Non-Metals and Their Compounds Practice Questions — JAMB UTME — Page 4

Question 61Question →

A water sample containing $0.010\text{ mol}$ of dissolved calcium hydrogentrioxocarbonate(IV), $\text{Ca(HCO}_3)_2$ , responsible for temporary hardness, is boiled until complete precipitation of calcium trioxocarbonate(IV) occurs. The resulting solid residue is isolated and treated with an excess of dilute hydrochloric acid. What volume of carbon(IV) oxide gas, measured at room temperature and pressure (RTP, molar volume of gas $= 24.0\text{ dm}^3\text{ mol}^{-1}$ ), is liberated during the acid treatment?

0.240\text{ dm}^3

0.224\text{ dm}^3

0.480\text{ dm}^3

0.120\text{ dm}^3

Show answer & explanation

Answer: $0.240\text{ dm}^3$

Answer

The volume of carbon(IV) oxide gas liberated during the acid treatment is

0.240\text{ dm}^3

.

Boiling

0.010\text{ mol}

of

\text{Ca(HCO}_3)_2

yields

0.010\text{ mol}

of

\text{CaCO}_3

precipitate. Reacting this precipitate with excess dilute acid yields

0.010\text{ mol}

of

\text{CO}_2

gas. At RTP,

0.010\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.240\text{ dm}^3

.

Step-by-Step Solution

1

Determine moles of

\text{CaCO}_3

precipitate formed during boiling.

The balanced thermal decomposition equation is

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

. Thus,

0.010\text{ mol}

of

\text{Ca(HCO}_3)_2

produces

0.010\text{ mol}

of

\text{CaCO}_3

.

Temporary hardness is removed by boiling, which decomposes hydrogentrioxocarbonate(IV) into insoluble trioxocarbonate(IV).

2

Determine moles of

\text{CO}_2

gas evolved from the reaction of

\text{CaCO}_3

with dilute

\text{HCl}

.

The balanced chemical equation is

\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

.

0.010\text{ mol}

of

\text{CaCO}_3

reacts to release

0.010\text{ mol}

of

\text{CO}_2

.

Trioxocarbonate(IV) salts react with acids in a 1:1 mole ratio with respect to carbon(IV) oxide gas produced.

3

Calculate the volume of

\text{CO}_2

gas evolved at RTP.

\text{Volume} = 0.010\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.240\text{ dm}^3$.

At room temperature and pressure, one mole of any gas occupies

24.0\text{ dm}^3

.

Key Concept

Thermal decomposition of temporary water hardness salts and acid reactions of trioxocarbonate(IV) salts

Question 62Question →

Ammonium trioxonitrate(V), $NH_4NO_3$ , decomposes on heating to produce dinitrogen(I) oxide and water vapor according to the balanced chemical equation:

NH_4NO_3(s) \rightarrow N_2O(g) + 2H_2O(g)

What volume of dinitrogen(I) oxide gas, measured at STP, is produced from the complete decomposition of $8.0\text{ g}$ of pure ammonium trioxonitrate(V)?
$[\text{Molar mass of } NH_4NO_3 = 80\text{ g mol}^{-1},\text{ Molar volume of gas at STP } = 22.4\text{ dm}^3\text{ mol}^{-1}]$

2.24\text{ dm}^3

2.40\text{ dm}^3

4.48\text{ dm}^3

6.72\text{ dm}^3

Show answer & explanation

Answer: $2.24\text{ dm}^3$

Answer

The volume of dinitrogen(I) oxide gas produced at STP is

2.24\text{ dm}^3

.

The correct response calculates the mole value of ammonium trioxonitrate(V) as

0.10\text{ mol}

(

8.0\text{ g} / 80\text{ g mol}^{-1}

). Using the

1:1

stoichiometric ratio from the balanced decomposition reaction,

0.10\text{ mol}

of dinitrogen(I) oxide is produced. At standard temperature and pressure (STP), one mole of any ideal gas occupies

22.4\text{ dm}^3

, giving

0.10 \times 22.4 = 2.24\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the number of moles of

NH_4NO_3

reacted

\text{Moles of } NH_4NO_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{8.0\text{ g}}{80\text{ g mol}^{-1}} = 0.10\text{ mol}

Converting given mass into moles is necessary to apply stoichiometric relationships.

2

Determine the mole ratio between

NH_4NO_3

and

N_2O

From the balanced equation,

1\text{ mol of } NH_4NO_3

produces

1\text{ mol of } N_2O

. Thus,

0.10\text{ mol of } NH_4NO_3

yields

0.10\text{ mol of } N_2O

.

Stoichiometry dictates the theoretical mole yield of the target product.

3

Calculate the volume of

N_2O

gas at STP

\text{Volume at STP} = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

Multiplying the mole amount of gas by the molar volume at STP (

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

) gives the volume.

Key Concept

Stoichiometric gas volume calculation at standard temperature and pressure (STP) for thermal decomposition reactions of nitrogen compounds.

Estimated Time:1m 30s

Question 63Question →

When a $16.0\text{ g}$ sample of pure ammonium trioxonitrate(V), $NH_4NO_3$ , undergoes complete thermal decomposition, it produces dinitrogen monoxide gas, $N_2O$ , and water vapor. Assuming standard temperature and pressure (STP, where molar gas volume = $22.4\text{ dm}^3\text{mol}^{-1}$ ), what is the volume of the oxide of nitrogen collected, and how does its rate of diffusion compare to that of carbon(IV) oxide, $CO_2$ , under identical conditions? (Molar masses: $N = 14\text{ g/mol}$ , $O = 16\text{ g/mol}$ , $H = 1\text{ g/mol}$ , $C = 12\text{ g/mol}$ )

4.48\text{ dm}^3

, and its rate of diffusion is equal to that of

CO_2

4.80\text{ dm}^3

, and its rate of diffusion is equal to that of

CO_2

4.48\text{ dm}^3

, and it diffuses

\sqrt{2}

times faster than

CO_2

8.96\text{ dm}^3

, and its rate of diffusion is slower than that of

CO_2

Show answer & explanation

Answer: $4.48\text{ dm}^3$ , and its rate of diffusion is equal to that of $CO_2$

Answer

4.48\text{ dm}^3

, and its rate of diffusion is equal to that of

CO_2

Thermal decomposition of

16.0\text{ g}

(

0.20\text{ mol}

) of

NH_4NO_3

produces

0.20\text{ mol}

of

N_2O

. At STP,

0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

. Both

N_2O

and

CO_2

have a molar mass of

44\text{ g/mol}

, so by Graham's law of diffusion, their relative rates of diffusion are identical.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of ammonium trioxonitrate(V)

NH_4NO_3(s) \rightarrow N_2O(g) + 2H_2O(g)

Establishes the stoichiometric mole ratio between

NH_4NO_3

and

N_2O

, which is

1 : 1

.

2

Calculate the molar mass and number of moles of

NH_4NO_3

reacted

Molar mass of

NH_4NO_3 = 2(14) + 4(1) + 3(16) = 80\text{ g/mol}

. Number of moles =

\frac{16.0\text{ g}}{80\text{ g/mol}} = 0.20\text{ mol}

.

Determines the mole quantity of reactant available.

3

Determine the volume of

N_2O

produced at STP

Moles of

N_2O = 0.20\text{ mol}

. Volume at STP =

0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

.

Applies standard molar volume of gas at STP.

4

Compare the rates of diffusion of

N_2O

and

CO_2

using Graham's law

Molar mass of

N_2O = 2(14) + 16 = 44\text{ g/mol}

. Molar mass of

CO_2 = 12 + 2(16) = 44\text{ g/mol}

. Ratio of diffusion rates =

\sqrt{\frac{44}{44}} = 1

.

Gases with equal molar masses diffuse at equal rates under identical conditions.

Key Concept

Thermal decomposition of nitrogen oxides, molar volume at STP, and Graham's law of diffusion

Estimated Time:2m 0s

Question 64Question →

A $50.0\text{ cm}^3$ gaseous mixture consisting of equal volumes of carbon(II) oxide and carbon(IV) oxide is passed through an excess solution of concentrated potassium hydroxide. What volume of gas remains unabsorbed, and which oxide of carbon was removed from the mixture?

25.0\text{ cm}^3

, with carbon(IV) oxide being removed

25.0\text{ cm}^3

, with carbon(II) oxide being removed

50.0\text{ cm}^3

, with neither gas being removed

0.0\text{ cm}^3

, with both gases being removed

Show answer & explanation

Answer: $25.0\text{ cm}^3$ , with carbon(IV) oxide being removed

Answer

25.0\text{ cm}^3

of gas remains unabsorbed, with carbon(IV) oxide being the component removed by potassium hydroxide solution.

Carbon(IV) oxide (

\text{CO}_2

) is an acidic oxide that dissolves and reacts with concentrated alkali solutions such as potassium hydroxide to form potassium trioxocarbonate(IV) and water. Since the mixture contains

25.0\text{ cm}^3

of

\text{CO}_2

and

25.0\text{ cm}^3

of neutral carbon(II) oxide (

\text{CO}

), only

\text{CO}_2

is absorbed, leaving

25.0\text{ cm}^3

of

\text{CO}

gas remaining.

Step-by-Step Solution

1

Determine the initial volume of each gas component in the mixture.

The mixture has

50.0\text{ cm}^3

total volume with equal proportions, giving

25.0\text{ cm}^3

of

\text{CO}

and

25.0\text{ cm}^3

of

\text{CO}_2

.

Equal volume proportions mean each gas makes up half of the total volume.

2

Analyze the reaction of each gas with concentrated potassium hydroxide (

\text{KOH}

) solution.

\text{CO}_2

reacts via

2\text{KOH(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{K}_2\text{CO}_3\text{(aq)} + \text{H}_2\text{O(l)}

, absorbing all

25.0\text{ cm}^3

of

\text{CO}_2

.

\text{CO}

does not react.

\text{CO}_2

is an acidic oxide which neutralizes alkali solutions, whereas

\text{CO}

is a neutral oxide.

3

Calculate the unabsorbed volume of gas.

25.0\text{ cm}^3

of

\text{CO}

remains as unabsorbed gas.

Subtracting the absorbed volume of

\text{CO}_2

(

25.0\text{ cm}^3

) from total volume (

50.0\text{ cm}^3

) leaves

25.0\text{ cm}^3

.

Key Concept

Chemical Behavior of Oxides of Carbon toward Alkalis

Question 65Question →

Arrange the following sequential stages involved in the industrial isolation of pure nitrogen gas from atmospheric air via fractional distillation in the correct chronological order, from ambient air intake to nitrogen gas collection.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological order of the process is: 1. Removal of dust, moisture, and carbon(IV) oxide; 2. Compression and Joule-Thomson expansion to produce liquid air; 3. Feeding liquid air into a fractionating column and gradual warming; 4. Selective vaporization and collection of nitrogen gas at -196 °C.

The industrial preparation of nitrogen gas relies on fractional distillation of liquid air. Ambient air must first be purified of water vapor and carbon(IV) oxide to prevent cryogenic equipment blockages caused by solid ice formation. The clean air is compressed under high pressure (around 200 atmospheres) and subjected to rapid Joule-Thomson expansion, which cools it repeatedly until it condenses into liquid air at approximately

-200\text{ }^\circ\text{C}

. When liquid air is fed into a fractionating column and warmed gradually, nitrogen gas boils off first at

-196\text{ }^\circ\text{C}

(

77\text{ K}

) due to having a lower boiling point than argon (

-186\text{ }^\circ\text{C}

) and oxygen (

-183\text{ }^\circ\text{C}

).

Step-by-Step Solution

1

Identify the initial purification stage required before gas liquefaction.

Air is scrubbed to remove dust, water vapor, and

CO_2

.

Freezing points of water (

0\text{ }^\circ\text{C}

) and carbon(IV) oxide (

-78.5\text{ }^\circ\text{C}

) are much higher than liquefaction temperature, meaning they would solidify and clog cryogenic tubes if not removed first.

2

Determine the phase change process of purified gaseous air into liquid air.

Purified air is compressed to

\sim 200\text{ atm}

and allowed to expand through a fine nozzle.

The Joule-Thomson expansion causes progressive cooling until air liquefies around

-200\text{ }^\circ\text{C}

.

3

Analyze the fractionating column feed and thermal gradient.

Liquid air enters the fractionating column and is warmed slowly.

Gradual heating drives components with lower boiling points to vaporize first.

4

Compare boiling points to establish which component distills first.

Nitrogen boils off at

-196\text{ }^\circ\text{C}

, followed by argon (

-186\text{ }^\circ\text{C}

) and oxygen (

-183\text{ }^\circ\text{C}

).

Nitrogen has the lowest boiling point, so it boils off first as a gas at the top of the column.

Key Concept

Fractional Distillation of Liquid Air for Industrial Production of Nitrogen

Question 66Question →

Under identical conditions of temperature and pressure, a neutral oxide of nitrogen diffuses $1.21$ times faster than carbon(IV) oxide ( $CO_2$ ). Which of the following is a characteristic chemical property of this oxide of nitrogen? ( $\text{Relative atomic masses: } C = 12, N = 14, O = 16$ )

It reacts rapidly with atmospheric oxygen at room temperature to form reddish-brown fumes.

It relights a glowing wooden splint almost as vigorously as oxygen gas.

It dissolves readily in water at STP to produce an acidic solution of trioxonitrate(V) acid.

It readily condenses near room temperature into a pale yellow liquid undergoing reversible dimerization.

Show answer & explanation

Answer: It reacts rapidly with atmospheric oxygen at room temperature to form reddish-brown fumes.

Answer

The gas reacts rapidly with atmospheric oxygen at room temperature to form reddish-brown fumes.

Using Graham's law of diffusion, the molar mass of the unknown gas is calculated as

\frac{44}{1.21^2} \approx 30.0\text{ g mol}^{-1}

, which corresponds uniquely to nitrogen(II) oxide (

NO

). A defining chemical test for

NO

is its immediate oxidation by air to give reddish-brown fumes of nitrogen(IV) oxide (

NO_2

).

Step-by-Step Solution

1

Calculate the molar mass of carbon(IV) oxide (

CO_2

).

M(CO_2) = 12 + 2(16) = 44\text{ g mol}^{-1}

.

Graham's law requires the molar mass of the reference gas.

2

Apply Graham's law of diffusion to determine the molar mass of the unknown nitrogen oxide (

M_x

).

\frac{R_x}{R_{CO_2}} = \sqrt{\frac{M(CO_2)}{M_x}} \implies 1.21 = \sqrt{\frac{44}{M_x}} \implies (1.21)^2 = \frac{44}{M_x} \implies M_x = \frac{44}{1.4641} \approx 30.0\text{ g mol}^{-1}

.

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

3

Identify the formula of the nitrogen oxide with a molar mass of

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

.

For nitrogen(II) oxide (

NO

),

M(NO) = 14 + 16 = 30\text{ g mol}^{-1}

.

Matching the calculated molar mass to known oxides of nitrogen (

N_2O = 44

,

NO = 30

,

NO_2 = 46

).

4

Determine the key chemical property of nitrogen(II) oxide (

NO

).

Nitrogen(II) oxide (

NO

) is a colorless neutral gas that reacts spontaneously with oxygen in air to form brown nitrogen(IV) oxide (

2NO + O_2 \rightarrow 2NO_2

).

To select the correct property corresponding to

NO

.

Key Concept

Graham's Law of Diffusion and Chemical Properties of Oxides of Nitrogen

Estimated Time:2m 0s

Question 67Question →

During the industrial separation of liquefied air to obtain pure nitrogen gas, nitrogen vaporizes at $-196^\circ\text{C}$ while oxygen remains in the liquid state at $-183^\circ\text{C}$ . Which of the following statements correctly explains why nitrogen distills over first?

Nitrogen has a lower boiling point than oxygen, so it vaporizes at a lower temperature.

Nitrogen has a higher boiling point than oxygen, causing it to evaporate before liquid oxygen.

Nitrogen diffuses faster than oxygen because it is less dense, forcing it out of the fractionating column first.

Nitrogen's boiling point appears lower only because negative Celsius temperatures convert inversely to the Kelvin scale.

Show answer & explanation

Answer: Nitrogen has a lower boiling point than oxygen, so it vaporizes at a lower temperature.

Answer

Nitrogen has a lower boiling point than oxygen, so it vaporizes at a lower temperature.

Nitrogen has a lower boiling point (

-196^\circ\text{C}

or

77\text{ K}

) than oxygen (

-183^\circ\text{C}

or

90\text{ K}

). In fractional distillation, liquid mixtures are separated based on differences in boiling points, and the component with the lower boiling point boils off and distills over first.

Step-by-Step Solution

1

Compare the boiling point values on the Celsius and Kelvin temperature scales.

Nitrogen boils at

-196^\circ\text{C}

(

77\text{ K}

) while oxygen boils at

-183^\circ\text{C}

(

90\text{ K}

).

On the Celsius scale,

-196

is smaller (lower) than

-183

.

2

Apply the physical principle of fractional distillation.

As liquefied air warms, nitrogen reaches its boiling point first and converts into gas.

The component with the lower boiling point always distills over first during fractional distillation.

Key Concept

Fractional Distillation of Liquid Air

Question 68Question →

Under identical conditions of temperature and pressure, a $100\text{ cm}^3$ sample of nitrogen(II) oxide ( $NO$ ) diffuses through a porous container in $20\text{ seconds}$ . If $100\text{ cm}^3$ of an unknown oxide of nitrogen diffuses through the exact same porous container in $35\text{ seconds}$ , what is the molecular formula of the unknown oxide of nitrogen? $[N = 14, O = 16]$

N_2O_4

NO_2

N_2O

N_2O_5

Show answer & explanation

Answer: $N_2O_4$

Answer

The molecular formula of the unknown oxide is

N_2O_4

.

The correct answer is derived using Graham's Law of Diffusion (

\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

). Given that nitrogen(II) oxide (

NO

) has a molar mass of

30\text{ g/mol}

and diffuses in

20\text{ s}

, an oxide diffusing in

35\text{ s}

must have a molar mass of

30 \times (35/20)^2 = 91.875 \approx 92\text{ g/mol}

. The formula matching this molar mass is dinitrogen tetroxide (

N_2O_4

).

Step-by-Step Solution

1

Calculate the molar mass of the reference gas, nitrogen(II) oxide (

NO

).

M_1(NO) = 14 + 16 = 30\text{ g/mol}

.

Graham's law requires the molar mass of the reference gas to determine the unknown gas mass.

2

Apply Graham's Law of Diffusion in terms of time taken for equal volumes of gas to diffuse.

\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

, where

t_1 = 20\text{ s}

,

t_2 = 35\text{ s}

, and

M_1 = 30\text{ g/mol}

.

Rate of diffusion is inversely proportional to diffusion time for fixed volume, so

\frac{\text{Rate}_1}{\text{Rate}_2} = \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

.

3

Solve for the unknown molar mass

M_2

.

\frac{35}{20} = 1.75 \implies (1.75)^2 = \frac{M_2}{30} \implies 3.0625 = \frac{M_2}{30} \implies M_2 = 91.875 \approx 92\text{ g/mol}

.

Squaring the time ratio isolates the molar mass ratio.

4

Identify the formula of the nitrogen oxide with a molar mass of

92\text{ g/mol}

.

N_2O_4

:

2(14) + 4(16) = 28 + 64 = 92\text{ g/mol}

.

Matching the calculated molar mass to the chemical formula of oxides of nitrogen.

Key Concept

Graham's Law of Diffusion applied to Oxides of Nitrogen

Estimated Time:2m 0s

Question 69Question →

Arrange the following sequential transformations of nitrogen in the biological nitrogen cycle, beginning with atmospheric molecular nitrogen ( $N_2$ ) and concluding with its re-release into the atmosphere:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of nitrogen cycle transformations is: Nitrogen fixation (

N_2 \rightarrow NH_4^+

), Nitrosification (

NH_4^+ \rightarrow NO_2^-

), Nitration (

NO_2^- \rightarrow NO_3^-

), Assimilation (

NO_3^- \rightarrow \text{organic plant proteins}

), and Denitrification (

NO_3^- \rightarrow N_2

).

The biological nitrogen cycle proceeds in a specific logical sequence. First, atmospheric nitrogen gas (

N_2

) must undergo nitrogen fixation to become ammonium ions (

NH_4^+

). Next, nitrification occurs in two steps: nitrosification converts ammonium (

NH_4^+

) to nitrite (

NO_2^-

) via *Nitrosomonas*, followed by nitration converting nitrite (

NO_2^-

) to nitrate (

NO_3^-

) via *Nitrobacter*. Plants then assimilate these nitrate ions (

NO_3^-

) into organic plant proteins. Finally, denitrifying bacteria reduce residual soil nitrates (

NO_3^-

) back into atmospheric nitrogen gas (

N_2

) under anaerobic conditions.

Step-by-Step Solution

1

Identify the initial entry point of atmospheric nitrogen into the biosphere.

Gaseous nitrogen (

N_2

) is fixed into ammonium ions (

NH_4^+

) by nitrogen-fixing bacteria (e.g., *Rhizobium*, *Azotobacter*).

Atmospheric

N_2

has a strong triple covalent bond (

N \equiv N

) and cannot be directly utilized by higher plants without preliminary biological fixation.

2

Determine the initial oxidation stage of nitrification (nitrosification).

Ammonium ions (

NH_4^+

) are oxidized to nitrite ions (

NO_2^-

) by *Nitrosomonas*.

Nitrification occurs in two distinct microbial steps, starting with ammonium conversion to nitrite.

3

Determine the second oxidation stage of nitrification (nitration).

Nitrite ions (

NO_2^-

) are oxidized to nitrate ions (

NO_3^-

) by *Nitrobacter*.

Nitrate (

NO_3^-

) is the most readily absorbed and utilized form of inorganic nitrogen for plants.

4

Trace the biological uptake of bioavailable soil nitrogen.

Plants absorb

NO_3^-

ions and assimilate them into plant proteins and nucleic acids.

Inorganic nitrate is reduced inside plant tissues and converted into organic amino acids.

5

Identify the pathway responsible for returning gaseous nitrogen to the atmosphere.

Denitrifying bacteria (e.g., *Pseudomonas denitrificans*) convert unabsorbed soil nitrates (

NO_3^-

) into atmospheric

N_2

gas under anaerobic conditions.

Denitrification closes the global nitrogen loop by replenishing free atmospheric

N_2

.

Key Concept

Nitrogen Cycle Transformations and Microorganisms

Estimated Time:2m 0s

Question 70Question →

Complete the following passage regarding the redox behavior of hydrogen sulfide gas when reacted with an iron(III) compound by filling in the blanks with the correct chemical terms.

Fill in the blanks below

When hydrogen sulfide gas is bubbled through an acidified solution of iron(III) chloride, the characteristic yellow color of the solution changes to due to the reduction of iron(III) ions to iron(II) ions, while a fine precipitate of elemental sulfur is deposited.

Show answer & explanation

Answer

The solution turns pale green (or green) due to the formation of iron(II) ions (

Fe^{2+}

), accompanied by the deposition of a yellow precipitate of elemental sulfur (

S

).

Hydrogen sulfide gas (

H_2S

) acts as a strong reducing agent. When passed into an acidified solution of iron(III) chloride (

FeCl_3

), it reduces yellow

Fe^{3+}

ions to light green

Fe^{2+}

ions, while

H_2S

itself gets oxidized to form a yellow precipitate of elemental sulfur (

S

). The overall ionic equation for the reaction is:

2Fe^{3+}_{(aq)} + H_2S_{(g)} \rightarrow 2Fe^{2+}_{(aq)} + 2H^+_{(aq)} + S_{(s)}

.

Step-by-Step Solution

1

Identify the role of hydrogen sulfide (

H_2S

) in the reaction.

H_2S

acts as a reducing agent in acidic solution.

Sulfur in

H_2S

has an oxidation state of

-2

and undergoes oxidation to elemental sulfur (

0

oxidation state).

2

Determine the change in oxidation state of iron.

Iron(III) ions (

Fe^{3+}

) are reduced to iron(II) ions (

Fe^{2+}

).

Iron(III) chloride solution is yellow/brown due to

Fe^{3+}_{(aq)}

, whereas iron(II) ions (

Fe^{2+}_{(aq)}

) impart a light green/pale green color to the aqueous solution.

3

Identify the physical appearance of oxidized sulfur product.

Elemental sulfur precipitates out of the solution.

Insoluble sulfur particles form a yellow suspension or precipitate in the solution.

Key Concept

Reducing property of Hydrogen Sulfide (

H_2S

)

Question 71Question →

Rhombic sulfur ( $\alpha$ -sulfur) and monoclinic sulfur ( $\beta$ -sulfur) are two well-known crystalline allotropes of sulfur. Which of the following statements correctly describes the stability and thermal transformation relationship between these two allotropes at standard atmospheric pressure?

Above

96^\circ\text{C}

, monoclinic sulfur is the stable crystalline form, whereas below

96^\circ\text{C}

, rhombic sulfur is the stable crystalline form.

Above

96^\circ\text{C}

, rhombic sulfur is the stable crystalline form, whereas below

96^\circ\text{C}

, monoclinic sulfur is the stable crystalline form.

Rhombic sulfur and monoclinic sulfur are chemical isomers that possess identical crystal lattices but different chemical formulas.

Monoclinic sulfur is an amorphous allotrope obtained by pouring boiling sulfur into cold water.

Show answer & explanation

Answer: Above $96^\circ\text{C}$ , monoclinic sulfur is the stable crystalline form, whereas below $96^\circ\text{C}$ , rhombic sulfur is the stable crystalline form.

Answer

Above

96^\circ\text{C}

, monoclinic sulfur is the stable crystalline form, whereas below

96^\circ\text{C}

, rhombic sulfur is the stable crystalline form.

The transition temperature for the two main crystalline allotropes of sulfur is

96^\circ\text{C}

. Below this temperature, rhombic sulfur (

\alpha

-sulfur) is the thermodynamically stable form. When heated above

96^\circ\text{C}

, rhombic sulfur slowly transforms into monoclinic sulfur (

\beta

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

119^\circ\text{C}

.

Step-by-Step Solution

1

Identify the nature of the allotropes of sulfur.

Both rhombic sulfur (

\alpha

-sulfur) and monoclinic sulfur (

\beta

-sulfur) are crystalline allotropes made up of crown-shaped

S_8

molecules.

Allotropes are different structural forms of the same element in the same physical state.

2

Determine the transition temperature between rhombic and monoclinic sulfur.

The transition temperature is

96^\circ\text{C}

. At this temperature, both allotropes coexist in equilibrium.

Temperature determines which crystal lattice structure is thermodynamically stable.

3

Evaluate stability ranges above and below the transition temperature.

Rhombic sulfur is stable below

96^\circ\text{C}

, while monoclinic sulfur is stable between

96^\circ\text{C}

and its melting point (

119^\circ\text{C}

).

Heating rhombic sulfur above

96^\circ\text{C}

slowly converts it to monoclinic sulfur needles.

Key Concept

Transition temperature and allotropy of crystalline sulfur

Estimated Time:1m 0s

Question 72Question →

A sample of $9.0\text{ g}$ of anhydrous ethanedioic acid, $\text{H}_2\text{C}_2\text{O}_4$ , is completely dehydrated using excess concentrated tetraoxosulfate(VI) acid. The evolved gaseous mixture is passed through an excess concentrated sodium hydroxide solution to absorb carbon(IV) oxide. What volume of pure carbon(II) oxide gas is collected at STP? [Molar mass of $\text{H}_2\text{C}_2\text{O}_4 = 90\text{ g mol}^{-1}$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{ mol}^{-1}$ ]

2.24\text{ dm}^3

2.40\text{ dm}^3

4.48\text{ dm}^3

4.80\text{ dm}^3

Show answer & explanation

Answer: $2.24\text{ dm}^3$

Answer

The volume of pure carbon(II) oxide collected at STP is

2.24\text{ dm}^3

.

Dehydration of

9.0\text{ g}

(

0.10\text{ mol}

) of ethanedioic acid produces

0.10\text{ mol}

of

\text{CO}

and

0.10\text{ mol}

of

\text{CO}_2

. Concentrated sodium hydroxide absorbs all the

\text{CO}_2

, leaving only

0.10\text{ mol}

of neutral

\text{CO}

gas. At STP (

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

),

0.10\text{ mol}

of

\text{CO}

occupies

2.24\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the number of moles of ethanedioic acid (

\text{H}_2\text{C}_2\text{O}_4

).

\text{Moles} = \frac{9.0\text{ g}}{90\text{ g mol}^{-1}} = 0.10\text{ mol}

.

Determining initial mole quantity from given mass and molar mass.

2

Write the balanced chemical equation for the dehydration of ethanedioic acid.

\text{H}_2\text{C}_2\text{O}_4(\text{s}) \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4} \text{CO}(\text{g}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

.

1 mole of ethanedioic acid yields 1 mole of

\text{CO}

gas and 1 mole of

\text{CO}_2

gas.

3

Determine the mole amount of carbon(II) oxide collected after passing through sodium hydroxide.

Sodium hydroxide reacts with acidic oxide

\text{CO}_2

to form

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

, leaving

0.10\text{ mol}

of unreacted neutral

\text{CO}

gas to be collected.

Sodium hydroxide selectively absorbs carbon(IV) oxide, allowing pure carbon(II) oxide gas to pass through.

4

Calculate the volume of pure

\text{CO}

gas at STP.

\text{Volume} = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3$.

Converting moles of gas to volume at standard temperature and pressure.

Key Concept

Dehydration of ethanedioic acid yields an equimolar mixture of carbon(II) oxide and carbon(IV) oxide. Sodium hydroxide selectively absorbs carbon(IV) oxide, allowing pure carbon(II) oxide to be collected.

Question 73Question →

When excess hydrogen sulfide gas (

H_2S

) reacts with sulfur(IV) oxide (

SO_2

) according to the balanced chemical equation:

2H_2S(g) + SO_2(g) \rightarrow 3S(s) + 2H_2O(l)

What mass of elemental sulfur is precipitated when

5.6\text{ dm}^3

of

SO_2

gas at STP reacts completely?
(Molar volume of gas at STP =

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

; Relative atomic mass:

S = 32

)

24.0\text{ g}

8.0\text{ g}

22.4\text{ g}

16.0\text{ g}

Show answer & explanation

Answer: $24.0\text{ g}$

Answer

The mass of elemental sulfur precipitated is

24.0\text{ g}

.

First, find the moles of

SO_2

at STP:

\frac{5.6\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.25\text{ mol}

. According to the balanced equation,

1\text{ mole of } SO_2

produces

3\text{ moles of } S

. Thus,

0.25\text{ mol of } SO_2

produces

0.75\text{ mol of } S

. Multiplying by the relative atomic mass of sulfur (

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

) gives

0.75 \times 32 = 24.0\text{ g}

.

Step-by-Step Solution

1

Calculate the number of moles of

SO_2

gas reacting at STP.

\text{Moles of } SO_2 = \frac{\text{Volume at STP}}{\text{Molar volume at STP}} = \frac{5.6\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.25\text{ mol}

.

Gas volume at STP is converted to moles using the standard molar volume of

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

.

2

Determine the mole ratio between

SO_2

and

S

from the balanced reaction equation.

From

2H_2S(g) + SO_2(g) \rightarrow 3S(s) + 2H_2O(l)

,

1\text{ mol of } SO_2

yields

3\text{ mol of } S

.

Stoichiometric coefficients give the exact molar equivalence between reactants and products.

3

Calculate the moles of

S

formed.

\text{Moles of } S = 0.25\text{ mol } SO_2 \times 3 = 0.75\text{ mol } S.

Multiplying moles of

SO_2

by the stoichiometric factor of 3 gives the moles of produced sulfur.

4

Convert the moles of sulfur to mass in grams.

\text{Mass of } S = 0.75\text{ mol} \times 32\text{ g mol}^{-1} = 24.0\text{ g}.

Mass is obtained by multiplying number of moles by relative atomic mass.

Key Concept

Gas stoichiometry at STP and redox reaction between hydrogen sulfide and sulfur(IV) oxide

Estimated Time:1m 30s

Question 74Question →

During the industrial production of water gas, steam is blown over red-hot coke at elevated temperatures ( $1000^\circ\text{C}$ ). Why must the steam supply be periodically interrupted to blow air over the coke bed instead?

The reaction between steam and coke is endothermic, which continuously cools the coke bed below the required reaction temperature.

Steam acts as a homogeneous catalyst that shifts the position of dynamic equilibrium back toward reactants.

Air is required to convert amorphous carbon into graphite isomers before steam can react.

The gaseous water gas product expands beyond the molar volume of 22.4 dm³ per mole, stopping the flow of steam.

Show answer & explanation

Answer: The reaction between steam and coke is endothermic, which continuously cools the coke bed below the required reaction temperature.

Answer

The steam supply must be interrupted because the reaction between steam and red-hot coke (

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

) is strongly endothermic. This lowers the temperature of the coke bed below

1000^\circ\text{C}

, requiring air to be blown in so that carbon burns exothermically in oxygen (

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

) to reheat the bed.

The reaction

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

is endothermic, absorbing heat from the red-hot coke. To maintain the high temperatures necessary for continuous reaction, the steam blow is periodically stopped and an air blow is introduced to exothermically burn some coke and reheat the bed.

Step-by-Step Solution

1

Analyze the thermochemical nature of the water gas reaction.

The formation of water gas (

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

) from coke and steam absorbs heat (

\Delta H > 0

).

Endothermic reactions remove heat from the reaction vessel, causing the solid coke bed temperature to drop.

2

Determine the industrial necessity of alternating steam and air runs.

Blowing air causes exothermic combustion of carbon, restoring the red-hot temperature necessary for water gas synthesis.

High temperatures (

1000^\circ\text{C}

) are required to maintain high reaction rates and favor product yield.

Key Concept

Industrial production and energetics of water gas synthesis from coal/coke

Estimated Time:1m 0s

Question 75Question →

When

6.62\text{ g}

of lead(II) trioxonitrate(V),

\text{Pb(NO}_3)_2

, is heated strongly until complete decomposition occurs according to the equation:

2\text{Pb(NO}_3)_2(s) \rightarrow 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)

What is the total volume of gaseous products evolved at standard temperature and pressure (STP)?
[

M_{\text{r}}(\text{Pb(NO}_3)_2) = 331\text{ g mol}^{-1}

; Molar volume of gas at STP

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

]

1.12\text{ dm}^3

1.20\text{ dm}^3

0.896\text{ dm}^3

0.448\text{ dm}^3

Show answer & explanation

Answer: $1.12\text{ dm}^3$

Answer

The total volume of gaseous products evolved at STP is

1.12\text{ dm}^3

.

Decomposition of

0.02\text{ mol}

of lead(II) trioxonitrate(V) yields

0.04\text{ mol}

of

\text{NO}_2

and

0.01\text{ mol}

of

\text{O}_2

, making

0.05\text{ mol}

of total gas. At STP,

0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the amount in moles of

\text{Pb(NO}_3)_2

decomposed

\text{Moles of Pb(NO}_3)_2 = \frac{6.62\text{ g}}{331\text{ g mol}^{-1}} = 0.02\text{ mol}

Dividing given mass by relative formula mass gives the mole amount.

2

Determine total moles of gaseous products using stoichiometric ratios

From the balanced equation,

2\text{ mol of Pb(NO}_3)_2

yields

4\text{ mol of NO}_2(g) + 1\text{ mol of O}_2(g) = 5\text{ mol of gas}

. Total gas moles

= 0.02 \times \frac{5}{2} = 0.05\text{ mol}

.

Both

\text{NO}_2

and

\text{O}_2

are gases at STP, so their mole quantities must be summed.

3

Calculate total gas volume at STP

\text{Volume}

= 0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

Multiplying total gaseous moles by the molar volume at STP (

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

) gives the required volume.

Key Concept

Thermal decomposition of metallic trioxonitrate(V) salts and gas stoichiometry at STP

Estimated Time:1m 30s

Question 76Question →

A sample of $4.80\text{ g}$ of pure rhombic sulfur is completely burned in excess oxygen gas at room temperature and pressure (RTP). What is the volume of sulfur(IV) oxide gas liberated at RTP, and what is the change in the oxidation state of sulfur during this combustion process?

(Relative atomic mass: $S = 32.0$ ; Molar volume of gas at RTP = $24.0\text{ dm}^3\text{ mol}^{-1}$ )

3.60\text{ dm}^3

and an increase from 0 to

+4

3.36\text{ dm}^3

and an increase from 0 to

+4

3.60\text{ dm}^3

and an increase from 0 to

+6

1.80\text{ dm}^3

and an increase from

-2

to

+4

Show answer & explanation

Answer: $3.60\text{ dm}^3$ and an increase from 0 to $+4$

Answer

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

3.60\text{ dm}^3

, and the oxidation state of sulfur increases from 0 to

+4

.

Burning

4.80\text{ g}

(

0.15\text{ mol}

) of sulfur produces

0.15\text{ mol}

of

SO_2

gas. Multiplying by the molar volume at RTP (

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

) gives

3.60\text{ dm}^3

. In elemental rhombic sulfur, sulfur has an oxidation state of 0, which increases to

+4

in

SO_2

.

Step-by-Step Solution

1

Calculate the amount of sulfur reacted in moles.

\text{Moles of } S = \frac{4.80\text{ g}}{32.0\text{ g mol}^{-1}} = 0.15\text{ mol}

Dividing the mass of sulfur by its relative atomic mass gives the mole amount.

2

Write the balanced chemical equation and determine the mole ratio.

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

(Mole ratio of

S : SO_2 = 1 : 1

, so

0.15\text{ mol}

of

SO_2

is formed).

Direct combustion of sulfur yields sulfur(IV) oxide.

3

Calculate the volume of

SO_2

gas produced at RTP.

\text{Volume} = 0.15\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 3.60\text{ dm}^3

At room temperature and pressure (RTP),

1\text{ mol}

of any gas occupies

24.0\text{ dm}^3

.

4

Determine the change in oxidation state of sulfur.

Elemental sulfur

S(s)

has an oxidation number of 0. In

SO_2

, oxygen is

-2

, so

x + 2(-2) = 0 \implies x = +4

. The change is from 0 to

+4

.

Free uncombined elements have an oxidation state of 0.

Key Concept

Stoichiometry of gas liberation at RTP and oxidation state changes during non-metal combustion

Question 77Question →

Graphite and diamond are two crystalline allotropes of carbon that exhibit vastly different physical properties. Which of the following statements correctly accounts for why graphite conducts electricity whereas diamond is an electrical insulator?

Graphite consists of carbon atoms that are structurally isomeric to diamond, allowing free molecules to move through the crystal layers.

Each carbon atom in graphite forms three covalent bonds using

sp^2

hybridization, leaving one delocalized electron per atom free to move along the hexagonal layers.

Diamond is composed of heavier isotopes of carbon whose extra neutrons trap valence electrons within the atomic nucleus.

Graphite contains mobile carbon ions that dissociate under an applied electric field, whereas diamond contains neutral molecules.

Show answer & explanation

Answer: Each carbon atom in graphite forms three covalent bonds using $sp^2$ hybridization, leaving one delocalized electron per atom free to move along the hexagonal layers.

Answer

Each carbon atom in graphite forms three covalent bonds using

sp^2

hybridization, leaving one delocalized electron per atom free to move along the hexagonal layers.

In graphite, each carbon atom forms three covalent

\sigma

-bonds via

sp^2

hybridization. The fourth valence electron resides in an unhybridized

p

-orbital and becomes delocalized across the hexagonal layer, allowing electrical current to flow. In contrast, diamond features

sp^3

hybridization where all four valence electrons are tightly bound in localized covalent bonds, leaving no mobile electrons.

Step-by-Step Solution

1

Analyze the bonding and hybridization of carbon in diamond.

In diamond, each carbon atom undergoes

sp^3

hybridization and forms four strong covalent bonds directed tetrahedrally toward adjacent carbon atoms.

All four valence electrons per carbon atom are involved in localized single covalent bonds, leaving no free electrons to conduct electricity.

2

Analyze the bonding and hybridization of carbon in graphite.

In graphite, each carbon atom undergoes

sp^2

hybridization to form three strong covalent bonds within a hexagonal planar sheet.

This leaves one unhybridized

p

-orbital electron per carbon atom, which forms a delocalized

\pi

-electron system capable of moving freely along the layers.

3

Correlate structural features to electrical conductivity.

The presence of delocalized valence electrons in graphite enables electrical conduction, whereas the localized

sp^3

bonding in diamond makes it an insulator.

Electrical conduction in solid non-metallic elements requires mobile charge carriers such as delocalized electrons.

Key Concept

Structural bonding and hybridization of carbon allotropes (Graphite vs Diamond)

Estimated Time:1m 0s

Question 78Question →

Match each sulfur-related compound or allotrope on the left with its correct physical characteristic, qualitative test, or stability range on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Rhombic sulfur matches the octahedral allotrope stable up to 96°C; Hydrogen sulfide matches the rotten-egg smelling gas turning lead(II) ethanoate paper black; Sulfur(IV) oxide matches the choking gas turning acidified potassium dichromate(VI) green; Monoclinic sulfur matches the needle-shaped allotrope stable between 96°C and 119°C.

Rhombic sulfur (

\alpha

-sulfur) is octahedral and stable below

96^\circ\text{C}

. Monoclinic sulfur (

\beta

-sulfur) is needle-shaped and stable between

96^\circ\text{C}

and

119^\circ\text{C}

. Hydrogen sulfide (

H_2S

) is a gas with a rotten-egg odor that reacts with

Pb^{2+}

to form black insoluble

PbS

. Sulfur(IV) oxide (

SO_2

) has a pungent choking smell and reduces orange dichromate(VI) solutions to green

Cr^{3+}

ions.

Step-by-Step Solution

1

Analyze the crystal structures and thermal stability range of sulfur allotropes.

Rhombic sulfur is octahedral and stable below 96°C. Monoclinic sulfur is needle-shaped and stable between 96°C and 119°C.

96°C is the transition temperature at which the two crystalline allotropes exist in equilibrium.

2

Identify the qualitative gas tests and chemical properties of

H_2S

and

SO_2

.

H_2S

forms black

PbS

precipitate with lead(II) ethanoate paper.

SO_2

reduces orange dichromate(VI) to green

Cr^{3+}

ions.

Both gases are reducing agents, but

H_2S

forms insoluble sulfides while

SO_2

undergoes specific color-change redox reactions with dichromate ions.

Key Concept

Physical and chemical properties of sulfur allotropes, hydrogen sulfide, and sulfur(IV) oxide

Question 79Question →

In the Contact Process for the industrial manufacture of tetraoxosulfate(VI) acid, sulfur(IV) oxide gas reacts with oxygen gas according to the equation: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$ . What volume of oxygen gas, measured in $\text{dm}^3$ at stp, is required to completely react with $56\text{ dm}^3$ of $SO_2$ gas at stp?

Show answer & explanation

Answer: 28

Answer

The volume of oxygen gas required at stp is 28 dm³.

According to the balanced chemical equation

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

, 2 volumes of

SO_2

gas require 1 volume of

O_2

gas for complete oxidation. Therefore,

56\text{ dm}^3

of

SO_2

requires half its volume in oxygen, which equals

28\text{ dm}^3

.

Step-by-Step Solution

1

Determine the stoichiometric ratio between

SO_2

and

O_2

from the balanced equation.

2 volumes of

SO_2(g)

react with 1 volume of

O_2(g)

.

By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number volume ratios under the same conditions of temperature and pressure.

2

Compute the required volume of

O_2

gas for

56\text{ dm}^3

of

SO_2

.

\text{Volume of } O_2 = \frac{56\text{ dm}^3}{2} = 28\text{ dm}^3

.

Since the ratio of

SO_2

to

O_2

is

2:1

, the volume of oxygen gas required is half the volume of sulfur(IV) oxide gas.

Key Concept

Stoichiometric Volume Calculations in Gas Reactions (Contact Process)

Question 80Question →

When excess dry ammonia gas is passed over

24.0\text{ g}

of heated copper(II) oxide (

CuO

), the oxide is completely reduced to copper metal according to the equation:

2NH_3(g) + 3CuO(s) \rightarrow 3Cu(s) + N_2(g) + 3H_2O(l)

What is the volume of nitrogen gas, in

\text{dm}^3

, evolved at s.t.p.?
(

Cu = 64.0

,

O = 16.0

, molar volume of gas at s.t.p.

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

)

Show answer & explanation

Answer: 2.24

Answer

The volume of nitrogen gas evolved at s.t.p. is

2.24\text{ dm}^3

.

According to the balanced chemical equation

2NH_3(g) + 3CuO(s) \rightarrow 3Cu(s) + N_2(g) + 3H_2O(l)

, 3 moles (

240.0\text{ g}

) of copper(II) oxide produce 1 mole (

22.4\text{ dm}^3

at s.t.p.) of nitrogen gas. Hence,

24.0\text{ g}

of

CuO

yields

\frac{24.0}{240.0} \times 22.4\text{ dm}^3 = 2.24\text{ dm}^3

of nitrogen gas.

Step-by-Step Solution

1

Calculate the molar mass of

CuO

Molar mass of

CuO = 64.0 + 16.0 = 80.0\text{ g mol}^{-1}

Required to convert the given mass of reactant to moles.

2

Find the number of moles of

CuO

Moles of

CuO = \frac{24.0\text{ g}}{80.0\text{ g mol}^{-1}} = 0.30\text{ mol}

Determines the exact amount of copper(II) oxide reacting.

3

Apply stoichiometry to find moles of

N_2

gas produced

Moles of

N_2 = 0.30\text{ mol } CuO \times \frac{1\text{ mol } N_2}{3\text{ mol } CuO} = 0.10\text{ mol } N_2

The mole ratio between

CuO

and

N_2

in the balanced chemical equation is

3:1

.

4

Convert moles of

N_2

to volume at s.t.p.

Volume of

N_2 = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

at s.t.p.

Key Concept

Reduction of metallic oxides by ammonia gas and gas stoichiometry at STP

Non-Metals and Their Compounds

Answer

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