Chemical Equations and Stoichiometric Calculations (Mass-Mass, Mass-Volume, Mole Relations) Practice Questions — JAMB UTME

A sample containing

8.0\text{ g}

of impure calcium trioxocarbonate(IV) reacts completely with excess dilute hydrochloric acid according to the equation:

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

If

1.344\text{ dm}^3

of carbon(IV) oxide gas measured at s.t.p. is liberated, what is the percentage purity of the calcium trioxocarbonate(IV) sample?

[\text{Ca} = 40, \text{C} = 12, \text{O} = 16, \text{Molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]

75.0\%

70.0\%

25.0\%

60.0\%

Show answer & explanation

Answer: $75.0\%$

Answer

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

75.0\%

.

First, find the moles of carbon(IV) oxide produced at s.t.p.:

1.344 / 22.4 = 0.060\text{ mol}

. According to the balanced chemical equation,

1\text{ mol}

of calcium trioxocarbonate(IV) reacts to yield

1\text{ mol}

of carbon(IV) oxide. Therefore,

0.060\text{ mol}

of pure calcium trioxocarbonate(IV) reacted. The mass of pure calcium trioxocarbonate(IV) is

0.060 \times 100\text{ g mol}^{-1} = 6.0\text{ g}

. Calculating percentage purity gives

(6.0\text{ g} / 8.0\text{ g}) \times 100\% = 75.0\%

.

Step-by-Step Solution

1

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

\text{Moles of CO}_2 = \frac{1.344\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.060\text{ mol}

Molar gas volume at s.t.p. equals

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

.

2

Determine the mole relation and calculate the mass of pure calcium trioxocarbonate(IV).

Molar mass of

\text{CaCO}_3 = 40 + 12 + (3 \times 16) = 100\text{ g mol}^{-1}

. From the

1:1

stoichiometric ratio,

\text{moles of pure CaCO}_3 = 0.060\text{ mol}

.

\text{Mass} = 0.060\text{ mol} \times 100\text{ g mol}^{-1} = 6.0\text{ g}

.

The balanced chemical equation shows a 1:1 mole ratio between calcium trioxocarbonate(IV) and carbon(IV) oxide.

3

Calculate the percentage purity of the sample.

\text{Percentage purity} = \frac{\text{Mass of pure CaCO}_3}{\text{Total mass of impure sample}} \times 100\% = \frac{6.0\text{ g}}{8.0\text{ g}} \times 100\% = 75.0\%

Percentage purity expresses the mass fraction of pure active component relative to total sample mass.

Key Concept

Mass-volume stoichiometric calculations and percentage purity determination

Estimated Time:2m 0s

Question 2Question →

What volume of hydrogen gas measured at s.t.p. is evolved when

13.0\text{ g}

of pure zinc completely reacts with excess dilute hydrochloric acid according to the equation below?

\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)

[\text{Zn} = 65, \text{Molar gas volume at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]

4.48\text{ dm}^3

4.80\text{ dm}^3

8.96\text{ dm}^3

2.24\text{ dm}^3

Show answer & explanation

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

Answer

4.48\text{ dm}^3

The balanced chemical equation indicates a 1:1 molar relationship between zinc and hydrogen gas. Reacting

13.0\text{ g}

of zinc corresponds to

\frac{13.0}{65} = 0.20\text{ mol}

of zinc, yielding

0.20\text{ mol}

of hydrogen gas. Multiplying by the molar volume at s.t.p. (

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

) gives

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

.

Step-by-Step Solution

1

Calculate the number of moles of zinc reacted

\text{Moles of Zn} = \frac{13.0\text{ g}}{65\text{ g mol}^{-1}} = 0.20\text{ mol}

Converting given mass of reactant to moles using relative atomic mass.

2

Determine the moles of hydrogen gas produced from the balanced equation

\text{Mole ratio of Zn to H}_2 = 1 : 1

, so

\text{moles of H}_2 = 0.20\text{ mol}

Applying mole ratio constraints from the chemical equation.

3

Calculate the volume of hydrogen gas at s.t.p.

\text{Volume of H}_2 = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 4.48\text{ dm}^3

Multiplying moles of gas by standard molar volume at s.t.p.

Key Concept

Mass-Volume stoichiometric calculation at standard temperature and pressure (s.t.p.)

Estimated Time:1m 0s

Question 3Question →

What volume of hydrogen gas, measured at STP, is liberated when

5.4\text{ g}

of aluminium react completely with excess hydrochloric acid according to the following equation?

2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{AlCl}_3(aq) + 3\text{H}_2(g)

[Relative atomic mass:

\text{Al} = 27

; Molar volume of gas at STP

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

]

4.48\text{ dm}^3

6.72\text{ dm}^3

7.20\text{ dm}^3

13.44\text{ dm}^3

Show answer & explanation

Answer: $6.72\text{ dm}^3$

Answer

The volume of hydrogen gas liberated at STP is

6.72\text{ dm}^3

.

The option specifying

6.72\text{ dm}^3

is correct because

5.4\text{ g}

of aluminium corresponds to

0.20\text{ mol}

. Based on the stoichiometric ratio

2\text{Al}:3\text{H}_2

,

0.20\text{ mol}

of aluminium liberates

0.30\text{ mol}

of hydrogen gas. Multiplying

0.30\text{ mol}

by the molar gas volume at STP (

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

) yields exactly

6.72\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the amount in moles of aluminium that reacted

\text{Moles of Al} = \frac{5.4\text{ g}}{27\text{ g mol}^{-1}} = 0.20\text{ mol}

Converting mass of reactant to moles is required to apply the stoichiometric mole ratio.

2

Determine the moles of hydrogen gas produced using the balanced chemical equation

\text{Moles of H}_2 = 0.20\text{ mol Al} \times \frac{3\text{ mol H}_2}{2\text{ mol Al}} = 0.30\text{ mol H}_2

The balanced chemical equation shows that

2\text{ mol}

of

\text{Al}

produce

3\text{ mol}

of

\text{H}_2

.

3

Calculate the volume of hydrogen gas produced at STP

\text{Volume of H}_2 = 0.30\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 6.72\text{ dm}^3

At STP, one mole of any ideal gas occupies

22.4\text{ dm}^3

.

Key Concept

Mass-Volume Stoichiometric Calculations at STP

Question 4Question →

Consider the thermal decomposition of potassium trioxochlorate(V) represented by the balanced equation:

2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)

What mass of

KClO_3

is required to produce

6.72\text{ dm}^3

of oxygen gas measured at STP?

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

24.5\text{ g}

12.25\text{ g}

36.75\text{ g}

22.87\text{ g}

Show answer & explanation

Answer: $24.5\text{ g}$

Answer

24.5\text{ g}

of

KClO_3

is required.

According to the balanced chemical equation,

2\text{ moles}

of

KClO_3

(

245.0\text{ g}

) produce

3\text{ moles}

of

O_2

(

67.2\text{ dm}^3

at STP). By direct proportion,

6.72\text{ dm}^3

of

O_2

requires

\frac{6.72}{67.2} \times 245.0 = 24.5\text{ g}

of

KClO_3

.

Step-by-Step Solution

1

Calculate the molar mass of

KClO_3

and the total mass of 2 moles of

KClO_3

.

Molar mass of

KClO_3 = 39.0 + 35.5 + 3(16.0) = 122.5\text{ g mol}^{-1}

. Mass of

2\text{ moles} = 2 \times 122.5 = 245.0\text{ g}

.

The balanced chemical equation shows

2\text{ moles}

of

KClO_3

undergo decomposition.

2

Calculate the volume of 3 moles of

O_2

at STP.

Volume of

3\text{ moles of } O_2 = 3 \times 22.4\text{ dm}^3 = 67.2\text{ dm}^3

.

At STP,

1\text{ mole}

of any gas occupies

22.4\text{ dm}^3

.

3

Set up a proportion to find the mass of

KClO_3

needed to yield

6.72\text{ dm}^3

of

O_2

.

\text{Mass of } KClO_3 = \frac{6.72\text{ dm}^3}{67.2\text{ dm}^3} \times 245.0\text{ g} = 24.5\text{ g}.

Direct stoichiometric ratio relates mass of reactant to volume of gaseous product.

Key Concept

Mass-Volume Stoichiometry at STP

Question 5Question →

When

6.62\text{ g}

of lead(II) trioxonitrate(V) is completely decomposed by heating according to the balanced chemical equation:

2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)

What is the total volume of gaseous products liberated at STP?
[

1\text{ mole of gas at STP} = 22.4\text{ dm}^3

; relative atomic masses:

Pb = 207

,

N = 14

,

O = 16

]

0.896\text{ dm}^3

1.12\text{ dm}^3

1.20\text{ dm}^3

0.224\text{ dm}^3

Show answer & explanation

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

Answer

1.12\text{ dm}^3

The correct answer is

1.12\text{ dm}^3

. Decomposing

6.62\text{ g}

(

0.02\text{ mol}

) of

Pb(NO_3)_2

yields

0.04\text{ mol}

of

NO_2

and

0.01\text{ mol}

of

O_2

, totaling

0.05\text{ mol}

of 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 molar mass of lead(II) trioxonitrate(V),

Pb(NO_3)_2

.

\text{Molar Mass} = 207 + 2 \times (14 + 3 \times 16) = 331\text{ g mol}^{-1}

.

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

2

Determine the amount (in moles) of

Pb(NO_3)_2

reacted.

n(Pb(NO_3)_2) = \frac{6.62\text{ g}}{331\text{ g mol}^{-1}} = 0.02\text{ mol}

.

Quantitative stoichiometric relations require knowing the exact mole quantity of the reactant.

3

Identify the total mole ratio of gaseous products to reactant from the balanced chemical equation.

2\text{ moles } Pb(NO_3)_2 \rightarrow 4\text{ moles } NO_2(g) + 1\text{ mole } O_2(g) = 5\text{ moles of total gas}

. Total gas mole ratio

= \frac{5}{2} = 2.5

.

Both

NO_2

and

O_2

are gases at STP, so both contribute to the total volume evolved.

4

Calculate the total moles and volume of gas liberated at STP.

\text{Total moles of gas} = 0.02 \times 2.5 = 0.05\text{ mol}

.

\text{Total volume at STP} = 0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

.

Multiplying total gaseous moles by the standard molar gas volume gives the total volume at STP.

Key Concept

Mass-Volume Stoichiometric Calculation for Reaction Systems Yielding Multiple Gaseous Products

Question 6Question →

What volume of hydrogen gas, measured at STP, is produced when $4.8\text{ g}$ of magnesium ribbon reacts completely with excess dilute tetraoxosulfate(VI) acid according to the chemical equation below?

Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g)

$[Mg = 24, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]$

Show answer & explanation

Answer: 4.48

Answer

The volume of hydrogen gas produced at STP is

4.48\text{ dm}^3

.

From the stoichiometric relationship in the balanced reaction

Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g)

,

1\text{ mol}

of

Mg

(

24\text{ g}

) produces

1\text{ mol}

of

H_2

gas (

22.4\text{ dm}^3

at STP). For

4.8\text{ g}

of

Mg

, the number of moles is

\frac{4.8}{24} = 0.20\text{ mol}

. Multiplying by the molar gas volume gives

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

of

H_2

gas.

Step-by-Step Solution

1

Calculate the amount in moles of magnesium (

Mg

) reacted.

n(Mg) = \frac{4.8\text{ g}}{24\text{ g mol}^{-1}} = 0.20\text{ mol}

Dividing given mass by relative atomic mass yields the number of moles.

2

Determine the amount in moles of hydrogen gas (

H_2

) produced.

n(H_2) = 0.20\text{ mol}

The balanced chemical equation shows a

1:1

stoichiometric molar ratio between

Mg

and

H_2

.

3

Calculate the volume of

H_2

gas at STP.

V(H_2) = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

At standard temperature and pressure (STP), one mole of any gas occupies

22.4\text{ dm}^3

.

Key Concept

Mass-Volume stoichiometric calculation at STP

Question 7Question →

Consider the balanced chemical equation for the complete combustion of ethane gas:

2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)

What volume of oxygen gas, measured at STP, is required to react completely with

0.5\text{ mol}

of ethane? Complete the statement below with the calculated numerical value.

Fill in the blanks below

The volume of oxygen gas required at STP is

\text{dm}^3

. (Molar gas volume at STP =

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

)

Show answer & explanation

Answer

39.2 dm³ (or 39.2)

According to the balanced equation

2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)

,

2\text{ moles}

of ethane require

7\text{ moles}

of oxygen gas for complete combustion. Therefore,

0.5\text{ mol}

of ethane requires

\frac{0.5 \times 7}{2} = 1.75\text{ mol}

of oxygen gas. At standard temperature and pressure (STP),

1\text{ mole}

of gas occupies

22.4\text{ dm}^3

. Thus, the volume of oxygen gas required is

1.75\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 39.2\text{ dm}^3

.

Step-by-Step Solution

1

Identify the stoichiometric mole ratio between ethane and oxygen from the balanced equation

From

2C_2H_6 + 7O_2

, the mole ratio of

C_2H_6

to

O_2

is

2 : 7

.

Stoichiometric coefficients represent the relative mole proportions of reactants.

2

Calculate the moles of oxygen gas needed for 0.5 mol of ethane

Moles of

O_2 = 0.5\text{ mol} \times \frac{7}{2} = 1.75\text{ mol}

Multiplying the given amount of ethane by the stoichiometric factor gives the required moles of oxygen.

3

Convert the calculated moles of oxygen gas to volume at STP

Volume of

O_2 = 1.75\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 39.2\text{ dm}^3

One mole of any ideal gas occupies

22.4\text{ dm}^3

at STP.

Key Concept

Mole-Volume Stoichiometric Calculations at STP

Question 8Question →

Sodium hydrogentrioxocarbonate(IV) decomposes upon heating according to the balanced chemical equation:

2NaHCO_3(s) \rightarrow Na_2CO_3(s) + H_2O(l) + CO_2(g)

What mass of sodium trioxocarbonate(IV) (

Na_2CO_3

) is produced by the complete decomposition of

16.8\text{ g}

of

NaHCO_3

?

[Na = 23, C = 12, O = 16, H = 1]

10.6\text{ g}

21.2\text{ g}

4.4\text{ g}

42.4\text{ g}

Show answer & explanation

Answer: $10.6\text{ g}$

Answer

The mass of sodium trioxocarbonate(IV) produced is

10.6\text{ g}

.

The complete decomposition of

16.8\text{ g}

(

0.20\text{ mol}

) of

NaHCO_3

produces

0.10\text{ mol}

of

Na_2CO_3

according to the

2:1

mole ratio in the balanced equation. Multiplying

0.10\text{ mol}

by the molar mass of

Na_2CO_3

(

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

) gives

10.6\text{ g}

.

Step-by-Step Solution

1

Calculate the molar masses of

NaHCO_3

and

Na_2CO_3

Molar mass of

NaHCO_3 = 23 + 1 + 12 + 3(16) = 84\text{ g mol}^{-1}

. Molar mass of

Na_2CO_3 = 2(23) + 12 + 3(16) = 106\text{ g mol}^{-1}

.

Molar masses are required to convert between mass and mole amounts.

2

Determine the number of moles of

NaHCO_3

reacted

\text{Moles of } NaHCO_3 = \frac{16.8\text{ g}}{84\text{ g mol}^{-1}} = 0.20\text{ mol}.

Stoichiometric relations are calculated using mole amounts rather than raw masses.

3

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

Na_2CO_3

From

2NaHCO_3 \rightarrow Na_2CO_3

, ratio is

2:1

. Therefore, \text{moles of } Na_2CO_3 = \frac{0.20}{2} = 0.10\text{ mol}.

Two moles of

NaHCO_3

produce one mole of

Na_2CO_3

.

4

Calculate the mass of

Na_2CO_3

produced

\text{Mass of } Na_2CO_3 = 0.10\text{ mol} \times 106\text{ g mol}^{-1} = 10.6\text{ g}.

Multiplying moles of product by its molar mass yields the theoretical yield mass.

Key Concept

Mass-Mass Stoichiometric Calculations

Estimated Time:1m 30s

Question 9Question →

What volume of chlorine gas, measured at STP, is required for the complete reaction with a given mass of iron metal?

Fill in the blanks below

When

11.2\text{ g}

of iron reacts completely with excess dry chlorine gas according to the balanced equation:

2Fe(s) + 3Cl_2(g) \rightarrow 2FeCl_3(s)

the volume of chlorine gas consumed at STP is

\text{dm}^3

.
[Relative atomic mass:

\text{Fe} = 56

; Molar gas volume at STP

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

]

Show answer & explanation

Answer

The volume of chlorine gas consumed at STP is

6.72\text{ dm}^3

.

To find the volume of

Cl_2

gas consumed at STP, first determine the amount of

Fe

in moles:

\text{Moles of } Fe = \frac{11.2\text{ g}}{56\text{ g mol}^{-1}} = 0.20\text{ mol}

. From the balanced equation

2Fe(s) + 3Cl_2(g) \rightarrow 2FeCl_3(s)

,

2\text{ moles of } Fe

react with

3\text{ moles of } Cl_2

. Therefore,

0.20\text{ mol of } Fe

requires

0.20 \times \frac{3}{2} = 0.30\text{ mol of } Cl_2

. At STP,

1\text{ mole of gas}

occupies

22.4\text{ dm}^3

, so the volume of

Cl_2

is

0.30\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 6.72\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the number of moles of iron reacted

\text{Moles of } Fe = \frac{11.2\text{ g}}{56\text{ g mol}^{-1}} = 0.20\text{ mol}

Convert the mass of iron to moles using its relative atomic mass.

2

Determine the moles of chlorine gas (

Cl_2

) required using the stoichiometric mole ratio

\text{Moles of } Cl_2 = 0.20\text{ mol } Fe \times \frac{3\text{ mol } Cl_2}{2\text{ mol } Fe} = 0.30\text{ mol}

The balanced chemical equation shows that

2\text{ moles of } Fe

react with

3\text{ moles of } Cl_2

(a

2:3

mole ratio).

3

Calculate the volume of

Cl_2

gas consumed at STP

\text{Volume of } Cl_2 = 0.30\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 6.72\text{ dm}^3

Multiply the number of moles of

Cl_2

by the molar volume of a gas at STP.

Key Concept

Mass-Volume Stoichiometric Calculations at STP

Estimated Time:2m 0s

Question 10Question →

Complete the statement by calculating the required volume of gas at STP and the mass of metal produced in the following reduction reaction.

Fill in the blanks below

When

16.0\text{ g}

of iron(III) oxide (

Fe_2O_3

) is completely reduced by carbon monoxide gas according to the equation:

Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)

the volume of carbon monoxide gas consumed at STP is , and the mass of iron metal produced is .

[Relative atomic masses:

\text{Fe} = 56

,

\text{O} = 16

; Molar volume of gas at STP

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

]

Show answer & explanation

Answer

The volume of carbon monoxide consumed at STP is

6.72\text{ dm}^3

and the mass of iron metal produced is

11.2\text{ g}

.

Based on the stoichiometry of the reaction

Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)

,

1\text{ mol}

(

160\text{ g}

) of

Fe_2O_3

reacts with

3\text{ mol}

(

0.30\text{ mol}

for

16.0\text{ g}

) of

CO

gas, yielding a volume of

6.72\text{ dm}^3

at STP, and produces

2\text{ mol}

(

0.20\text{ mol}

for

16.0\text{ g}

) of

Fe

, corresponding to

11.2\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of

Fe_2O_3

and determine the number of moles of

Fe_2O_3

present.

Molar mass of

Fe_2O_3 = (2 \times 56) + (3 \times 16) = 160\text{ g mol}^{-1}

. Moles of

Fe_2O_3 = \frac{16.0\text{ g}}{160\text{ g mol}^{-1}} = 0.10\text{ mol}

.

Converting given mass to moles is required for stoichiometric mole-ratio calculations.

2

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

CO(g)

required at STP.

From the balanced equation,

1\text{ mol } Fe_2O_3

reacts with

3\text{ mol } CO

.
Moles of

CO = 3 \times 0.10\text{ mol} = 0.30\text{ mol}

.
Volume of

CO

at STP

= 0.30\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 6.72\text{ dm}^3

.

Gas volume at STP is obtained by multiplying moles of gas by the molar volume (

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

).

3

Use the mole ratio from the balanced equation to calculate the mass of iron (

Fe

) produced.

From the equation,

1\text{ mol } Fe_2O_3

produces

2\text{ mol } Fe

.
Moles of

Fe = 2 \times 0.10\text{ mol} = 0.20\text{ mol}

.
Mass of

Fe = 0.20\text{ mol} \times 56\text{ g mol}^{-1} = 11.2\text{ g}

.

Mass of product is calculated by multiplying its moles by its relative atomic mass.

Key Concept

Mass-Mass and Mass-Volume Stoichiometric Calculations

Question 11Question →

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 12Question →

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 13Question →

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

Chemical Equations and Stoichiometric Calculations (Mass-Mass, Mass-Volume, Mole Relations)

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