Chemical Combination and Stoichiometry Practice Questions — JAMB UTME — Page 4

Consider the catalytic oxidation of ammonia gas represented by the balanced chemical equation below:

4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)

If

6.8\text{ g}

of ammonia gas reacts completely with excess oxygen gas at STP, calculate the volume of nitrogen(II) oxide gas and the volume of steam produced.

[H = 1, N = 14, O = 16, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{ mol}^{-1}]

Fill in the missing values in the statement below.

Fill in the blanks below

The volume of nitrogen(II) oxide gas (

NO

) produced at STP is dm³, and the volume of steam (

H_2O

) produced at STP is dm³.

Show answer & explanation

Answer

The volume of nitrogen(II) oxide produced at STP is 8.96 dm³ and the volume of steam produced at STP is 13.44 dm³.

Molar mass of ammonia is 17 g/mol, meaning 6.8 g equals 0.4 mol of ammonia. According to the balanced equation, 4 moles of ammonia produce 4 moles of nitrogen(II) oxide gas and 6 moles of steam. Thus, 0.4 mol of ammonia yields 0.4 mol of nitrogen(II) oxide gas (0.4 * 22.4 = 8.96 dm³) and 0.6 mol of steam (0.6 * 22.4 = 13.44 dm³).

Step-by-Step Solution

1

Calculate the molar mass of ammonia (

NH_3

)

Molar mass of

NH_3 = 14 + (3 \times 1) = 17\text{ g mol}^{-1}

Required to convert the given mass of reactant into moles.

2

Calculate the number of moles of

NH_3

reacted

\text{Moles of } NH_3 = \frac{6.8\text{ g}}{17\text{ g mol}^{-1}} = 0.4\text{ mol}

Stoichiometric relations in chemical equations are expressed in mole ratios.

3

Determine the moles of

NO(g)

and

H_2O(g)

produced using stoichiometric coefficients

From the balanced equation,

4\text{ mol } NH_3 \rightarrow 4\text{ mol } NO

, so

0.4\text{ mol } NH_3 \rightarrow 0.4\text{ mol } NO

.
Also,

4\text{ mol } NH_3 \rightarrow 6\text{ mol } H_2O(g)

, so

\text{Moles of } H_2O = \frac{6}{4} \times 0.4 = 0.6\text{ mol}

.

The coefficients in the balanced equation define the molar ratio between reactants and products.

4

Convert the moles of each gaseous product to volume at STP using molar gas volume

\text{Volume of } NO = 0.4\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 8.96\text{ dm}^3

\text{Volume of } H_2O = 0.6\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 13.44\text{ dm}^3

At STP,

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

.

Key Concept

Mass-Volume Stoichiometry at STP

Estimated Time:2m 0s

Question 62Question →

Potassium trioxonitrate(V) decomposes upon heating according to the balanced chemical equation:

2KNO_3(s) \rightarrow 2KNO_2(s) + O_2(g)

What volume of oxygen gas measured at STP is produced by the complete thermal decomposition of

50.5\text{ g}

of

KNO_3

?

[K = 39, N = 14, O = 16\text{; Molar gas volume at STP } = 22.4\text{ dm}^3\text{ mol}^{-1}]

2.8\text{ dm}^3

5.6\text{ dm}^3

6.0\text{ dm}^3

11.2\text{ dm}^3

Show answer & explanation

Answer: $5.6\text{ dm}^3$

Answer

The correct volume of oxygen gas produced at STP is

5.6\text{ dm}^3

.

The complete decomposition of

50.5\text{ g}

(

0.5\text{ mol}

) of

KNO_3

yields

0.25\text{ mol}

of

O_2

gas according to the

2:1

stoichiometric mole ratio. Multiplying

0.25\text{ mol}

by the standard molar gas volume at STP (

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

) gives

5.6\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

KNO_3

Molar mass of

KNO_3 = 39 + 14 + (3 \times 16) = 101\text{ g mol}^{-1}

Molar mass is needed to convert the given mass of reactant into moles.

2

Calculate the number of moles of

KNO_3

reacted

\text{Moles of } KNO_3 = \frac{50.5\text{ g}}{101\text{ g mol}^{-1}} = 0.5\text{ mol}

Determines the exact mole amount of reactant supplied.

3

Use the mole ratio from the balanced equation to find moles of

O_2

produced

\text{Moles of } O_2 = \frac{1}{2} \times 0.5\text{ mol} = 0.25\text{ mol}

The equation shows that

2\text{ moles}

of

KNO_3

yield

1\text{ mole}

of

O_2

.

4

Convert moles of

O_2

to volume at STP

\text{Volume of } O_2 = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.6\text{ dm}^3

Molar gas volume at standard temperature and pressure (STP) is

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

.

Key Concept

Mass-Volume Stoichiometric Calculations at STP

Estimated Time:1m 30s

Question 63Question →

Phosphorus reacts with oxygen gas to produce phosphorus(V) oxide according to the balanced chemical equation:

4\text{P}_{(s)} + 5\text{O}_{2(g)} \rightarrow \text{P}_4\text{O}_{10(s)}

If a mixture containing

12.4\text{ g}

of phosphorus and

20.0\text{ g}

of oxygen gas is allowed to react to completion, what is the mass of the excess reactant remaining unreacted in grams?

[\text{P} = 31.0, \text{O} = 16.0]

Show answer & explanation

Answer: 4

Answer

The mass of unreacted excess oxygen gas remaining is

4.0\text{ g}

.

The correct calculation shows that

0.40\text{ mol}

of phosphorus requires

0.50\text{ mol}

of oxygen gas for complete reaction according to the

4:5

mole ratio in

4\text{P} + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}

. Subtracting the

0.50\text{ mol}

consumed from the initial

0.625\text{ mol}

leaves

0.125\text{ mol}

of unreacted oxygen gas, which equals

4.0\text{ g}

.

Step-by-Step Solution

1

Calculate the moles of phosphorus and oxygen gas present initially.

\text{Moles of P} = \frac{12.4\text{ g}}{31.0\text{ g/mol}} = 0.40\text{ mol}

;

\text{Moles of O}_2 = \frac{20.0\text{ g}}{32.0\text{ g/mol}} = 0.625\text{ mol}

.

Molar mass of

\text{P}

is

31.0\text{ g/mol}

and molar mass of

\text{O}_2

is

2 \times 16.0 = 32.0\text{ g/mol}

.

2

Determine the limiting reactant by comparing the required mole ratio to the available mole ratio.

\text{P}

is the limiting reactant, and

\text{O}_2

is the excess reactant.

From the balanced equation,

4\text{ moles of P}

require

5\text{ moles of O}_2

, so

1\text{ mole of P}

requires

1.25\text{ moles of O}_2

. Thus,

0.40\text{ mol}

of

\text{P}

requires

0.40 \times 1.25 = 0.50\text{ mol}

of

\text{O}_2

. Since

0.625\text{ mol}

of

\text{O}_2

is available,

\text{O}_2

is in excess.

3

Calculate the unreacted moles of oxygen gas remaining.

\text{Excess moles of O}_2 = 0.625\text{ mol} - 0.50\text{ mol} = 0.125\text{ mol}.

Subtracting the reacted moles from the initial moles gives the remaining amount.

4

Convert the unreacted moles of oxygen gas to mass in grams.

\text{Mass of remaining O}_2 = 0.125\text{ mol} \times 32.0\text{ g/mol} = 4.0\text{ g}.

Multiplying the excess moles by the molar mass of

\text{O}_2

(

32.0\text{ g/mol}

) gives the mass in grams.

Key Concept

Limiting and excess reactant stoichiometry

Question 64Question →

What is the volume, in $\text{dm}^3$ , occupied by $6.80\text{ g}$ of hydrogen sulfide gas ( $\text{H}_2\text{S}$ ) measured at standard temperature and pressure (STP)? $[\text{H} = 1.0, \text{S} = 32.0, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]$

4.48\text{ dm}^3

4.80\text{ dm}^3

4.76\text{ dm}^3

2.24\text{ dm}^3

Show answer & explanation

Answer: $4.48\text{ dm}^3$

Answer

The volume occupied by

6.80\text{ g}

of hydrogen sulfide gas at STP is

4.48\text{ dm}^3

.

To find the volume of gas at STP, first determine the molar mass of

\text{H}_2\text{S}

:

(2 \times 1.0) + 32.0 = 34.0\text{ g mol}^{-1}

. Next, convert the mass to moles:

\frac{6.80\text{ g}}{34.0\text{ g mol}^{-1}} = 0.20\text{ mol}

. Finally, multiply the moles by the molar volume at STP (

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

):

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

.

Step-by-Step Solution

1

Calculate the molar mass of hydrogen sulfide (

\text{H}_2\text{S}

)

\text{Molar Mass} = (2 \times 1.0) + 32.0 = 34.0\text{ g mol}^{-1}

Molar mass is required to convert mass to moles.

2

Calculate the number of moles in

6.80\text{ g}

of

\text{H}_2\text{S}

\text{Moles} = \frac{6.80\text{ g}}{34.0\text{ g mol}^{-1}} = 0.20\text{ mol}

Dividing given mass by molar mass yields amount of substance in moles.

3

Calculate the volume occupied at STP

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

1 mole of any ideal gas occupies

22.4\text{ dm}^3

at STP.

Key Concept

Molar Volume of Gases at STP

Estimated Time:1m 30s

Question 65Question →

Calcium carbide reacts with water according to the balanced chemical equation:

\text{CaC}_{2(s)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{Ca(OH)}_{2(aq)} + \text{C}_2\text{H}_{2(g)}

If

32.0\text{ g}

of

\text{CaC}_2

is reacted with

21.6\text{ g}

of

\text{H}_2\text{O}

, calculate the mass of the excess reactant remaining unreacted upon completion of the reaction.

[\text{Relative atomic masses: } \text{Ca} = 40, \text{C} = 12, \text{O} = 16, \text{H} = 1]

Show answer & explanation

Answer: 3.6

Answer

3.6 g

Converting initial masses to mole values yields 0.50 mol of CaC₂ and 1.20 mol of H₂O. Based on the 1:2 mole ratio in the balanced equation, 0.50 mol of CaC₂ requires 1.00 mol of H₂O to react completely. This leaves 0.20 mol of H₂O unreacted. Converting 0.20 mol of H₂O to mass gives 0.20 mol × 18 g/mol = 3.6 g of excess reactant remaining.

Step-by-Step Solution

1

Calculate molar masses of CaC₂ and H₂O

Molar mass of CaC₂ = 64 g/mol; Molar mass of H₂O = 18 g/mol

Molar mass is required to convert given mass values into mole quantities.

2

Convert initial masses to moles

Moles of CaC₂ = 0.50 mol; Moles of H₂O = 1.20 mol

Stoichiometric relationships depend strictly on mole ratios rather than direct mass ratios.

3

Determine limiting and excess reactants using mole ratios

CaC₂ is the limiting reactant; H₂O is in excess

According to the balanced equation coefficient ratio (1:2), 0.50 mol of CaC₂ requires 1.00 mol of H₂O. Because 1.20 mol of H₂O is present, H₂O is in excess.

4

Calculate remaining unreacted mass of excess reactant

3.6 g of excess H₂O remaining

Unreacted moles of H₂O = 1.20 - 1.00 = 0.20 mol. Mass = 0.20 mol × 18 g/mol = 3.6 g.

Key Concept

Determining limiting and excess reagents in chemical reactions and calculating unreacted leftover mass using mole ratios from balanced equations.

Estimated Time:2m 0s

Question 66Question →

An atom of an element $Q$ has a mass equal to $2.5$ times the mass of a single carbon-12 atom ( $^{12}\text{C}$ ). What is the relative atomic mass of element $Q$ on the carbon-12 scale?

2.5

14.4

30

60

Show answer & explanation

Answer: $30$

Answer

The relative atomic mass of element

Q

on the carbon-12 scale is 30.

On the carbon-12 scale,

1\text{ atomic mass unit (amu)}

is defined as

\frac{1}{12}

of the mass of a single carbon-12 atom (

^{12}\text{C}

). This gives a single

^{12}\text{C}

atom a relative mass of

12

. If an atom of element

Q

is

2.5

times as heavy as one

^{12}\text{C}

atom, its relative atomic mass is

2.5 \times 12 = 30

.

Step-by-Step Solution

1

Define atomic mass unit on the carbon-12 scale

1 atomic mass unit (amu) =

\frac{1}{12}

of the mass of one carbon-12 atom

The standard reference point for relative atomic masses is defined as

\frac{1}{12}\text{th}

the mass of a single

^{12}\text{C}

atom, which means one carbon-12 atom equals

12\text{ amu}

.

2

Calculate the atomic mass of element

Q

in amu

\text{Mass of } Q = 2.5 \times 12\text{ amu} = 30\text{ amu}

Since one atom of element

Q

has a mass

2.5

times that of a carbon-12 atom, its absolute atomic mass relative to

1\text{ amu}

is

2.5 \times 12

.

3

Determine the relative atomic mass

\text{RAM} = \frac{30\text{ amu}}{1\text{ amu}} = 30

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

\frac{1}{12}\text{th}

the mass of carbon-12.

Key Concept

Relative Atomic Mass on the Carbon-12 Scale

Question 67Question →

A mixture containing

10.8\text{ g}

of aluminium powder is reacted with

16.0\text{ g}

of oxygen gas according to the balanced chemical equation:

4\text{Al}_{(s)} + 3\text{O}_{2(g)} \rightarrow 2\text{Al}_2\text{O}_{3(s)}

What is the mass in grams of the excess reactant remaining unreacted at the end of the reaction?

[\text{Al} = 27, \text{O} = 16]

Show answer & explanation

Answer: 6.4

Answer

The mass of the excess reactant (oxygen gas) remaining unreacted is 6.4 g.

To determine the unreacted mass of the excess reactant, first convert given masses to moles: 10.8 g of Al corresponds to 0.40 mol and 16.0 g of O₂ corresponds to 0.50 mol. Using the mole ratio from the balanced equation (4 moles Al : 3 moles O₂), 0.40 mol of Al reacts completely with 0.30 mol of O₂. Thus, Al is the limiting reactant and O₂ is in excess. The unreacted amount of O₂ is 0.50 mol - 0.30 mol = 0.20 mol. Converting 0.20 mol of O₂ back to mass using its molar mass of 32 g/mol yields 6.4 g.

Step-by-Step Solution

1

Calculate the mole amounts of reactants provided

n(Al) = 0.40 mol, n(O₂) = 0.50 mol

Converting masses to moles using molar masses (Al = 27 g/mol, O₂ = 32 g/mol) is necessary for stoichiometric comparison.

2

Determine the theoretical moles of oxygen needed to react with all aluminium

n(O₂) required = 0.30 mol

From the balanced equation, 4 moles of Al require 3 moles of O₂, so 0.40 mol Al requires 0.40 × (3/4) = 0.30 mol O₂.

3

Identify the excess reactant and compute remaining moles

O₂ is in excess by 0.20 mol

Available O₂ (0.50 mol) exceeds required O₂ (0.30 mol), leaving 0.50 - 0.30 = 0.20 mol of O₂ unreacted.

4

Convert remaining moles of excess reactant back to mass

Mass of excess O₂ = 6.4 g

Multiplying 0.20 mol by the molar mass of O₂ (32 g/mol) gives the unreacted mass of oxygen.

Key Concept

Limiting and excess reactant calculations based on stoichiometric coefficients and mole conversions

Estimated Time:2m 0s

Question 68Question →

A

4.20\text{ g}

sample of impure sodium hydrogentrioxocarbonate(IV),

\text{NaHCO}_3

, was thermally decomposed according to the equation:

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

If

0.448\text{ dm}^3

of carbon(IV) oxide gas was collected at s.t.p., what is the percentage purity of the

\text{NaHCO}_3

sample?

[\text{Na} = 23, \text{H} = 1, \text{C} = 12, \text{O} = 16, \text{molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]

Show answer & explanation

Answer: 80

Answer

The percentage purity of the sodium hydrogentrioxocarbonate(IV) sample is 80.0%.

The volume of CO₂ gas collected (

0.448\text{ dm}^3

) corresponds to

0.020\text{ mol}

at s.t.p. According to the balanced equation,

2\text{ mol}

of

\text{NaHCO}_3

decompose to form

1\text{ mol}

of

\text{CO}_2

, meaning

0.040\text{ mol}

of pure

\text{NaHCO}_3

reacted. Multiplying by the molar mass of

\text{NaHCO}_3

(

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

) gives

3.36\text{ g}

of pure compound, which represents

80.0\%

of the original

4.20\text{ g}

sample.

Step-by-Step Solution

1

Calculate the moles of carbon(IV) oxide gas evolved at s.t.p.

0.020 mol of CO₂

Dividing the volume of gas collected by the molar volume of a gas at s.t.p. gives the chemical amount in moles.

2

Determine the moles of pure NaHCO₃ using the mole ratio from the balanced chemical equation.

0.040 mol of NaHCO₃

The reaction stoichiometry shows a 2:1 mole ratio between NaHCO₃ and CO₂.

3

Calculate the mass of pure NaHCO₃ by multiplying its moles by its molar mass (84 g/mol).

3.36 g of pure NaHCO₃

Mass is obtained by converting moles to grams using molar mass.

4

Divide the mass of pure NaHCO₃ by the initial mass of the impure sample (4.20 g) and multiply by 100%.

80.0%

Percentage purity expresses the proportion of active compound relative to the total mass of the sample.

Key Concept

Percentage purity calculation based on gas stoichiometry

Estimated Time:1m 30s

Question 69Question →

A

2.50\text{ g}

sample of impure iron(III) oxide,

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

, was completely reduced by excess carbon(II) oxide gas to yield

1.40\text{ g}

of pure iron metal according to the equation:

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

What is the percentage purity of the iron(III) oxide sample? [

\text{Fe} = 56, \text{O} = 16, \text{C} = 12

]

80.0\%

70.0\%

56.0\%

40.0\%

Show answer & explanation

Answer: $80.0\%$

Answer

The percentage purity of the iron(III) oxide sample is 80.0%.

According to the balanced chemical equation,

1\text{ mole}

of

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

(

160\text{ g}

) yields

2\text{ moles}

of

\text{Fe}

(

112\text{ g}

). To obtain

1.40\text{ g}

of pure iron metal, the mass of pure

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

required is

1.40 \times \frac{160}{112} = 2.00\text{ g}

. The percentage purity is calculated by taking the mass of pure

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

divided by the total sample mass (

2.50\text{ g}

) multiplied by

100

, yielding

80.0\%

.

Step-by-Step Solution

1

Calculate the molar mass of iron(III) oxide (

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

) and total mass of iron produced per mole.

Molar mass of

\text{Fe}_2\text{O}_3 = 2(56) + 3(16) = 112 + 48 = 160\text{ g/mol}

. Mass of

2\text{ moles}

of

\text{Fe} = 2 \times 56 = 112\text{ g}

.

Stoichiometry shows

1\text{ mole}

of

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

(

160\text{ g}

) produces

2\text{ moles}

of

\text{Fe}

(

112\text{ g}

).

2

Determine the mass of pure

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

in the sample required to produce

1.40\text{ g}

of

\text{Fe}

.

\text{Mass of pure Fe}_2\text{O}_3 = 1.40\text{ g Fe} \times \frac{160\text{ g Fe}_2\text{O}_3}{112\text{ g Fe}} = 2.00\text{ g}

.

Mass proportions allow determination of the mass of reacting pure compound.

3

Calculate the percentage purity of the sample.

\text{Percentage purity} = \left(\frac{2.00\text{ g}}{2.50\text{ g}}\right) \times 100 = 80.0\%

.

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

Key Concept

Determining percentage purity using stoichiometric mass calculations

Estimated Time:1m 30s

Question 70Question →

Zinc metal reacts with hydrochloric acid according to the balanced chemical equation:

\text{Zn}_{(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{ZnCl}_{2(aq)} + \text{H}_{2(g)}

If

13.0\text{ g}

of zinc is added to a solution containing

7.3\text{ g}

of hydrochloric acid, what mass of zinc remains unreacted after the reaction goes to completion?
(

\text{Zn} = 65.0,\text{ H} = 1.0,\text{ Cl} = 35.5

)

6.5\text{ g}

5.7\text{ g}

0.0\text{ g}

3.25\text{ g}

Show answer & explanation

Answer: $6.5\text{ g}$

Answer

The mass of zinc remaining unreacted is

6.5\text{ g}

.

Converting the given masses into moles shows that 0.20 mol of zinc and 0.20 mol of hydrochloric acid are present. Since 1 mole of zinc requires 2 moles of hydrochloric acid, 0.20 mol of hydrochloric acid reacts with only 0.10 mol of zinc. Hydrochloric acid is completely consumed, leaving 0.10 mol (6.5 g) of zinc unreacted.

Step-by-Step Solution

1

Calculate the molar masses of the reactants

Molar mass of

\text{Zn} = 65.0\text{ g/mol}

; Molar mass of

\text{HCl} = 1.0 + 35.5 = 36.5\text{ g/mol}

.

Molar masses are required to convert the given masses to mole quantities.

2

Determine the initial mole quantities of each reactant

Moles of

\text{Zn} = \frac{13.0\text{ g}}{65.0\text{ g/mol}} = 0.20\text{ mol}

; Moles of

\text{HCl} = \frac{7.3\text{ g}}{36.5\text{ g/mol}} = 0.20\text{ mol}

.

Chemical reactions occur according to mole ratios, not mass ratios.

3

Identify the limiting reactant and calculate the moles of zinc consumed

From the equation,

1\text{ mol of Zn}

reacts with

2\text{ mol of HCl}

. Thus,

0.20\text{ mol of HCl}

requires

\frac{0.20}{2} = 0.10\text{ mol of Zn}

.

\text{HCl}

is the limiting reactant.

The limiting reactant determines the extent of the reaction.

4

Calculate the unreacted moles and mass of zinc

Unreacted moles of

\text{Zn} = 0.20\text{ mol} - 0.10\text{ mol} = 0.10\text{ mol}

. Unreacted mass of

\text{Zn} = 0.10\text{ mol} \times 65.0\text{ g/mol} = 6.5\text{ g}

.

Subtracting consumed moles from initial moles gives the remaining amount.

Key Concept

Limiting and Excess Reactants

Question 71Question →

Propane burns completely in oxygen gas according to the following balanced chemical equation:

C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)

What volume of oxygen gas, measured at STP, is required for the complete combustion of

4.4\text{ g}

of propane?

[\text{H} = 1.0, \text{C} = 12.0; \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{ mol}^{-1}]

2.24\text{ dm}^3

6.72\text{ dm}^3

11.2\text{ dm}^3

12.0\text{ dm}^3

Show answer & explanation

Answer: $11.2\text{ dm}^3$

Answer

The volume of oxygen gas required at STP is

11.2\text{ dm}^3

.

The option stating

11.2\text{ dm}^3

is correct.

4.4\text{ g}

of propane corresponds to

0.10\text{ mol}

. According to the balanced equation,

1\text{ mol}

of

C_3H_8

reacts with

5\text{ mol}

of

O_2

, so

0.10\text{ mol}

of propane requires

0.50\text{ mol}

of

O_2

. At STP,

0.50\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of propane (

C_3H_8

).

Molar mass of

C_3H_8 = (3 \times 12.0) + (8 \times 1.0) = 36.0 + 8.0 = 44.0\text{ g mol}^{-1}

.

Molar mass is needed to convert the given mass of reactant into moles.

2

Determine the number of moles of propane in

4.4\text{ g}

.

Moles of

C_3H_8 = \frac{4.4\text{ g}}{44.0\text{ g mol}^{-1}} = 0.10\text{ mol}

.

Stoichiometric calculations rely on mole ratios from the balanced chemical equation.

3

Use the mole ratio from the balanced equation to find the required moles of

O_2

.

Mole ratio

C_3H_8 : O_2 = 1 : 5

. Moles of

O_2 = 0.10\text{ mol} \times 5 = 0.50\text{ mol}

.

Every

1\text{ mole}

of propane requires

5\text{ moles}

of oxygen gas for complete combustion.

4

Calculate the volume of

O_2

gas at STP.

Volume of

O_2 = 0.50\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3

.

Multiply the calculated number of moles of gas by the molar volume at STP (

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

).

Key Concept

Mass-Volume Stoichiometric Calculations at STP

Estimated Time:1m 30s

Question 72Question →

A sample of a pure iron oxide synthesized in a laboratory contains $5.60\text{ g}$ of iron and $2.40\text{ g}$ of oxygen. According to the Law of Definite Proportions, what is the mass of iron, in grams, present in a $20.0\text{ g}$ sample of the same iron oxide collected from a natural deposit?

Show answer & explanation

Answer: 14

Answer

The mass of iron present in the 20.0 g sample of iron oxide is 14.0 g.

According to the Law of Definite Proportions, a pure chemical compound always contains the same elements combined together in the exact same proportion by mass, regardless of its source or method of preparation. In the laboratory sample,

8.00\text{ g}

of iron oxide contains

5.60\text{ g}

of iron, giving an iron mass composition of

70\%

. Therefore, a

20.0\text{ g}

natural sample of the same compound must also contain

70\%

iron by mass, which equals

14.0\text{ g}

.

Step-by-Step Solution

1

Calculate total mass of the laboratory sample

5.60\text{ g} + 2.40\text{ g} = 8.00\text{ g}

The total mass of the compound is the sum of the constituent element masses (Law of Conservation of Mass).

2

Find the mass percentage/fraction of iron

\frac{5.60\text{ g}}{8.00\text{ g}} = 0.70

(or

70\%

)

By the Law of Definite Proportions, the mass ratio of elements in a pure compound is constant.

3

Calculate mass of iron in the 20.0 g natural sample

0.70 \times 20.0\text{ g} = 14.0\text{ g}

Applying the constant mass composition percentage to the new sample mass.

Key Concept

Law of Definite Proportions (Constant Composition)

Estimated Time:1m 30s

Question 73Question →

A mixture of $15\text{ cm}^3$ of ethene ( $C_2H_4$ ) and $50\text{ cm}^3$ of oxygen gas was exploded at constant temperature and pressure. If the resulting mixture was cooled to room temperature, what is the total volume of the residual gas in $\text{cm}^3$ ?

Show answer & explanation

Answer: 35

Answer

35 cm³

According to Gay-Lussac's Law,

15\text{ cm}^3

of ethene reacts with

45\text{ cm}^3

of oxygen to produce

30\text{ cm}^3

of carbon(IV) oxide gas. Since

50\text{ cm}^3

of oxygen was initially present,

5\text{ cm}^3

of unreacted oxygen gas remains. Upon cooling to room temperature, water condenses to liquid, leaving a total residual gaseous volume of

5\text{ cm}^3 + 30\text{ cm}^3 = 35\text{ cm}^3

.

Step-by-Step Solution

1

Write the balanced equation for the complete combustion of ethene gas.

C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)

By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number volume ratios equal to their stoichiometric coefficients at constant temperature and pressure.

2

Determine the volume of oxygen required and the unreacted excess volume.

Volume of

O_2

consumed =

3 \times 15\text{ cm}^3 = 45\text{ cm}^3

. Remaining unreacted

O_2 = 50\text{ cm}^3 - 45\text{ cm}^3 = 5\text{ cm}^3

.

Since

15\text{ cm}^3

of

C_2H_4

requires only

45\text{ cm}^3

of

O_2

, oxygen is in excess and ethene is the limiting reactant.

3

Calculate the volume of carbon(IV) oxide gas produced.

Volume of

CO_2

produced =

2 \times 15\text{ cm}^3 = 30\text{ cm}^3

.

1 volume of

C_2H_4

produces 2 volumes of

CO_2

gas.

4

Calculate the total volume of the residual gaseous mixture after cooling to room temperature.

Total residual gaseous volume =

5\text{ cm}^3 \text{ (excess } O_2\text{)} + 30\text{ cm}^3 \text{ (produced } CO_2\text{)} = 35\text{ cm}^3

.

Water formed is in the liquid state at room temperature and contributes negligible volume to the gaseous mixture.

Key Concept

Gay-Lussac's Law of Combining Volumes and Residual Gas Volume Calculations

Estimated Time:1m 30s

Question 74Question →

A mixture of

10\text{ cm}^3

of hydrogen sulfide (

H_2S

) gas and

40\text{ cm}^3

of oxygen gas was sparked to react completely at constant temperature and pressure according to the equation:

2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(g) + 2H_2O(l)

Assuming the water formed condenses into liquid, what is the total volume of the residual gas mixture?

35\text{ cm}^3

25\text{ cm}^3

10\text{ cm}^3

45\text{ cm}^3

Show answer & explanation

Answer: $35\text{ cm}^3$

Answer

The total volume of the residual gas mixture is

35\text{ cm}^3

.

According to Gay-Lussac's law,

2\text{ cm}^3

of

H_2S

reacts with

3\text{ cm}^3

of

O_2

to form

2\text{ cm}^3

of

SO_2

gas. Thus,

10\text{ cm}^3

of

H_2S

consumes

15\text{ cm}^3

of

O_2

and yields

10\text{ cm}^3

of

SO_2

. The remaining excess

O_2

is

40\text{ cm}^3 - 15\text{ cm}^3 = 25\text{ cm}^3

. Combining the unreacted oxygen (

25\text{ cm}^3

) and produced sulfur(IV) oxide (

10\text{ cm}^3

) gives a total residual gas volume of

35\text{ cm}^3

.

Step-by-Step Solution

1

Determine the stoichiometric volume ratios from the balanced chemical equation

2 volumes of

H_2S(g)

react with 3 volumes of

O_2(g)

to produce 2 volumes of

SO_2(g)

.

According to Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number volume ratios at constant temperature and pressure.

2

Calculate the volume of oxygen required to react with

10\text{ cm}^3

of

H_2S

Volume of

O_2

used =

10\text{ cm}^3 \times \frac{3}{2} = 15\text{ cm}^3

.

Since

10\text{ cm}^3

of

H_2S

is available, it is the limiting reactant and requires

15\text{ cm}^3

of

O_2

.

3

Find the volume of unreacted excess oxygen

Unreacted

O_2 = 40\text{ cm}^3 - 15\text{ cm}^3 = 25\text{ cm}^3

.

Subtracting the reacted oxygen volume from the initial volume gives the excess.

4

Calculate the volume of gaseous

SO_2

produced

Volume of

SO_2 = 10\text{ cm}^3 \times \frac{2}{2} = 10\text{ cm}^3

.

The mole ratio of

H_2S

to

SO_2

is

2:2

, so

10\text{ cm}^3

of

H_2S

produces

10\text{ cm}^3

of

SO_2

gas. Water is liquid so its volume is neglected.

5

Calculate total residual gas volume

Total residual volume =

25\text{ cm}^3 \text{ (excess } O_2) + 10\text{ cm}^3 \text{ (produced } SO_2) = 35\text{ cm}^3

.

The residual gas consists of both unreacted excess reactant and gaseous product.

Key Concept

Gay-Lussac's Law of Combining Volumes

Question 75Question →

Fill in the blanks with the correct numerical values based on volume stoichiometry at constant temperature and pressure.

Fill in the blanks below

When

40\text{ cm}^3

of nitrogen(II) oxide gas (

NO

) is reacted with

30\text{ cm}^3

of oxygen gas (

O_2

) according to the equation

2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}

, the volume of unreacted oxygen gas remaining is

\text{cm}^3

and the total volume of the resulting gaseous mixture is

\text{cm}^3

.

Show answer & explanation

Answer

The volume of unreacted oxygen gas remaining is

10\text{ cm}^3

and the total volume of the resulting gaseous mixture is

50\text{ cm}^3

.

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

2\text{ volumes}

of

NO

react with

1\text{ volume}

of

O_2

to yield

2\text{ volumes}

of

NO_2

. Therefore,

40\text{ cm}^3

of

NO

reacts with

20\text{ cm}^3

of

O_2

, leaving

10\text{ cm}^3

of unreacted

O_2

(

30\text{ cm}^3 - 20\text{ cm}^3

). The reaction produces

40\text{ cm}^3

of

NO_2

gas. Summing the product volume (

40\text{ cm}^3

) and unreacted excess gas (

10\text{ cm}^3

) gives a total residual gas volume of

50\text{ cm}^3

.

Step-by-Step Solution

1

Determine combining volume ratios from the balanced chemical equation.

The mole/volume ratio is

2\text{ vol } NO : 1\text{ vol } O_2 : 2\text{ vol } NO_2

.

By Gay-Lussac's Law of Combining Volumes, gases at the same temperature and pressure react in volumes that bear simple whole-number ratios to one another and to gaseous products.

2

Identify the limiting reactant and calculate volumes consumed and produced.

40\text{ cm}^3

of

NO

requires

\frac{1}{2} \times 40 = 20\text{ cm}^3

of

O_2

, producing

40\text{ cm}^3

of

NO_2

.

NO

is completely used up.

40\text{ cm}^3

of

NO

is the limiting reactant because

30\text{ cm}^3

of

O_2

is available, which is more than the required

20\text{ cm}^3

.

3

Calculate remaining unreacted oxygen and total residual gas volume.

Excess

O_2 = 30\text{ cm}^3 - 20\text{ cm}^3 = 10\text{ cm}^3

. Total residual volume =

40\text{ cm}^3\,(NO_2) + 10\text{ cm}^3\,(O_2) = 50\text{ cm}^3

.

The total final volume equals the volume of gaseous product formed plus any unreacted excess gas.

Key Concept

Gay-Lussac's Law of Combining Volumes and Avogadro's Law

Estimated Time:1m 30s

Question 76Question →

An element $X$ forms two distinct gaseous oxides, Oxide A and Oxide B. Quantitative analysis reveals that $14.0\text{ g}$ of $X$ combines with $16.0\text{ g}$ of oxygen in Oxide A, whereas $14.0\text{ g}$ of $X$ combines with $32.0\text{ g}$ of oxygen in Oxide B. What is the ratio of the masses of oxygen combining with a fixed mass of element $X$ , and which law of chemical combination does this illustrate?

1:2

; Law of Multiple Proportions

1:2

; Law of Definite Proportions

7:8

; Law of Conservation of Mass

2:1

; Law of Combining Volumes

Show answer & explanation

Answer: $1:2$ ; Law of Multiple Proportions

Answer

The ratio of oxygen masses is

1:2

, which illustrates the Law of Multiple Proportions.

For a constant mass of element

X

(

14.0\text{ g}

), the mass of oxygen in Oxide A (

16.0\text{ g}

) and Oxide B (

32.0\text{ g}

) forms a simple whole-number ratio of

16:32 = 1:2

. This directly satisfies John Dalton's Law of Multiple Proportions.

Step-by-Step Solution

1

Identify the fixed mass of element

X

in both compounds

Mass of element

X = 14.0\text{ g}

in both Oxide A and Oxide B

To apply the Law of Multiple Proportions, the mass of one element must be held constant

2

Determine the masses of oxygen combined with the fixed mass of element

X

Oxide A has

16.0\text{ g}

of oxygen, Oxide B has

32.0\text{ g}

of oxygen

These are the given masses of oxygen reacting with

14.0\text{ g}

of

X

3

Calculate the simple ratio between the masses of oxygen

\frac{16.0}{32.0} = \frac{1}{2} \text{ or } 1:2

Dividing both masses by the common factor

16.0\text{ g}

yields a simple whole-number ratio

4

Match the observed phenomenon to the appropriate chemical law

Law of Multiple Proportions

When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers

Key Concept

Law of Multiple Proportions

Estimated Time:1m 30s

Question 77Question →

Phosphorus forms two distinct chlorides, X and Y. Analysis shows that $3.10\text{ g}$ of phosphorus combines with $10.65\text{ g}$ of chlorine to form compound X, whereas $3.10\text{ g}$ of phosphorus combines with $17.75\text{ g}$ of chlorine to form compound Y. Which law of chemical combination is illustrated by these experimental data, and what is the simple mass ratio of chlorine reacting with the fixed mass of phosphorus?

Law of Multiple Proportions with a ratio of 3 : 5

Law of Definite Proportions with a ratio of 3 : 5

Law of Multiple Proportions with a ratio of 2 : 3

Law of Conservation of Mass with a ratio of 1 : 2

Show answer & explanation

Answer: Law of Multiple Proportions with a ratio of 3 : 5

Answer

Law of Multiple Proportions with a ratio of 3 : 5

The option specifying the Law of Multiple Proportions with a ratio of 3 : 5 is correct because the mass of phosphorus is held constant at

3.10\text{ g}

, while the masses of chlorine in compounds X and Y are

10.65\text{ g}

and

17.75\text{ g}

respectively. Simplifying the ratio

\frac{10.65}{17.75}

yields

\frac{3}{5}

(

3 : 5

). Because two elements form different compounds whose mass ratios reduce to simple integers, the observation demonstrates John Dalton's Law of Multiple Proportions.

Step-by-Step Solution

1

Identify the fixed mass and variable mass values from the given data.

Fixed mass of phosphorus =

3.10\text{ g}

in both compounds. Mass of chlorine in compound X =

10.65\text{ g}

. Mass of chlorine in compound Y =

17.75\text{ g}

.

The Law of Multiple Proportions compares the varying masses of one element combining with a constant mass of another.

2

Calculate the simple whole-number ratio of the masses of chlorine.

\frac{10.65}{17.75} = \frac{3}{5} \implies 3 : 5

Dividing both chlorine masses by their greatest common factor (

3.55

) gives the integer ratio

3 : 5

.

3

Determine which law of chemical combination corresponds to this relationship.

Law of Multiple Proportions

When two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a ratio of small whole numbers.

Key Concept

Law of Multiple Proportions

Chemical Combination and Stoichiometry

Answer

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