All practice questions

Question 8201Question →

Match each calcium-based compound or reagent listed on the left with its corresponding industrial process, chemical behavior, or metallurgical function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Calcium fluoride addition matches lowering the electrolyte melting point and enhancing conductivity; Calcium oxide hydration matches forming slaked lime used in water softening; Calcium hydroxide reaction with chlorine matches forming bleaching powder; Plaster of Paris rehydration matches forming dihydrate gypsum with volume expansion.

Each calcium compound or reagent is paired strictly according to its industrial function or chemical reaction behavior as specified in the JAMB UTME syllabus.

Step-by-Step Solution

1

Analyze the metallurgical role of calcium fluoride in calcium extraction

Identified

\text{CaF}_2

as a flux that lowers the melting point of fused

\text{CaCl}_2

from

800^\circ\text{C}

to

600^\circ\text{C}

.

Electrolysis of pure

\text{CaCl}_2

requires high temperature where molten calcium would dissolve in the electrolyte, so

\text{CaF}_2

is added as a flux.

2

Evaluate the chemical properties of calcium oxide and its slaked product

Slaking

\text{CaO}

gives

\text{Ca(OH)}_2

, which precipitates

\text{Ca(HCO}_3)_2

in Clark's method.

Calcium hydroxide reacts with hydrogen carbonate ions to precipitate insoluble

\text{CaCO}_3

, removing temporary hardness.

3

Determine the industrial reaction between slaked lime and chlorine

Chlorination of dry

\text{Ca(OH)}_2

yields bleaching powder,

\text{CaOCl}_2 \cdot \text{H}_2\text{O}

.

This specific gas-solid reaction forms active bleaching agents used in water treatment and industrial oxidation.

4

Examine the hydration chemistry of calcium sulfate hemihydrate

Plaster of Paris absorbs water to form gypsum with a slight increase in solid volume.

The crystallization process of gypsum yields interlocking monoclinic crystals that expand slightly to fill mold details perfectly.

Key Concept

Extraction, reactions, and industrial applications of alkaline earth metal (calcium) compounds.

Question 8202Question →

Complete the following statement describing the bonding and structural change during the dimerization of aluminium chloride.

Fill in the blanks below

In the formation of the dimer

Al_2Cl_6

from monomeric

AlCl_3

, a chlorine atom from one molecule donates a lone pair of electrons to the electron-deficient aluminium atom of another, resulting in a total of coordinate bonds in the dimer. This coordination causes the electron pair geometry around each aluminium atom to become .

Show answer & explanation

Answer

Blank 1: 2 (or two); Blank 2: tetrahedral

In the dimerization of

AlCl_3

into

Al_2Cl_6

, two bridging chlorine atoms act as Lewis bases by donating lone pairs to the electron-deficient aluminium atoms, creating 2 coordinate bonds. Because each aluminium atom now shares 4 electron pairs, the hybridization changes from

sp^2

to

sp^3

, resulting in a tetrahedral geometry around each aluminium atom.

Step-by-Step Solution

1

Analyze the electron configuration and geometry of monomeric

AlCl_3

.

Aluminium has 3 valence electrons and forms 3 single covalent bonds with three chlorine atoms, leaving aluminium with 6 valence electrons (an incomplete octet) and a trigonal planar geometry with

sp^2

hybridization.

Identifying the electron deficiency in monomeric

AlCl_3

explains why it undergoes dimerization.

2

Determine the number of coordinate (dative) bonds formed in the

Al_2Cl_6

dimer.

To complete its octet, each aluminium atom accepts one lone pair of electrons from a chlorine atom belonging to the neighboring

AlCl_3

molecule. Since there are two aluminium atoms being bridged, exactly 2 coordinate covalent bonds are formed in

Al_2Cl_6

.

Two bridging chlorine atoms each contribute a lone pair to form two separate dative bonds.

3

Determine the resulting molecular geometry around each aluminium atom in

Al_2Cl_6

.

Each aluminium atom becomes surrounded by 4 electron pair bonds (3 single covalent bonds + 1 coordinate bond), shifting its hybridization from

sp^2

to

sp^3

and yielding a tetrahedral geometry.

According to VSEPR theory, four bonding pairs arrange themselves to minimize repulsion in a 3D tetrahedral orientation.

Key Concept

Dative Covalent Bonding and Geometric Transition in Dimeric Aluminium Chloride (

Al_2Cl_6

)

Estimated Time:1m 30s

Question 8203Question →

Match each set of thermodynamic conditions for a chemical reaction on the left with its corresponding temperature-dependent spontaneity behavior on the right, based on the Gibbs free energy relationship $\Delta G = \Delta H - T\Delta S$ .

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Endothermic with

\Delta S > 0

matches Spontaneous only at high temperatures (

T > \frac{\Delta H}{\Delta S}

).
2. Exothermic with

\Delta S < 0

matches Spontaneous only at low temperatures (

T < \frac{\Delta H}{\Delta S}

).
3. Endothermic with

\Delta S < 0

matches Non-spontaneous at all temperatures.
4. Exothermic with

\Delta S > 0

matches Spontaneous at all temperatures.

Each pair is matched by evaluating the sign of

\Delta G = \Delta H - T\Delta S

. An endothermic reaction with positive entropy change requires high temperature to make

T\Delta S > \Delta H

. An exothermic reaction with negative entropy change requires low temperature to keep

|\Delta H| > T|\Delta S|

. An endothermic reaction with negative entropy change yields positive

\Delta G

at all temperatures. An exothermic reaction with positive entropy change yields negative

\Delta G

at all temperatures.

Step-by-Step Solution

1

Recall the fundamental thermodynamic criterion for spontaneity

A reaction is spontaneous when Gibbs free energy change is negative (

\Delta G < 0

), given by

\Delta G = \Delta H - T\Delta S

.

Temperature

T

in Kelvin is always positive, so the sign of

\Delta G

depends on the combination of signs of

\Delta H

and

\Delta S

.

2

Analyze Case 1:

\Delta H > 0

and

\Delta S > 0

\Delta G = (+\text{value}) - T(+\text{value})

. For

\Delta G < 0

,

T\Delta S > \Delta H \implies T > \frac{\Delta H}{\Delta S}

.

The reaction is driven by entropy increase at elevated temperatures.

3

Analyze Case 2:

\Delta H < 0

and

\Delta S < 0

\Delta G = (-\text{value}) - T(-\text{value}) = -|\Delta H| + T|\Delta S|

. For

\Delta G < 0

,

|\Delta H| > T|\Delta S| \implies T < \frac{\Delta H}{\Delta S}

.

The reaction is enthalpy-driven and requires low temperatures so that the positive

-T\Delta S

term does not outweigh

\Delta H

.

4

Analyze Case 3:

\Delta H > 0

and

\Delta S < 0

\Delta G = (+\text{value}) - T(-\text{value}) = (+\text{value}) + T(+\text{value}) > 0

always.

Both enthalpy and entropy factors oppose spontaneity.

5

Analyze Case 4:

\Delta H < 0

and

\Delta S > 0

\Delta G = (-\text{value}) - T(+\text{value}) < 0

always.

Both enthalpy release and entropy increase favor spontaneity under all conditions.

Key Concept

Dependence of Gibbs Free Energy and Reaction Spontaneity on the Signs of Enthalpy and Entropy Changes

Question 8204Question →

During the industrial extraction of iron from hematite in the blast furnace, distinct chemical reactions occur across different temperature zones. Arrange the following key reaction stages in sequential order from the top of the furnace (coolest zone, approx. $250^\circ\text{C}-400^\circ\text{C}$ ) down to the hearth/tuyere region at the bottom (hottest zone, approx. $1500^\circ\text{C}-1900^\circ\text{C}$ ):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequential order from top (coolest zone) to bottom (hottest zone) is: (1) Reduction of hematite to triiron tetroxide, (2) Reduction of triiron tetroxide to iron(II) oxide, (3) Reduction of iron(II) oxide to metallic iron, (4) Combination of calcium oxide with silica to form slag, and (5) Exothermic combustion of coke with preheated air.

The blast furnace operates with a temperature gradient rising from top to bottom. Near the top (

200^\circ\text{C}-400^\circ\text{C}

), hematite (

\text{Fe}_2\text{O}_3

) is reduced to

\text{Fe}_3\text{O}_4

. As the mixture descends to warmer regions (

500^\circ\text{C}-700^\circ\text{C}

),

\text{Fe}_3\text{O}_4

is reduced to

\text{FeO}

. Deeper down (

800^\circ\text{C}-1000^\circ\text{C}

),

\text{FeO}

is reduced to spongy iron (

\text{Fe}

). At around

1000^\circ\text{C}-1200^\circ\text{C}

, limestone-derived

\text{CaO}

reacts with

\text{SiO}_2

to form molten slag (

\text{CaSiO}_3

). Finally, at the base near the tuyeres (

1500^\circ\text{C}-1900^\circ\text{C}

), coke reacts exothermically with oxygen to fuel the entire process.

Step-by-Step Solution

1

Analyze the thermal gradient inside the blast furnace

Temperatures increase from top (

200^\circ\text{C}

) to bottom hearth region (

>1500^\circ\text{C}

).

Cold raw materials enter from the top while hot air blasts enter from the tuyeres at the bottom.

2

Identify top-zone reactions (

200^\circ\text{C}-400^\circ\text{C}

)

3\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow 2\text{Fe}_3\text{O}_4 + \text{CO}_2

Hematite is initially converted to magnetite at relatively low temperatures by upflowing carbon(II) oxide.

3

Identify upper-middle zone reactions (

500^\circ\text{C}-700^\circ\text{C}

)

\text{Fe}_3\text{O}_4 + \text{CO} \rightarrow 3\text{FeO} + \text{CO}_2

Further reduction converts magnetite into iron(II) oxide as the charge descends.

4

Identify lower-middle zone reactions (

800^\circ\text{C}-1000^\circ\text{C}

)

\text{FeO} + \text{CO} \rightarrow \text{Fe} + \text{CO}_2

Complete reduction of iron(II) oxide to spongy metallic iron takes place here.

5

Identify slag formation zone (

1000^\circ\text{C}-1200^\circ\text{C}

)

\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3

Limestone decomposes into calcium oxide, which reacts with acidic silicon dioxide impurities to form molten calcium trioxosilicate(IV) slag.

6

Identify bottom tuyere region reactions (

1500^\circ\text{C}-1900^\circ\text{C}

)

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

Preheated air blown in through tuyeres reacts violently with coke to generate heat and carbon dioxide, which is subsequently reduced by hot coke to carbon monoxide.

Key Concept

Blast furnace temperature zones and sequential chemical reduction stages of iron ore

Question 8205Question →

A gaseous mixture containing carbon monoxide ( $\text{CO}$ ) and carbon dioxide ( $\text{CO}_2$ ) has a total mass of $10.0\text{ g}$ . If the mixture contains a total of $2.408 \times 10^{23}$ oxygen atoms, calculate the mass, in grams, of carbon dioxide ( $\text{CO}_2$ ) present in the mixture. $[\text{C} = 12.0, \text{O} = 16.0, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

Show answer & explanation

Answer: 4.4

Answer

The mass of carbon dioxide (

\text{CO}_2

) present in the mixture is

4.4\text{ g}

.

The correct calculation yields 4.4 g by converting the oxygen atom count to 0.40 moles of O atoms, formulating the system of equations for total mass (28x + 44y = 10.0) and total oxygen moles (x + 2y = 0.40), and solving for the mass of CO₂.

Step-by-Step Solution

1

Determine the molar masses of carbon monoxide and carbon dioxide.

\text{Molar mass of CO} = 12.0 + 16.0 = 28.0\text{ g mol}^{-1}

;

\text{Molar mass of CO}_2 = 12.0 + 2(16.0) = 44.0\text{ g mol}^{-1}

.

Molar masses are required to relate the mass of each component to its molar quantity.

2

Calculate the total number of moles of oxygen atoms in the mixture using Avogadro's constant.

n_{\text{O}} = \frac{2.408 \times 10^{23}}{6.02 \times 10^{23}\text{ mol}^{-1}} = 0.40\text{ mol of O atoms}

.

Avogadro's constant converts particle count to mole quantity.

3

Set up a system of linear equations representing the total mass and total moles of oxygen atoms.

Let

x = \text{moles of CO}

and

y = \text{moles of CO}_2

.
Equation 1 (Mass):

28x + 44y = 10.0

Equation 2 (Oxygen atoms):

x + 2y = 0.40

CO contains 1 O atom per molecule and CO₂ contains 2 O atoms per molecule.

4

Solve the system of linear equations for

y

(moles of

\text{CO}_2

).

From Equation 2,

x = 0.40 - 2y

. Substituting into Equation 1 gives

28(0.40 - 2y) + 44y = 10.0 \implies 11.2 - 56y + 44y = 10.0 \implies 12y = 1.2 \implies y = 0.10\text{ mol}

.

Algebraic substitution yields the mole quantity of carbon dioxide.

5

Calculate the mass of

\text{CO}_2

present in the sample.

\text{Mass of CO}_2 = y \times \text{Molar mass} = 0.10\text{ mol} \times 44.0\text{ g mol}^{-1} = 4.4\text{ g}

.

Multiplying moles of CO₂ by its molar mass gives the required mass in grams.

Key Concept

Mole Concept, Avogadro's Constant, and Gas Mixture Stoichiometry

Question 8206Question →

During the catalytic decomposition of hydrogen peroxide according to the equation:

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

a student reacts

100\text{ cm}^3

of a

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

solution of

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

. If the peroxide decomposes completely in

80\text{ s}

, what is the average rate of formation of oxygen gas in

\text{cm}^3\text{ s}^{-1}

at STP? (Molar volume of gas at STP =

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

)

Show answer & explanation

Answer: 5.6

Answer

The average rate of formation of oxygen gas at STP is

5.6\text{ cm}^3\text{ s}^{-1}

.

First, the amount of hydrogen peroxide in moles is calculated as

0.40\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.040\text{ mol}

. From the reaction stoichiometry,

2\text{ moles of H}_2\text{O}_2

produce

1\text{ mole of O}_2

, meaning

0.020\text{ mol of O}_2

is evolved. At STP,

0.020\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 448\text{ cm}^3

of oxygen. Dividing by the total time of

80\text{ s}

yields

5.6\text{ cm}^3\text{ s}^{-1}

.

Step-by-Step Solution

1

Calculate the total number of moles of H₂O₂ present in the reaction solution

Moles of

\text{H}_2\text{O}_2 = 0.40\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.040\text{ mol}

Concentration and volume determine the total amount of reactant available.

2

Determine the total moles of O₂ gas formed using stoichiometric ratios

Moles of

\text{O}_2 = \frac{0.040\text{ mol}}{2} = 0.020\text{ mol}

The balanced chemical equation shows a 2:1 mole ratio between

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

and

\text{O}_2

.

3

Convert moles of O₂ into volume in cm³ at STP

Volume of

\text{O}_2 = 0.020\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 448\text{ cm}^3

1 mole of any ideal gas occupies

22.4\text{ dm}^3

(

22400\text{ cm}^3

) at STP.

4

Divide total volume of O₂ produced by the elapsed reaction time

Rate of

\text{O}_2

formation

= \frac{448\text{ cm}^3}{80\text{ s}} = 5.6\text{ cm}^3\text{ s}^{-1}

Reaction rate with respect to gas product evolution is change in volume per unit time.

Key Concept

Stoichiometric rate of reaction and gas volume calculations

Question 8207Question →

Starch mucilage is classified as a colloidal system rather than a true solution. Which of the following physical properties distinguishes starch mucilage from a true solution of sodium chloride?

Ability to scatter a beam of light passing through the mixture

Rapid settling of dispersed particles under gravity upon standing

Retention of dispersed particles when passed through standard laboratory filter paper

Distinct separation of phases visible to the naked eye

Show answer & explanation

Answer: Ability to scatter a beam of light passing through the mixture

Answer

The ability to scatter a beam of light passing through the mixture (the Tyndall effect)

Colloidal systems such as starch mucilage contain particles ranging from

1\text{ nm}

to

100\text{ nm}

which scatter light (the Tyndall effect). In contrast, true solutions like sodium chloride contain ions or molecules smaller than

1\text{ nm}

that cannot scatter visible light.

Step-by-Step Solution

1

Compare particle size ranges of true solutions and colloidal systems.

True solutions have solute particles smaller than

1\text{ nm}

, whereas colloidal systems have dispersed particles in the range of

1\text{ nm}

to

100\text{ nm}

.

Particle size dictates whether light scattering occurs.

2

Identify the optical phenomenon associated with colloidal particle sizes.

Particles between

1\text{ nm}

and

100\text{ nm}

scatter light rays, making the beam's path visible (Tyndall effect).

This property clearly differentiates colloids like starch mucilage from true solutions like sodium chloride in water.

Key Concept

Tyndall Effect and Particle Size Differences in Mixtures

Estimated Time:45s

Question 8208Question →

In the reduction half-reaction

\text{Fe}^{3+}(\text{aq}) + n e^- \rightarrow \text{Fe}(\text{s})

what is the value of

n

required to balance the electrical charge?

1

2

3

6

Show answer & explanation

Answer: $3$

Answer

The value of

n

is

3

.

To reduce one

\text{Fe}^{3+}

ion with a

+3

charge to a neutral iron atom with a

0

charge, exactly

3

electrons (each carrying a charge of

-1

) must be added to the reactant side so that the net charge on both sides equals

0

.

Step-by-Step Solution

1

Determine the charge on the reactant side and product side

The left side has one

\text{Fe}^{3+}

ion with a charge of

+3

. The right side has neutral

\text{Fe}(\text{s})

with a charge of

0

.

Charge conservation must be satisfied in a balanced half-reaction.

2

Calculate the number of negative electrons (

e^-

) required to balance net charge

+3 + n(-1) = 0 \implies n = 3

.

Each electron carries a single negative charge (

-1

).

Key Concept

Charge conservation in half-reactions

Question 8209Question →

According to the collision theory of chemical reactions, which of the following conditions must be met for a collision between reactant molecules to result in a successful chemical reaction?

The colliding molecules must possess energy equal to or greater than the activation energy and collide with correct orientation.

The colliding molecules must increase the overall equilibrium constant to favor product formation.

The collision must reverse the sign of the overall enthalpy change of the reaction.

Every collision between reactant particles results in product formation regardless of energy.

Show answer & explanation

Answer: The colliding molecules must possess energy equal to or greater than the activation energy and collide with correct orientation.

Answer

The colliding molecules must possess energy equal to or greater than the activation energy and collide with correct orientation.

According to collision theory, a chemical reaction occurs only when colliding particles possess a minimum threshold energy known as the activation energy (

E_a

) and collide with the appropriate spatial orientation to allow new bonds to form.

Step-by-Step Solution

1

Recall the fundamental postulates of collision theory.

Collision theory states that for a reaction to occur, reactant particles must collide with each other.

Physical contact is necessary for bond breaking and bond formation.

2

Identify the two necessary conditions for an effective (successful) collision.

1. Minimum kinetic energy equal to or exceeding the activation energy (

E_a

). 2. Proper relative orientation of colliding species.

Particles with insufficient energy bounce off each other without reacting, and incorrect orientation prevents the proper alignment of reacting bonds.

Key Concept

Collision Theory Criteria for Effective Collisions

Question 8210Question →

Underground steel pipelines are often connected to sacrificial blocks of magnesium to prevent rusting. Which statement best explains the electrochemical principle behind this method of corrosion protection?

Carbon(II) oxide in the surrounding soil reduces any oxidized iron ions back to metallic iron, preventing corrosion.

Magnesium is more electropositive than iron, so it loses electrons preferentially and serves as a sacrificial anode.

Magnesium acts as the cathode where oxidation takes place, while the iron pipe functions as the protective anode.

The rate of iron oxidation is halted because magnesium acts as the limiting reactant based purely on its initial mass.

Show answer & explanation

Answer: Magnesium is more electropositive than iron, so it loses electrons preferentially and serves as a sacrificial anode.

Answer

Magnesium is more electropositive than iron, so it loses electrons preferentially and serves as a sacrificial anode.

Magnesium is higher than iron in the electrochemical series. When connected electrically in a moist environment, magnesium oxidizes preferentially by donating electrons to the iron structure, causing magnesium to dissolve as sacrificial anode while keeping iron uncorroded as the cathode.

Step-by-Step Solution

1

Compare the positions of magnesium and iron in the electrochemical series.

Magnesium (

Mg

) has a higher standard oxidation potential than iron (

Fe

).

Metals positioned higher in the activity series lose valence electrons more readily.

2

Determine which metal undergoes oxidation in the galvanic cell formed by moisture.

Magnesium undergoes oxidation at the sacrificial anode (

Mg \rightarrow Mg^{2+} + 2e^-

).

The more electropositive metal preferentially oxidizes, suppressing the oxidation of iron (

Fe \rightarrow Fe^{2+} + 2e^-

).

3

Identify the overall protection mechanism.

Cathodic protection (sacrificial protection) keeps the iron pipeline safe as long as magnesium is present to corrode sacrificial anodes.

Iron is forced to remain the cathode, maintaining its elemental state.

Key Concept

Cathodic (Sacrificial) Protection of Iron

Estimated Time:1m 0s

Question 8211Question →

During domestic water treatment, a household adds a measured amount of calcium hypochlorite (bleaching powder) to clear well water before consumption. What is the primary chemical purpose of adding this substance?

To destroy pathogenic microorganisms present in the water

To precipitate dissolved calcium and magnesium ions that cause hardness

To coagulate fine suspended clay particles into larger settles

To neutralize organic tastes and unpleasant odors in the water

Show answer & explanation

Answer: To destroy pathogenic microorganisms present in the water

Answer

Adding calcium hypochlorite serves to sterilize the water by killing pathogenic bacteria and microorganisms.

Calcium hypochlorite (

Ca(ClO)_2

) releases hypochlorous acid and chlorine when added to water. This powerful oxidizing environment kills bacteria, viruses, and other disease-causing pathogens, fulfilling the sterilization stage of water treatment.

Step-by-Step Solution

1

Identify the chemical reagent and its functional group

Calcium hypochlorite,

Ca(ClO)_2

, acts as a chlorine-releasing disinfecting agent.

Bleaching powder hydrolyzes in water to release active chlorine ions.

2

Distinguish municipal/domestic water treatment roles

Disinfection/sterilization kills microorganisms, whereas coagulation removes turbidity and precipitation removes hardness.

Each water treatment reagent plays a specific, distinct chemical role.

Key Concept

Chemical Sterilization in Water Purification

Estimated Time:1m 0s

Question 8212Question →

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

Question 8213Question →

In the reduction zone of a blast furnace, hematite (

\text{Fe}_2\text{O}_3

) reacts with gaseous carbon(II) oxide (

\text{CO}

) according to the balanced equation:

\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(l) + 3\text{CO}_2(g)

If a furnace charge containing $160.0\text{ kg}$ of pure $\text{Fe}_2\text{O}_3$ is reacted with $40.32\text{ m}^3$ of $\text{CO}$ gas measured at STP, which of the following correctly identifies the limiting reactant and the maximum mass of iron produced?
[ $\text{Fe} = 56$ , $\text{O} = 16$ , $\text{C} = 12$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{mol}^{-1}$ ]

\text{CO}

is the limiting reactant, yielding

67.2\text{ kg}

of iron.

\text{Fe}_2\text{O}_3

is the limiting reactant, yielding

112.0\text{ kg}

of iron.

Carbon (

\text{C}

) acts as the direct reducing agent in this zone, yielding

112.0\text{ kg}

of iron.

\text{CO}

is the limiting reactant, yielding

100.8\text{ kg}

of iron.

Show answer & explanation

Answer: $\text{CO}$ is the limiting reactant, yielding $67.2\text{ kg}$ of iron.

Answer

Carbon(II) oxide (

\text{CO}

) is the limiting reactant, yielding

67.2\text{ kg}

of iron.

Carbon(II) oxide (

\text{CO}

) is the limiting reactant because

1800\text{ mol}

of

\text{CO}

can only reduce

600\text{ mol}

of

\text{Fe}_2\text{O}_3

out of the available

1000\text{ mol}

. Based on the stoichiometric ratio of

3\text{ mol CO} : 2\text{ mol Fe}

,

1800\text{ mol}

of

\text{CO}

yields

1200\text{ mol}

of iron metal, which corresponds to

67.2\text{ kg}

.

Step-by-Step Solution

1

Calculate the moles of reactants provided

Moles of

\text{Fe}_2\text{O}_3 = \frac{160,000\text{ g}}{160\text{ g/mol}} = 1000\text{ mol}

. Moles of

\text{CO} = \frac{40,320\text{ dm}^3}{22.4\text{ dm}^3/\text{mol}} = 1800\text{ mol}

.

Converting quantities to moles allows stoichiometric comparison.

2

Determine the limiting reactant using the balanced equation stoichiometry

According to

\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2

,

1000\text{ mol}

of

\text{Fe}_2\text{O}_3

requires

3000\text{ mol}

of

\text{CO}

. Since only

1800\text{ mol}

of

\text{CO}

is available,

\text{CO}

is the limiting reactant.

The reactant that produces the lesser amount of product limits the reaction.

3

Calculate the theoretical yield of iron in kilograms

Moles of

\text{Fe}

produced

= 1800\text{ mol CO} \times \frac{2\text{ mol Fe}}{3\text{ mol CO}} = 1200\text{ mol Fe}

. Mass of

\text{Fe} = 1200\text{ mol} \times 56\text{ g/mol} = 67,200\text{ g} = 67.2\text{ kg}

.

Multiply moles of limiting reactant by the mole ratio and molar mass of iron.

Key Concept

Stoichiometric limiting reactant analysis in the blast furnace reduction of iron ore

Question 8214Question →

Automotive exhaust emissions significantly contribute to atmospheric pollution. Which of the following chemical reactions occurring inside a catalytic converter directly eliminates a primary pollutant that acts as a precursor to both photochemical smog and acid rain?

The reduction of nitrogen(II) oxide (

NO

) to nitrogen gas (

N_2

)

The oxidation of carbon(II) oxide (

CO

) to carbon(IV) oxide (

CO_2

)

The catalytic breakdown of chlorofluorocarbons (

CFCs

) into halide salts

The desulfurization of sulfur(IV) oxide (

SO_2

) by reaction with calcium oxide (

CaO

)

Show answer & explanation

Answer: The reduction of nitrogen(II) oxide ( $NO$ ) to nitrogen gas ( $N_2$ )

Answer

The reduction of nitrogen(II) oxide (

NO

) to nitrogen gas (

N_2

)

Nitrogen(II) oxide (

NO

) produced in internal combustion engines is a primary pollutant. Inside a catalytic converter, reduction catalysts convert

NO

into harmless nitrogen gas (

N_2

). This prevents

NO

from oxidizing in the atmosphere to

NO_2

, which is essential for driving the photochemical reactions that produce tropospheric ozone and PAN (photochemical smog) as well as nitric acid (

HNO_3

) in acid rain.

Step-by-Step Solution

1

Identify the primary pollutant responsible for both photochemical smog and acid rain

Nitrogen oxides (

NO_x

, primarily

NO

and

NO_2

) are key precursors for both atmospheric issues.

NO_2

reacts with sunlight and volatile organic compounds to generate tropospheric ozone (photochemical smog) and dissolves in moisture to form nitric acid (

HNO_3

).

2

Determine the role of the catalytic converter in treating nitrogen oxides

The reduction catalyst (platinum and rhodium) converts

NO

into harmless

N_2

gas:

2NO \rightarrow N_2 + O_2

.

By reducing

NO

before exhaust release, the formation of secondary atmospheric pollutants is prevented.

Key Concept

Role of catalytic converters in reducing

NO_x

precursors of photochemical smog and acid rain

Question 8215Question →

In the industrial refining of crude oil, catalytic cracking is widely used to convert long-chain heavy gas oil fractions into gasoline and light alkenes. Which of the following best describes the primary advantage of employing a zeolite catalyst in this process compared to thermal cracking?

It allows cracking to proceed at lower temperatures and pressures while yielding a higher proportion of branched-chain hydrocarbons

It shifts the chemical equilibrium to ensure total conversion of gas oil into straight-chain alkanes without producing alkenes

It increases the boiling points of the heavy gas oil molecules so that they condense faster in the fractionating column

It functions as a consumable limiting reactant that determines the maximum mole yield of gasoline produced

Show answer & explanation

Answer: It allows cracking to proceed at lower temperatures and pressures while yielding a higher proportion of branched-chain hydrocarbons

Answer

The primary advantage of catalytic cracking is that it reduces energy requirements by operating at lower temperatures and pressures while producing a higher yield of high-octane branched-chain hydrocarbons.

In industrial petroleum refining, catalytic cracking uses zeolites (aluminosilicate catalysts) to accelerate hydrocarbon breakdown at moderate temperatures (~500 °C) and relatively low pressures. In addition to reducing energy demands, zeolites selectively encourage the formation of branched-chain alkanes and aromatics, which greatly improves the octane rating of the resulting petrol.

Step-by-Step Solution

1

Evaluate the general role of catalysts in industrial chemistry

Catalysts provide an alternative reaction pathway with lower activation energy, enabling reactions to occur at significantly reduced operating temperatures and pressures.

Lowering pressure and temperature cuts operational energy costs in industrial plants.

2

Examine the specific structural yield of catalytic cracking using zeolites

Zeolite catalysts selectively promote isomerisation, giving products enriched in branched-chain alkanes and aromatic hydrocarbons.

Branched hydrocarbons have higher octane ratings, making the gasoline higher in quality for motor engines compared to product mixtures from thermal cracking.

Key Concept

Industrial Petroleum Refining and Catalytic Cracking

Question 8216Question →

An element $Z$ exists naturally as two isotopes, $^{24}Z$ and $^{26}Z$ . On the carbon-12 scale, one atomic mass unit ( $1\text{ amu}$ ) is defined as $\frac{1}{12}$ of the mass of a single $^{12}\text{C}$ atom, where the mass of one $^{12}\text{C}$ atom is $1.992 \times 10^{-23}\text{ g}$ . If the average absolute mass of one atom of element $Z$ is $4.0504 \times 10^{-23}\text{ g}$ , what is the percentage abundance of the heavier isotope, $^{26}Z$ ?

20\%

40\%

50\%

80\%

Show answer & explanation

Answer: $20\%$

Answer

The percentage abundance of the heavier isotope

^{26}Z

is

20\%

.

One atomic mass unit (

1\text{ amu}

) on the carbon-12 scale equals

\frac{1.992 \times 10^{-23}\text{ g}}{12} = 1.66 \times 10^{-24}\text{ g}

. Dividing the absolute mass of one atom of

Z

(

4.0504 \times 10^{-23}\text{ g}

) by

1.66 \times 10^{-24}\text{ g}

yields a relative atomic mass of

24.4

. Setting up the weighted average formula

24.4 = 24(1 - p) + 26p

, where

p

is the fractional abundance of

^{26}Z

, gives

2p = 0.4

, so

p = 0.20

or

20\%

.

Step-by-Step Solution

1

Calculate the value of

1\text{ amu}

in grams using the carbon-12 reference standard

1\text{ amu} = \frac{1.992 \times 10^{-23}\text{ g}}{12} = 1.66 \times 10^{-24}\text{ g}

Relative atomic mass is based on the carbon-12 scale, where

1\text{ amu}

equals one-twelfth the mass of a carbon-12 atom.

2

Determine the relative atomic mass (RAM) of element

Z

\text{RAM of } Z = \frac{4.0504 \times 10^{-23}\text{ g}}{1.66 \times 10^{-24}\text{ g}} = 24.4

Relative atomic mass is a dimensionless ratio of the average atomic mass of an element to

1\text{ amu}

.

3

Set up and solve the isotopic abundance equation for percentage

x

of

^{26}Z

24.4 = \frac{24(100 - x) + 26x}{100} \implies 2440 = 2400 - 24x + 26x \implies 2x = 40 \implies x = 20\%

The relative atomic mass of an element is the weighted average of the mass numbers of its naturally occurring isotopes.

Key Concept

Relative Atomic Mass calculation on the Carbon-12 scale and isotopic abundance determination

Estimated Time:2m 30s

Question 8217Question →

Two gaseous reactants, $\text{X}(g)$ and $\text{Y}(g)$ , undergo a bimolecular reaction inside a rigid container at constant volume. When the temperature of the reaction vessel is raised from $300\text{ K}$ to $310\text{ K}$ , the rate of reaction approximately doubles. According to collision theory, which statement correctly explains the primary molecular factor responsible for this marked increase in rate?

The fraction of colliding molecules with kinetic energy equal to or exceeding the activation energy (

E_a

) increases exponentially.

The total collision frequency between reactant molecules increases proportionally with the absolute temperature.

The activation energy (

E_a

) of the reaction is lowered as thermal energy increases in the system.

The molecular orientation requirements become negligible as thermal motion increases.

Show answer & explanation

Answer: The fraction of colliding molecules with kinetic energy equal to or exceeding the activation energy ( $E_a$ ) increases exponentially.

Answer

The primary factor responsible for the rate increase is that the fraction of colliding molecules possessing energy greater than or equal to the activation energy increases exponentially with temperature.

Increasing the temperature shifts the Maxwell-Boltzmann kinetic energy distribution toward higher energies. Because the proportion of molecules exceeding the activation energy is determined by the exponential factor

e^{-E_a/RT}

, a small temperature increase causes a dramatic, exponential increase in the fraction of collisions that are energetically successful.

Step-by-Step Solution

1

Analyze collision theory requirements for a chemical reaction to occur.

Effective collisions require both proper steric orientation and collision energy

E \ge E_a

.

Collision theory states that rate depends on collision frequency, steric factor, and the fraction of collisions exceeding activation energy.

2

Evaluate the effect of a

10\text{ K}

temperature rise on total collision frequency versus the fraction of energetic molecules.

Collision frequency increases only slightly (

∝ \sqrt{T}

, about

1.6\%

rise), whereas the fraction of molecules with

E \ge E_a

given by

e^{-E_a/RT}

increases exponentially.

The Maxwell-Boltzmann distribution curve shifts to higher kinetic energies, greatly expanding the area under the curve beyond

E_a

.

3

Select the option that identifies the exponential increase in energetic molecules as the governing cause.

The option describing an exponential increase in the fraction of energetic collisions is correct.

This exponential increase directly explains why a small temperature increase leads to a massive (often doubled) reaction rate.

Key Concept

Collision Theory and Temperature Dependence of Reaction Rate

Estimated Time:1m 30s

Question 8218Question →

In hot, concentrated alkaline solutions, chlorine gas undergoes a disproportionation redox reaction according to the unbalanced equation:

\text{Cl}_2(\text{g}) + \text{OH}^-(\text{aq}) \rightarrow \text{ClO}_3^-(\text{aq}) + \text{Cl}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})

When this equation is balanced using the smallest set of whole-number coefficients, what is the stoichiometric coefficient of hydroxide ions (

\text{OH}^-

) and the total number of moles of electrons transferred in the balanced equation?

6 hydroxide ions and 5 moles of electrons

6 hydroxide ions and 6 moles of electrons

12 hydroxide ions and 10 moles of electrons

3 hydroxide ions and 5 moles of electrons

Show answer & explanation

Answer: 6 hydroxide ions and 5 moles of electrons

Answer

The balanced equation requires 6 hydroxide ions and involves the transfer of 5 moles of electrons.

In the balanced redox equation

3\text{Cl}_2(\text{g}) + 6\text{OH}^-(\text{aq}) \rightarrow \text{ClO}_3^-(\text{aq}) + 5\text{Cl}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

, the stoichiometric coefficient of hydroxide ions is 6, and 5 moles of electrons are transferred per mole of reaction as written.

Step-by-Step Solution

1

Assign oxidation numbers to determine the oxidation and reduction half-reactions.

Elemental chlorine

\text{Cl}_2

has an oxidation number of

0

. In

\text{ClO}_3^-

, chlorine has an oxidation state of

+5

(oxidation). In

\text{Cl}^-

, chlorine has an oxidation state of

-1

(reduction).

Disproportionation involves the simultaneous oxidation and reduction of the same element.

2

Write and balance the oxidation half-reaction in basic medium.

\frac{1}{2}\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 3\text{H}_2\text{O} + 5e^-

One chlorine atom increases in oxidation state from

0

to

+5

, releasing

5e^-

. Six

\text{OH}^-

ions balance the charge and oxygen/hydrogen mass.

3

Write and balance the reduction half-reaction.

\frac{1}{2}\text{Cl}_2 + e^- \rightarrow \text{Cl}^-

One chlorine atom decreases in oxidation state from

0

to

-1

, accepting

1e^-

.

4

Equalize electron transfer between half-reactions and combine.

Multiply the reduction half-reaction by 5:

\frac{5}{2}\text{Cl}_2 + 5e^- \rightarrow 5\text{Cl}^-

. Combine with the oxidation half-reaction:

3\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 5\text{Cl}^- + 3\text{H}_2\text{O}

. Total electrons transferred

n_e = 5

.

The number of electrons lost in oxidation must equal the number gained in reduction.

Key Concept

Balancing Disproportionation Redox Reactions in Basic Medium

Question 8219Question →

In an energy profile diagram for a reversible chemical reaction, the potential energy of the reactants is $140\text{ kJ mol}^{-1}$ and the potential energy of the products is $60\text{ kJ mol}^{-1}$ . If the activation energy for the reverse uncatalyzed reaction is $210\text{ kJ mol}^{-1}$ , what is the activation energy of the reverse reaction in the presence of a catalyst that reduces the forward activation energy by $35\text{ kJ mol}^{-1}$ ?

175\text{ kJ mol}^{-1}

95\text{ kJ mol}^{-1}

245\text{ kJ mol}^{-1}

290\text{ kJ mol}^{-1}

Show answer & explanation

Answer: $175\text{ kJ mol}^{-1}$

Answer

The activation energy of the reverse reaction in the presence of the catalyst is

175\text{ kJ mol}^{-1}

.

A catalyst lowers the energy of the activated complex (peak of the curve) by a fixed amount without altering the ground state energies of reactants or products. Because the transition state is lowered by

35\text{ kJ mol}^{-1}

, both the forward and reverse activation energies decrease by exactly

35\text{ kJ mol}^{-1}

. Subtracting

35\text{ kJ mol}^{-1}

from the uncatalyzed reverse activation energy (

210\text{ kJ mol}^{-1}

) gives

175\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Determine the effect of a catalyst on the energy barrier

A catalyst provides an alternative pathway with a lower transition state energy, reducing the activation energy of both the forward and reverse reactions by the exact same value.

The difference between reactant/product energy levels (

Δ H

) remains constant, so

Δ E_a

is identical for forward and reverse directions.

2

Calculate the catalyzed reverse activation energy

E_{a,\text{reverse(cat)}} = E_{a,\text{reverse(uncat)}} - 35\text{ kJ mol}^{-1} = 210\text{ kJ mol}^{-1} - 35\text{ kJ mol}^{-1} = 175\text{ kJ mol}^{-1}

.

Directly subtracting the energy reduction brought by the catalyst yields the new reverse activation energy barrier.

Key Concept

Equal lowering of forward and reverse activation energy barriers by catalysts

Question 8220Question →

An aqueous solution of a sodium salt $X$ completely decolorizes acidified potassium tetraoxomanganate(VII) solution. Subsequent addition of barium chloride solution to the resulting reaction mixture yields a white precipitate that is insoluble in dilute hydrochloric acid. What role does salt $X$ play in the redox reaction, and what is the chemical formula of the sulfur-containing anion in $X$ ?

Salt

X

acts as a reducing agent, and the anion is

SO_3^{2-}

.

Salt

X

acts as an oxidizing agent, and the anion is

SO_3^{2-}

.

Salt

X

acts as a reducing agent, and the anion is

SO_4^{2-}

.

Salt

X

acts as an oxidizing agent, and the anion is

SO_4^{2-}

.

Show answer & explanation

Answer: Salt $X$ acts as a reducing agent, and the anion is $SO_3^{2-}$ .

Answer

Salt

X

acts as a reducing agent, and the anion is

SO_3^{2-}

.

In the reaction with acidified potassium tetraoxomanganate(VII), the manganate(VII) ion (

MnO_4^-

) is reduced from oxidation state +7 to +2, causing the purple solution to become colorless. The species that causes this reduction must be a reducing agent. Trioxosulfate(IV) ions (

SO_3^{2-}

) contain sulfur in the +4 oxidation state and are oxidized by

KMnO_4

to tetraoxosulfate(VI) ions (

SO_4^{2-}

). The formed

SO_4^{2-}

ions then react with

Ba^{2+}

from barium chloride to form a white insoluble precipitate of

BaSO_4

. Thus, salt

X

is a reducing agent containing

SO_3^{2-}

.

Step-by-Step Solution

1

Analyze the color change of acidified potassium tetraoxomanganate(VII).

The change from purple (

MnO_4^-

) to colorless (

Mn^{2+}

) indicates that

MnO_4^-

is reduced.

Manganese is reduced from oxidation state +7 to +2 by gaining electrons from a reducing agent.

2

Determine the role of salt

X

.

Salt

X

provides electrons to reduce

MnO_4^-

, meaning salt

X

is oxidized and acts as a reducing agent.

A substance that undergoes oxidation and causes another species to be reduced is a reducing agent.

3

Identify the sulfur-containing anion from the precipitation test.

The oxidation product of salt

X

reacts with

Ba^{2+}

to form insoluble

BaSO_4

. Therefore, the oxidized species formed in solution is

SO_4^{2-}

.

Trioxosulfate(IV) ion (

SO_3^{2-}

) has sulfur in oxidation state +4 and is oxidized by acidified

KMnO_4

to tetraoxosulfate(VI) ion (

SO_4^{2-}

), which gives a white precipitate of

BaSO_4

insoluble in HCl.

Key Concept

Oxidizing and Reducing Agents and Qualitative Redox Tests

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