specific color change is the standard laboratory confirmatory test for identifying

H_2S

gas.

Key Concept

Qualitative test for hydrogen sulfide gas using lead(II) ethanoate paper or solution

Question 44Question →

A water treatment facility needs to soften $100\text{ dm}^3$ of well water containing $0.005\text{ mol dm}^{-3}$ of dissolved calcium hydrogentrioxocarbonate(IV), $\text{Ca(HCO}_3)_2$ , using Clark's process. What is the minimum mass of calcium hydroxide, $\text{Ca(OH)}_2$ , required to completely precipitate the calcium ions responsible for this temporary hardness? $[\text{Ca} = 40,\text{ O} = 16,\text{ H} = 1]$

37.0\text{ g}

74.0\text{ g}

18.5\text{ g}

81.0\text{ g}

Show answer & explanation

Answer: $37.0\text{ g}$

Answer

The minimum mass of calcium hydroxide required is

37.0\text{ g}

.

The correct answer of

37.0\text{ g}

is obtained by finding the moles of dissolved

\text{Ca(HCO}_3)_2

(

0.005\text{ mol dm}^{-3} \times 100\text{ dm}^3 = 0.5\text{ mol}

) and applying the

1:1

stoichiometric ratio from the reaction equation

\text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3 + 2\text{H}_2\text{O}

. Multiplying

0.5\text{ mol}

by the molar mass of

\text{Ca(OH)}_2

(

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

) yields

37.0\text{ g}

.

Step-by-Step Solution

1

Calculate the amount in moles of dissolved calcium hydrogentrioxocarbonate(IV) in the water sample.

\text{Moles of Ca(HCO}_3)_2 = \text{Concentration} \times \text{Volume} = 0.005\text{ mol dm}^{-3} \times 100\text{ dm}^3 = 0.5\text{ mol}

.

Determining the exact molar quantity of solute is the first step in stoichiometric calculations.

2

Write the balanced chemical equation for Clark's process (slaked lime softening).

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

. The mole ratio of

\text{Ca(HCO}_3)_2

to

\text{Ca(OH)}_2

is

1:1

.

Clark's process uses calculated amounts of calcium hydroxide to convert soluble hydrogentrioxocarbonates into insoluble trioxocarbonate(IV) precipitates.

3

Calculate the molar mass of calcium hydroxide,

\text{Ca(OH)}_2

.

\text{Molar mass} = 40 + 2(16 + 1) = 74\text{ g mol}^{-1}

.

Molar mass is required to convert moles of reagent into mass in grams.

4

Determine the required mass of

\text{Ca(OH)}_2

.

\text{Mass} = \text{Moles} \times \text{Molar mass} = 0.5\text{ mol} \times 74\text{ g mol}^{-1} = 37.0\text{ g}

.

Multiplying the required moles by molar mass gives the required mass of slaked lime.

Key Concept

Removal of temporary water hardness using Clark's process (addition of calculated lime).

Estimated Time:2m 0s

Question 45Question →

Under identical conditions of temperature and pressure, a given volume of deuterium gas ( $\text{D}_2$ ) requires $40\text{ seconds}$ to diffuse through a porous membrane. How long will it take for an equal volume of protium gas ( $\text{H}_2$ ) to diffuse through the same membrane? (Relative atomic masses: $\text{H} = 1.0$ , $\text{D} = 2.0$ )

28.3\text{ s}

20.0\text{ s}

56.6\text{ s}

80.0\text{ s}

Show answer & explanation

Answer: $28.3\text{ s}$

Answer

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

28.3\text{ s}

.

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 protium gas (

\text{H}_2

,

M = 2.0\text{ g mol}^{-1}

) is lighter than deuterium gas (

\text{D}_2

,

M = 4.0\text{ g mol}^{-1}

), it diffuses faster, taking

t = 40 \times \sqrt{2/4} = 28.3\text{ s}

.

Step-by-Step Solution

1

Calculate the relative molecular masses of deuterium gas (

\text{D}_2

) and protium gas (

\text{H}_2

).

M(\text{D}_2) = 2 \times 2.0 = 4.0\text{ g mol}^{-1}

and

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

.

Molecular masses are needed to apply Graham's Law of diffusion.

2

State Graham's Law relating diffusion time to molar mass for a constant volume.

\frac{t_{\text{H}_2}}{t_{\text{D}_2}} = \sqrt{\frac{M(\text{H}_2)}{M(\text{D}_2)}}

.

The time required for diffusion of a fixed volume is directly proportional to the square root of the molar mass of the gas.

3

Substitute the given values into the formula and solve for

t_{\text{H}_2}

.

t_{\text{H}_2} = 40 \times \sqrt{\frac{2.0}{4.0}} = 40 \times \sqrt{0.5} \approx 40 \times 0.7071 = 28.3\text{ s}

.

Performing the calculation yields the exact diffusion time for protium gas.

Key Concept

Graham's Law of Diffusion applied to hydrogen isotopes

Question 46Question →

A $250\text{ dm}^3$ sample of hard water contains $0.012\text{ mol dm}^{-3}$ of dissolved magnesium tetraoxosulfate(VI), $\text{MgSO}_4$ . What mass, in grams, of anhydrous sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3$ , is required to completely precipitate all the magnesium ions as magnesium trioxocarbonate(IV) and soften the water? $[\text{Molar mass of Na}_2\text{CO}_3 = 106\text{ g mol}^{-1}]$

Show answer & explanation

Answer: 318

Answer

318 g of anhydrous sodium trioxocarbonate(IV) is required.

Permanent water hardness caused by soluble magnesium salts like magnesium tetraoxosulfate(VI) (

\text{MgSO}_4

) is removed by reaction with sodium trioxocarbonate(IV) (

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

). The balanced reaction

\text{MgSO}_4\text{(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow \text{MgCO}_3\text{(s)} + \text{Na}_2\text{SO}_4\text{(aq)}

shows a 1:1 molar ratio. A

250\text{ dm}^3

volume at

0.012\text{ mol dm}^{-3}

contains

3.0\text{ moles}

of

\text{MgSO}_4

, which requires

3.0\text{ moles}

of

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

. Multiplying by its molar mass (

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

) gives

318\text{ g}

.

Step-by-Step Solution

1

Calculate the total number of moles of magnesium tetraoxosulfate(VI) in the water sample.

n(\text{MgSO}_4) = 250\text{ dm}^3 \times 0.012\text{ mol dm}^{-3} = 3.0\text{ mol}

Molar amount is calculated by multiplying the solution volume by its molar concentration.

2

Write the balanced chemical equation for softening permanent hardness with sodium trioxocarbonate(IV).

\text{MgSO}_4(\text{aq}) + \text{Na}_2\text{CO}_3(\text{aq}) \rightarrow \text{MgCO}_3(\text{s}) + \text{Na}_2\text{SO}_4(\text{aq})

Soluble magnesium ions causing permanent hardness are removed by precipitation as insoluble magnesium trioxocarbonate(IV).

3

Calculate the mass of anhydrous sodium trioxocarbonate(IV) needed.

\text{Mass} = 3.0\text{ mol} \times 106\text{ g mol}^{-1} = 318\text{ g}

From the 1:1 stoichiometric ratio, 3.0 moles of sodium trioxocarbonate(IV) is required.

Key Concept

Quantitative removal of permanent water hardness using sodium trioxocarbonate(IV) (washing soda)

Question 47Question →

When dilute tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) is added to sodium peroxide ( $\text{Na}_2\text{O}_2$ ), hydrogen peroxide ( $\text{H}_2\text{O}_2$ ) is liberated. In contrast, treating lead(IV) oxide ( $\text{PbO}_2$ ) with dilute acids does not produce hydrogen peroxide. Which of the following best accounts for this chemical distinction?

Sodium peroxide contains the peroxide ion (

\text{O}_2^{2-}

), whereas lead(IV) oxide is a true dioxide containing discrete oxide ions (

\text{O}^{2-}

).

Lead(IV) oxide is an amphoteric oxide, whereas sodium peroxide is a neutral oxide.

Sodium peroxide acts purely as a dehydrating agent in acid solution, while lead(IV) oxide is a reducing agent.

Oxygen in sodium peroxide has an oxidation state of

-2

, whereas in lead(IV) oxide oxygen has an oxidation state of

-1

.

Show answer & explanation

Answer: Sodium peroxide contains the peroxide ion ( $\text{O}_2^{2-}$ ), whereas lead(IV) oxide is a true dioxide containing discrete oxide ions ( $\text{O}^{2-}$ ).

Answer

Sodium peroxide contains the peroxide ion (

\text{O}_2^{2-}

), whereas lead(IV) oxide is a true dioxide containing discrete oxide ions (

\text{O}^{2-}

).

Peroxides such as

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

contain the peroxide linkage (

-\text{O}-\text{O}-

or

\text{O}_2^{2-}

ion) where oxygen has an oxidation state of

-1

. When treated with dilute acids, the peroxide ion combines with hydrogen ions to yield hydrogen peroxide (

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

). Dioxides such as

\text{PbO}_2

contain standard oxide ions (

\text{O}^{2-}

) paired with a quadrivalent metal ion (

\text{Pb}^{4+}

); they do not contain the peroxide link and therefore cannot produce hydrogen peroxide upon reaction with dilute acids.

Step-by-Step Solution

1

Analyze the structural composition and oxidation states of oxygen in sodium peroxide (

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

).

In

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

, sodium is

+1

, so oxygen exists as the peroxide ion

\text{O}_2^{2-}

with an oxidation number of

-1

.

Peroxides are characterized by the oxygen-oxygen single bond (

-\text{O}-\text{O}-

) present in the

\text{O}_2^{2-}

group.

2

Analyze the structural composition and oxidation states of oxygen in lead(IV) oxide (

\text{PbO}_2

).

In

\text{PbO}_2

, lead is in the

+4

oxidation state, bonded to two separate oxide ions (

\text{O}^{2-}

), where oxygen has an oxidation number of

-2

.

Dioxides contain metal atoms in high oxidation states paired with simple oxide ions (

\text{O}^{2-}

), rather than peroxide ions.

3

Evaluate the chemical behavior of both compounds with dilute acids.

\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

(liberates

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

).

\text{PbO}_2

does not yield

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

because it lacks the

\text{O}_2^{2-}

link.

Only compounds containing the structural peroxide ion can form hydrogen peroxide upon acidification.

Key Concept

Distinction between peroxides (

\text{O}_2^{2-}

) and dioxides (

\text{O}^{2-}

)

Estimated Time:1m 15s

Question 48Question →

Heavy water is water in which the hydrogen atoms are replaced by the hydrogen isotope deuterium ( $^{2}_{1}\text{H}$ or $\text{D}$ ). Given that the atomic mass of deuterium is $2\text{ g mol}^{-1}$ and oxygen ( $^{16}_{8}\text{O}$ ) is $16\text{ g mol}^{-1}$ , what is the molar mass of heavy water ( $\text{D}_2\text{O}$ ) in $\text{g mol}^{-1}$ ?

Show answer & explanation

Answer: 20

Answer

The molar mass of heavy water (

\text{D}_2\text{O}

) is

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

.

Heavy water (

\text{D}_2\text{O}

) contains two atoms of deuterium (

^2\text{H}

, atomic mass =

2

) and one atom of oxygen (

^{16}\text{O}

, atomic mass =

16

). The molar mass is calculated as

(2 \times 2) + 16 = 20\text{ g mol}^{-1}

.

Step-by-Step Solution

1

Determine the molecular composition of heavy water

Heavy water (

\text{D}_2\text{O}

) contains 2 deuterium atoms and 1 oxygen atom.

Deuterium is an isotope of hydrogen containing one proton and one neutron, giving it a mass number of 2.

2

Sum the relative atomic masses of all constituent atoms

(2 \times 2) + 16 = 20\text{ g mol}^{-1}

Multiplying the mass of deuterium by two and adding the mass of one oxygen atom yields the molar mass of the compound.

Key Concept

Heavy Water and Deuterium Molar Mass

Question 49Question →

Match each of the following oxides on the left with its correct acid-base classification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium oxide (

\text{K}_2\text{O}

) matches Basic oxide; Sulfur dioxide (

\text{SO}_2

) matches Acidic oxide; Lead(II) oxide (

\text{PbO}

) matches Amphoteric oxide; Nitrogen(II) oxide (

\text{NO}

) matches Neutral oxide.

Potassium oxide is a basic metallic oxide; sulfur dioxide is an acidic non-metal oxide; lead(II) oxide reacts with both acids and alkalis (amphoteric); and nitrogen(II) oxide does not react with either acids or alkalis (neutral).

Step-by-Step Solution

1

Identify the nature of potassium oxide (

\text{K}_2\text{O}

).

As an alkali metal oxide, it dissolves in water to form potassium hydroxide (

\text{KOH}

) and reacts with acids to yield salts, classifying it as a basic oxide.

Basic oxides are metallic oxides that neutralize acids.

2

Identify the nature of sulfur dioxide (

\text{SO}_2

).

It is a covalent non-metal oxide that reacts with alkalis to form trioxosulfate(IV) salts, classifying it as an acidic oxide.

Acidic oxides (acid anhydrides) react with bases or water to yield acidic solutions/salts.

3

Identify the nature of lead(II) oxide (

\text{PbO}

).

It shows dual reactivity, dissolving in acids like

\text{HNO}_3

and bases like concentrated

\text{NaOH}

, classifying it as an amphoteric oxide.

Amphoteric oxides can behave as either acids or bases depending on the reactant.

4

Identify the nature of nitrogen(II) oxide (

\text{NO}

).

It displays no acid-base properties with aqueous acids or bases, classifying it as a neutral oxide.

Neutral oxides do not form salts when treated with acids or alkalis.

Key Concept

Classification of oxides based on acid-base character (basic, acidic, amphoteric, neutral)

Question 50Question →

An unknown solid metallic oxide $X$ dissolves in cold dilute tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) to form a solution that turns acidified potassium iodide ( $\text{KI}$ ) solution brown due to the liberation of iodine, without releasing any gas during the reaction. In contrast, lead(IV) oxide ( $\text{PbO}_2$ ) treated with cold dilute acid does not produce hydrogen peroxide ( $\text{H}_2\text{O}_2$ ). Which of the following correctly classifies oxide $X$ and accounts for its distinct behavior from lead(IV) oxide?

Oxide

X

is a peroxide containing the

\text{O}_2^{2-}

ion which yields

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

with dilute acid, whereas

\text{PbO}_2

is a dioxide containing simple

\text{O}^{2-}

ions.

Oxide

X

is an amphoteric oxide that oxidizes iodide ions, whereas

\text{PbO}_2

is a purely basic oxide that cannot participate in redox reactions.

Oxide

X

is an allotrope of oxygen, whereas

\text{PbO}_2

is a neutral oxide.

Oxide

X

is a basic oxide because it reacts with acid without gas evolution, whereas

\text{PbO}_2

is an acidic oxide.

Show answer & explanation

Answer: Oxide $X$ is a peroxide containing the $\text{O}_2^{2-}$ ion which yields $\text{H}_2\text{O}_2$ with dilute acid, whereas $\text{PbO}_2$ is a dioxide containing simple $\text{O}^{2-}$ ions.

Answer

Oxide X is a peroxide containing the

\text{O}_2^{2-}

ion which yields

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

with dilute acid, whereas

\text{PbO}_2

is a dioxide containing simple

\text{O}^{2-}

ions.

Peroxides (such as sodium peroxide or barium peroxide) contain the peroxide ion

\text{O}_2^{2-}

with oxygen in the

-1

oxidation state. When treated with cold dilute acids, they form hydrogen peroxide (

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

), which oxidizes

\text{I}^-

to brown

\text{I}_2

. In contrast, dioxides like lead(IV) oxide (

\text{PbO}_2

) contain standard oxide ions (

\text{O}^{2-}

) with lead in the

+4

oxidation state, so they do not produce

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

under cold dilute acid conditions.

Step-by-Step Solution

1

Analyze the chemical test result for Oxide X.

Oxide

X

reacts with cold dilute

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

to form a solution that oxidizes

\text{KI}

to

\text{I}_2

(brown solution).

This behavior is characteristic of peroxides (such as

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

or

\text{BaO}_2

), which react with dilute acids to produce hydrogen peroxide (

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

). Hydrogen peroxide is a powerful oxidizing agent that liberates iodine from acidified potassium iodide:

\text{H}_2\text{O}_2 + 2\text{H}^+ + 2\text{I}^- \rightarrow 2\text{H}_2\text{O} + \text{I}_2

.

2

Distinguish between peroxides and dioxides.

Peroxides contain the peroxide linkage

[-\text{O}-\text{O}-]^{2-}

(

\text{O}_2^{2-}

ion with oxygen in oxidation state

-1

), while true dioxides contain

\text{O}^{2-}

ions with the metal in oxidation state

+4

.

Dioxides like

\text{PbO}_2

and

\text{MnO}_2

do not contain peroxide ions and therefore do not form

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

when treated with cold dilute acids.

3

Select the correct classification matching the observed chemistry.

Oxide

X

is a peroxide, whereas

\text{PbO}_2

is a dioxide.

This structural difference explains why oxide

X

yields

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

capable of liberating iodine from

\text{KI}

, while

\text{PbO}_2

does not.

Key Concept

Distinction between peroxides (containing

\text{O}_2^{2-}

) and dioxides (containing

\text{O}^{2-}

) based on acid reactions and peroxide test

Estimated Time:2m 0s

Question 51Question →

A sample of $90\text{ cm}^3$ of pure oxygen gas ( $\text{O}_2$ ) is passed through an ozonizer where $20\%$ by volume of the oxygen is converted into ozone ( $\text{O}_3$ ). If the resulting gas mixture is reacted completely with excess carbon(II) oxide ( $\text{CO}$ ) to yield carbon(IV) oxide ( $\text{CO}_2$ ), what is the total volume of carbon(IV) oxide gas produced under the same conditions of temperature and pressure?

90\text{ cm}^3

168\text{ cm}^3

180\text{ cm}^3

216\text{ cm}^3

Show answer & explanation

Answer: $180\text{ cm}^3$

Answer

The total volume of carbon(IV) oxide gas produced is

180\text{ cm}^3

.

The total volume of carbon(IV) oxide produced is

180\text{ cm}^3

because the total number of oxygen atoms available for reaction remains conserved regardless of the conversion into ozone.

72\text{ cm}^3

of diatomic oxygen yields

144\text{ cm}^3

of carbon(IV) oxide and

12\text{ cm}^3

of triatomic ozone yields

36\text{ cm}^3

of carbon(IV) oxide.

Step-by-Step Solution

1

Calculate the volume of oxygen converted to ozone and the volume of ozone formed.

Volume of

\text{O}_2

converted =

20\% \times 90\text{ cm}^3 = 18\text{ cm}^3

. Remaining unreacted

\text{O}_2 = 90 - 18 = 72\text{ cm}^3

. According to

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

,

18\text{ cm}^3

of

\text{O}_2

produces

\frac{2}{3} \times 18\text{ cm}^3 = 12\text{ cm}^3

of

\text{O}_3

.

Ozonolysis causes a volume contraction where 3 volumes of

\text{O}_2

yield 2 volumes of

\text{O}_3

.

2

Calculate the volume of

\text{CO}_2

produced by the unreacted

\text{O}_2

.

Reaction equation:

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

. Volume of

\text{CO}_2

from

\text{O}_2 = 2 \times 72\text{ cm}^3 = 144\text{ cm}^3

.

By Gay-Lussac's Law of Combining Volumes, 1 volume of

\text{O}_2

produces 2 volumes of

\text{CO}_2

.

3

Calculate the volume of

\text{CO}_2

produced by the

\text{O}_3

.

Reaction equation:

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

. Volume of

\text{CO}_2

from

\text{O}_3 = 3 \times 12\text{ cm}^3 = 36\text{ cm}^3

.

Each molecule of

\text{O}_3

contains 3 oxygen atoms, reacting with 3 molecules of

\text{CO}

to yield 3 molecules of

\text{CO}_2

.

4

Sum the total volume of

\text{CO}_2

produced.

Total volume of

\text{CO}_2 = 144\text{ cm}^3 + 36\text{ cm}^3 = 180\text{ cm}^3

.

The total volume of carbon(IV) oxide formed is the sum of the volumes produced by both oxygen allotropes in the mixture.

Key Concept

Gay-Lussac's Law of Combining Volumes and Allotropic Conversion of Oxygen to Ozone

Estimated Time:2m 0s

Question 52Question →

Match each oxygen-containing compound listed on the left with its corresponding chemical classification or characteristic reaction behavior on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

\text{P}_4\text{O}_{10}

matches 'Acidic oxide that reacts with water to yield a triprotic acid';

\text{CaO}

matches 'Basic oxide that reacts exothermically with water to form an alkaline solution';

\text{CO}

matches 'Neutral oxide that fails to form salts when exposed to acids or alkalis';

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

matches 'Peroxide species that decomposes releasing oxygen gas'.

Each item matches its corresponding behavior based on oxide and oxygen compound classification:

\text{P}_4\text{O}_{10}

is an acidic oxide producing triprotic acid;

\text{CaO}

is a basic oxide producing an alkaline hydroxide;

\text{CO}

is a neutral oxide;

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

is a peroxide species that liberates oxygen gas upon decomposition.

Step-by-Step Solution

1

Classify

\text{P}_4\text{O}_{10}

based on its reaction with water.

\text{P}_4\text{O}_{10}

is a non-metal oxide (phosphorus(V) oxide) which reacts with water to yield

\text{H}_3\text{PO}_4

, a triprotic acid.

Non-metal oxides in high oxidation states act as acid anhydrides.

2

Classify

\text{CaO}

based on its acid-base character.

\text{CaO}

is an alkaline earth metal oxide that reacts exothermically with water to form the alkaline base

\text{Ca(OH)}_2

.

Metallic oxides typically display basic chemical behavior.

3

Determine the chemical reactivity of

\text{CO}

.

\text{CO}

is a non-metal oxide that does not react with acids or alkalis to form salts, making it a neutral oxide.

Certain low-oxidation non-metal oxides are neutral.

4

Identify the nature of

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

.

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

contains the peroxide group with oxygen in the

-1

oxidation state and decomposes to liberate oxygen gas.

Peroxides undergo decomposition to yield oxygen and water.

Key Concept

Classification of Oxides (Acidic, Basic, Neutral) and Peroxides

Question 53Question →

A $10\text{ dm}^3$ sample of well water contains $0.002\text{ mol dm}^{-3}$ of dissolved calcium hydrogentrioxocarbonate(IV), $\text{Ca(HCO}_3)_2$ , and $0.003\text{ mol dm}^{-3}$ of dissolved calcium tetraoxosulfate(VI), $\text{CaSO}_4$ . A student boils the water sample thoroughly to soften it and filters off any precipitate formed. What is the molar concentration of $\text{Ca}^{2+}$ ions remaining in the filtered solution?

0.003\text{ mol dm}^{-3}

0.000\text{ mol dm}^{-3}

0.002\text{ mol dm}^{-3}

0.005\text{ mol dm}^{-3}

Show answer & explanation

Answer: $0.003\text{ mol dm}^{-3}$

Answer

0.003\text{ mol dm}^{-3}

Boiling converts soluble

\text{Ca(HCO}_3)_2

into insoluble

\text{CaCO}_3

, which is removed by filtration. Since

\text{CaSO}_4

causes permanent hardness and does not decompose upon boiling, its concentration of

0.003\text{ mol dm}^{-3}

remains dissolved in the filtered water.

Step-by-Step Solution

1

Identify the types of water hardness present in the sample.

Dissolved

\text{Ca(HCO}_3)_2

causes temporary hardness, while dissolved

\text{CaSO}_4

causes permanent hardness.

Hydrogentrioxocarbonates of calcium and magnesium cause temporary hardness, whereas tetraoxosulfates and chlorides cause permanent hardness.

2

Determine the chemical effect of boiling on temporary hardness.

Thermal decomposition occurs:

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

.

Soluble calcium hydrogentrioxocarbonate(IV) decomposes on heating into insoluble calcium trioxocarbonate(IV), which precipitates out of solution.

3

Determine the chemical effect of boiling on permanent hardness.

\text{CaSO}_4

remains dissolved and unaffected by boiling.

Permanent hardness cannot be removed by boiling because calcium tetraoxosulfate(VI) does not undergo thermal decomposition at

100^\circ\text{C}

.

4

Calculate the residual concentration of

\text{Ca}^{2+}

ions after filtration.

[\text{Ca}^{2+}] = 0.003\text{ mol dm}^{-3}

.

All

0.002\text{ mol dm}^{-3}

of

\text{Ca}^{2+}

from temporary hardness is precipitated and filtered out, leaving only the

0.003\text{ mol dm}^{-3}

of

\text{Ca}^{2+}

contributed by

\text{CaSO}_4

.

Key Concept

Distinction between temporary and permanent water hardness and their behavior upon thermal treatment (boiling).

Question 54Question →

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

Zinc oxide (

\text{ZnO}

)

Calcium oxide (

\text{CaO}

)

Sulfur(IV) oxide (

\text{SO}_2

)

Carbon(II) oxide (

\text{CO}

)

Show answer & explanation

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

Answer

Zinc oxide (

\text{ZnO}

) is the correct answer because it is an amphoteric oxide.

Zinc oxide (

\text{ZnO}

) is an amphoteric oxide. It exhibits both basic and acidic properties, reacting with acids like hydrochloric acid to form zinc chloride and with strong bases like sodium hydroxide to form sodium zincate.

Step-by-Step Solution

1

Classify the given oxides based on their acid-base behavior

Calcium oxide is basic, Sulfur(IV) oxide is acidic, Carbon(II) oxide is neutral, and Zinc oxide is amphoteric.

Amphoteric oxides exhibit dual properties, reacting with both acids and bases.

2

Identify the oxide that reacts with both dilute hydrochloric acid and sodium hydroxide

Zinc oxide reacts with

\text{HCl}

to form zinc chloride and water, and with

\text{NaOH}

to form sodium zincate and water.

Only amphoteric oxides (such as

\text{ZnO}

,

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

, and

\text{PbO}

) form salts with both acids and strong alkalis.

Key Concept

Classification of Oxides (Amphoteric Oxides)

Question 55Question →

What volume of dry oxygen gas measured at STP is produced when $12.25\text{ g}$ of potassium trioxochlorate(V), $\text{KClO}_3$ , is completely decomposed by heating in the presence of manganese(IV) oxide catalyst?

$[\text{K} = 39.0, \text{Cl} = 35.5, \text{O} = 16.0, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]$

2.24\text{ dm}^3

3.36\text{ dm}^3

3.60\text{ dm}^3

6.72\text{ dm}^3

Show answer & explanation

Answer: $3.36\text{ dm}^3$

Answer

3.36\text{ dm}^3

Thermal decomposition of potassium trioxochlorate(V) follows

2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2

. Since

12.25\text{ g}

of

\text{KClO}_3

corresponds to

0.10\text{ mol}

, the reaction yields

0.15\text{ mol}

of

\text{O}_2

. At STP,

0.15\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 3.36\text{ dm}^3

of oxygen gas.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of potassium trioxochlorate(V).

2\text{KClO}_3\text{(s)} \rightarrow 2\text{KCl(s)} + 3\text{O}_2\text{(g)}

Establishing the correct stoichiometric mole ratio between reactant and gaseous product is essential for calculations.

2

Calculate the molar mass of

\text{KClO}_3

and determine the number of moles reacted.

Molar mass of

\text{KClO}_3 = 39.0 + 35.5 + (3 \times 16.0) = 122.5\text{ g mol}^{-1}

. Moles of

\text{KClO}_3 = \frac{12.25\text{ g}}{122.5\text{ g mol}^{-1}} = 0.10\text{ mol}

.

Mass must be converted to moles to utilize equation stoichiometry.

3

Determine the moles of

\text{O}_2

produced using the mole ratio.

Moles of

\text{O}_2 = 0.10\text{ mol KClO}_3 \times \frac{3\text{ mol O}_2}{2\text{ mol KClO}_3} = 0.15\text{ mol O}_2

.

Two moles of

\text{KClO}_3

yield three moles of

\text{O}_2

.

4

Calculate the volume of

\text{O}_2

produced at STP.

\text{Volume of } \text{O}_2 = 0.15\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 3.36\text{ dm}^3$.

Multiplying the amount of gas in moles by the molar gas volume at STP gives the volume.

Key Concept

Laboratory preparation of oxygen gas via catalytic thermal decomposition of potassium trioxochlorate(V) and stoichiometric volume calculations at STP.

Question 56Question →

Water hardness is classified as either temporary or permanent depending on the dissolved salts present. Which of the following chemical compounds causes temporary hardness in water that can be removed simply by boiling?

Calcium hydrogentrioxocarbonate(IV)

Calcium tetraoxosulfate(VI)

Magnesium chloride

Calcium chloride

Show answer & explanation

Answer: Calcium hydrogentrioxocarbonate(IV)

Answer

Calcium hydrogentrioxocarbonate(IV)

The correct compound is Calcium hydrogentrioxocarbonate(IV). Temporary water hardness is uniquely caused by dissolved hydrogentrioxocarbonate(IV) salts of calcium or magnesium. Heating or boiling the water decomposes this soluble compound into insoluble calcium trioxocarbonate(IV) precipitate, water, and carbon(IV) oxide gas, thereby removing the calcium ions responsible for hardness.

Step-by-Step Solution

1

Identify the cause of temporary water hardness.

Temporary hardness is caused by dissolved hydrogentrioxocarbonate(IV) salts of calcium and magnesium, such as

\text{Ca(HCO}_3)_2

.

Hydrogentrioxocarbonate(IV) ions decompose thermally when heated.

2

Analyze the chemical effect of boiling on calcium hydrogentrioxocarbonate(IV).

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

Boiling converts the soluble hydrogentrioxocarbonate(IV) into insoluble calcium trioxocarbonate(IV), which precipitates out as fur or scale.

Key Concept

Temporary Water Hardness and Removal by Boiling

Estimated Time:45s

Question 57Question →

A student investigates two unlabelled water samples, $X$ and $Y$ . Boiling sample $X$ eliminates its soap-consuming property, allowing it to lather readily, whereas sample $Y$ continues to form scum with soap after boiling. Which solute present in sample $Y$ accounts for its persistent hardness?

Calcium tetraoxosulfate(VI)

Calcium hydrogentrioxocarbonate(IV)

Magnesium hydrogentrioxocarbonate(IV)

Sodium chloride

Show answer & explanation

Answer: Calcium tetraoxosulfate(VI)

Answer

Calcium tetraoxosulfate(VI) accounts for the persistent (permanent) water hardness in sample Y.

Permanent water hardness is caused by dissolved calcium tetraoxosulfate(VI) or magnesium chloride. These salts do not decompose upon heating, so boiling does not precipitate the metal ions responsible for consuming soap.

Step-by-Step Solution

1

Classify the type of hardness based on response to boiling.

Sample X exhibits temporary hardness because boiling removes its soap-consuming capability. Sample Y exhibits permanent hardness because boiling fails to remove its scum-forming property.

Temporary hardness is caused by hydrogen trioxocarbonates which decompose on boiling, whereas permanent hardness is caused by soluble sulfate or chloride salts of calcium or magnesium.

2

Identify the specific chemical species responsible for permanent hardness in sample Y.

Calcium tetraoxosulfate(VI) (

\text{CaSO}_4

) remains dissolved after boiling.

Tetraoxosulfate(VI) ions do not undergo thermal decomposition at boiling temperatures to form insoluble precipitates.

Key Concept

Temporary vs. Permanent Water Hardness

Estimated Time:1m 0s

Question 58Question →

In an industrial furnace operation, producer gas is synthesized by passing air over red-hot coke. Which of the following pairs correctly identifies the main combustible constituent and the primary non-combustible diluent present in the resulting gas mixture?

Carbon(II) oxide and nitrogen

Carbon(II) oxide and hydrogen

Carbon(IV) oxide and nitrogen

Carbon(IV) oxide and hydrogen

Show answer & explanation

Answer: Carbon(II) oxide and nitrogen

Answer

The main combustible constituent of producer gas is carbon(II) oxide (

\text{CO}

) and the primary non-combustible diluent is nitrogen (

\text{N}_2

).

When air is passed over red-hot coke, oxygen reacts with carbon to form carbon(II) oxide (

\text{CO}

), which is the primary combustible fuel component. The unreactive nitrogen gas (

\text{N}_2

) present in air passes through the bed of coke without reacting, forming the main non-combustible diluent (approximately two-thirds of the total volume).

Step-by-Step Solution

1

Identify the reactants involved in the production of producer gas.

Air (containing

\text{O}_2

and

\text{N}_2

) is passed over incandescent coke (

\text{C}

).

Producer gas manufacture relies on passing air over red-hot carbon.

2

Determine the chemical reaction and resulting gaseous products.

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

, while atmospheric

\text{N}_2

remains unreacted.

Incomplete oxidation of coke produces

\text{CO}

, while

\text{N}_2

from air dilutes the product mixture.

3

Classify the role of each component in the product gas mixture.

\text{CO}

is combustible and serves as the fuel, whereas

\text{N}_2

is inert/non-combustible.

\text{CO}

can undergo further oxidation to

\text{CO}_2

releasing energy, whereas

\text{N}_2

does not burn.

Key Concept

Composition and industrial production of producer gas

Estimated Time:1m 0s

Question 59Question →

An industrial fuel gas mixture derived from coal processing contains $40.0\%$ carbon(II) oxide ( $\text{CO}$ ), $40.0\%$ hydrogen gas ( $\text{H}_2$ ), and $20.0\%$ carbon(IV) oxide ( $\text{CO}_2$ ) by volume. Assuming air contains $20.0\%$ oxygen ( $\text{O}_2$ ) by volume, what is the total volume of air required at s.t.p. for the complete combustion of $100.0\text{ dm}^3$ of this fuel gas mixture?

200.0\text{ dm}^3

40.0\text{ dm}^3

400.0\text{ dm}^3

500.0\text{ dm}^3

Show answer & explanation

Answer: $200.0\text{ dm}^3$

Answer

The total volume of air required at s.t.p. for complete combustion is

200.0\text{ dm}^3

.

In the

100.0\text{ dm}^3

sample of fuel gas, the combustible gases are

40.0\text{ dm}^3

of

\text{CO}

and

40.0\text{ dm}^3

of

\text{H}_2

, while

\text{CO}_2

is already fully oxidized and non-combustible. According to reaction stoichiometry,

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

and

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

,

40.0\text{ dm}^3

of

\text{CO}

requires

20.0\text{ dm}^3

of

\text{O}_2

and

40.0\text{ dm}^3

of

\text{H}_2

requires

20.0\text{ dm}^3

of

\text{O}_2

. The total pure oxygen needed is

40.0\text{ dm}^3

. Since air is

20.0\%

oxygen by volume, the required volume of air is

40.0\text{ dm}^3 \div 0.20 = 200.0\text{ dm}^3

.

Step-by-Step Solution

1

Determine the volumes of combustible and non-combustible components in the

100.0\text{ dm}^3

fuel gas mixture.

Volume of

\text{CO} = 40.0\text{ dm}^3

, volume of

\text{H}_2 = 40.0\text{ dm}^3

, and volume of

\text{CO}_2 = 20.0\text{ dm}^3

(non-combustible).

By Gay-Lussac's Law of Combining Volumes, volume percentage corresponds directly to gas volume at s.t.p.

2

Write the balanced chemical equations and determine oxygen requirements for each combustible component.

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

requires

20.0\text{ dm}^3

of

\text{O}_2

;

2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)

requires

20.0\text{ dm}^3

of

\text{O}_2

. Total pure

\text{O}_2 = 40.0\text{ dm}^3

.

Each combustible gas (

\text{CO}

and

\text{H}_2

) reacts with oxygen in a

2:1

mole (and volume) ratio.

3

Calculate the total volume of air needed to supply the required

40.0\text{ dm}^3

of pure oxygen.

\text{Volume of air} = \frac{40.0\text{ dm}^3}{0.20} = 200.0\text{ dm}^3

.

Air contains only

20.0\%

oxygen by volume, so five times the volume of oxygen is needed in total air.

Key Concept

Industrial Fuel Gas Stoichiometry and Gas Volumetric Calculations

Question 60Question →

During the destructive distillation of coal, the volatile gaseous fraction is collected and purified to yield coal gas. A $50.0\text{ dm}^3$ sample of this coal gas contains $50.0\%$ hydrogen ( $\text{H}_2$ ), $30.0\%$ methane ( $\text{CH}_4$ ), and $20.0\%$ carbon(II) oxide ( $\text{CO}$ ) by volume. What is the total volume of pure oxygen gas required at STP for the complete combustion of this sample?

47.5\text{ dm}^3

32.5\text{ dm}^3

60.0\text{ dm}^3

95.0\text{ dm}^3

Show answer & explanation

Answer: $47.5\text{ dm}^3$

Answer

The total volume of pure oxygen required at STP for complete combustion is

47.5\text{ dm}^3

.

By Gay-Lussac's law of combining volumes, gases react in simple whole-number volume ratios. In a

50.0\text{ dm}^3

coal gas mixture, the volumes of

\text{H}_2

,

\text{CH}_4

, and

\text{CO}

are

25.0\text{ dm}^3

,

15.0\text{ dm}^3

, and

10.0\text{ dm}^3

respectively. According to their balanced combustion equations,

\text{H}_2

requires half its volume in

\text{O}_2

(

12.5\text{ dm}^3

),

\text{CH}_4

requires twice its volume in

\text{O}_2

(

30.0\text{ dm}^3

), and

\text{CO}

requires half its volume in

\text{O}_2

(

5.0\text{ dm}^3

). Summing these gives

12.5 + 30.0 + 5.0 = 47.5\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the individual volumes of each constituent gas in the

50.0\text{ dm}^3

coal gas mixture.

Volume of

\text{H}_2 = 0.500 \times 50.0\text{ dm}^3 = 25.0\text{ dm}^3

; Volume of

\text{CH}_4 = 0.300 \times 50.0\text{ dm}^3 = 15.0\text{ dm}^3

; Volume of

\text{CO} = 0.200 \times 50.0\text{ dm}^3 = 10.0\text{ dm}^3

.

By Gay-Lussac's Law, volume percentages directly represent mole fractions at constant temperature and pressure.

2

Write the balanced chemical equations for the complete combustion of each component.

(1)

2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}

(2) \text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}

(3)

2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)}$.

Stoichiometric coefficients define the combining volume ratios of reactants.

3

Determine the volume of

\text{O}_2

needed for each gas component.

For

\text{H}_2

:

V(\text{O}_2) = \frac{1}{2} \times 25.0\text{ dm}^3 = 12.5\text{ dm}^3

.
For

\text{CH}_4

:

V(\text{O}_2) = 2 \times 15.0\text{ dm}^3 = 30.0\text{ dm}^3

.
For

\text{CO}

:

V(\text{O}_2) = \frac{1}{2} \times 10.0\text{ dm}^3 = 5.0\text{ dm}^3

.

Applying Gay-Lussac's law of combining volumes to each combustion reaction.

4

Sum the required volumes of oxygen for all three components.

Total

V(\text{O}_2) = 12.5\text{ dm}^3 + 30.0\text{ dm}^3 + 5.0\text{ dm}^3 = 47.5\text{ dm}^3

.

The total oxygen needed is the additive sum of individual combustion demands.

Key Concept

Combustion Stoichiometry of Industrial Fuel Gases

Estimated Time:2m 0s

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