Chemical Combination and Stoichiometry Practice Questions — JAMB UTME — Page 2

Question 21Question →

What is the percentage by mass of water of crystallization in copper(II) tetraoxosulfate(VI) pentahydrate ( $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ )? $[\text{Cu} = 64, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

7.2\%

36.0\%

56.3\%

64.0\%

Show answer & explanation

Answer: $36.0\%$

Answer

The percentage by mass of water of crystallization in

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

is

36.0\%

.

The total molar mass of

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

is

250\text{ g/mol}

and the mass contributed by the five water molecules is

90\text{ g/mol}

. Dividing

90

by

250

and multiplying by

100\%

yields

36.0\%

.

Step-by-Step Solution

1

Calculate the molar mass of water (

\text{H}_2\text{O}

) and the total mass of five moles of water.

Molar mass of

\text{H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}

. For

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

, mass

= 5 \times 18 = 90\text{ g/mol}

.

Water of crystallization in the formula consists of five water molecules per formula unit.

2

Calculate the total molar mass of hydrated copper(II) tetraoxosulfate(VI) (

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

).

Molar mass

= 64 + 32 + (4 \times 16) + 90 = 64 + 32 + 64 + 90 = 250\text{ g/mol}

.

The percentage composition must be based on the complete formula weight of the hydrated compound.

3

Calculate the percentage by mass of water of crystallization.

\text{Percentage of } \text{H}_2\text{O} = \left(\frac{90}{250}\right) \times 100\% = 36.0\%

.

Percentage composition by mass is the mass of the component divided by the total molar mass of the compound multiplied by 100.

Key Concept

Percentage Water of Crystallization in Hydrated Salts

Question 22Question →

A $14.3\text{ g}$ sample of washing soda crystals, hydrated sodium trioxocarbonate(IV) ( $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ ), is completely dissolved in water. Calculate the total number of moles of oxygen atoms present in this sample. [Relative atomic masses: $\text{Na} = 23$ , $\text{C} = 12$ , $\text{O} = 16$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: 0.65

Answer

The total number of moles of oxygen atoms present in the sample is 0.65 mol.

The molar mass of

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

is

286\text{ g/mol}

. A mass of

14.3\text{ g}

corresponds to

\frac{14.3}{286} = 0.05\text{ mol}

of the compound. Since each formula unit contains 13 oxygen atoms (3 from the trioxocarbonate anion and 10 from the water of crystallization), the total moles of oxygen atoms present is

0.05 \times 13 = 0.65\text{ mol}

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

.

Molar mass =

286\text{ g/mol}

.

Summing the atomic masses of all constituent atoms, including the 10 water molecules of crystallization:

(2 \times 23) + 12 + (3 \times 16) + 10 \times (18) = 286\text{ g/mol}

.

2

Calculate the number of moles of

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

present in the sample.

Moles of compound =

0.05\text{ mol}

.

Dividing the given sample mass by its molar mass:

\frac{14.3\text{ g}}{286\text{ g/mol}} = 0.05\text{ mol}

.

3

Determine the stoichiometric ratio of oxygen atoms per mole of hydrated compound.

13 moles of O atoms per 1 mole of

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

.

Each formula unit contains 3 oxygen atoms from the trioxocarbonate group and 10 oxygen atoms from the ten water of crystallization molecules (

3 + 10 = 13

).

4

Multiply the moles of compound by the number of oxygen atoms per formula unit.

Total moles of O atoms =

0.65\text{ mol}

.

Calculating total oxygen moles:

0.05\text{ mol} \times 13 = 0.65\text{ mol}

.

Key Concept

The Mole Concept, Molar Mass, and Stoichiometric Ratios in Hydrated Salts

Question 23Question →

What is the total number of ions present in a $9.5\text{ g}$ sample of pure magnesium chloride ( $\text{MgCl}_2$ )? $[\text{Mg} = 24, \text{Cl} = 35.5, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

1.81 \times 10^{23}

6.02 \times 10^{22}

1.20 \times 10^{23}

2.24 \times 10^{23}

Show answer & explanation

Answer: $1.81 \times 10^{23}$

Answer

1.81 \times 10^{23}

ions

The sample contains

0.1\text{ mol}

of

\text{MgCl}_2

. Because each formula unit of

\text{MgCl}_2

dissociates into 3 ions (

1\text{ Mg}^{2+}

and

2\text{ Cl}^-

), there are

0.3\text{ mol}

of total ions. Multiplying

0.3\text{ mol}

by Avogadro's constant (

6.02 \times 10^{23}

) yields

1.81 \times 10^{23}

ions.

Step-by-Step Solution

1

Calculate the molar mass of magnesium chloride (

\text{MgCl}_2

)

Molar mass

= 24 + 2(35.5) = 95\text{ g/mol}

Molar mass is required to convert the given mass into moles.

2

Determine the number of moles of

\text{MgCl}_2

in

9.5\text{ g}

Moles

= 9.5\text{ g} / 95\text{ g/mol} = 0.1\text{ mol}

Applying the mole formula:

\text{moles} = \text{mass} / \text{molar mass}

.

3

Determine the total moles of ions released upon dissociation

\text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^-

, giving 3 moles of ions per mole of salt. Total ion moles

= 0.1 \times 3 = 0.3\text{ mol}

Each formula unit of magnesium chloride contains one magnesium ion and two chloride ions.

4

Calculate the total number of individual ions using Avogadro's constant

Total ions

= 0.3 \times 6.02 \times 10^{23} = 1.806 \times 10^{23} \approx 1.81 \times 10^{23}

ions

Multiplying the total moles of ions by Avogadro's constant yields particle count.

Key Concept

The Mole Concept and Avogadro's Constant applied to Ionic Dissociation

Estimated Time:1m 30s

Question 24Question →

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 25Question →

Calculate the mass, in grams, of nitrogen contained in a $16.4\text{ g}$ sample of pure calcium trioxonitrate(V), $\text{Ca(NO}_3)_2$ . $[\text{Ca} = 40, \text{N} = 14, \text{O} = 16]$

Show answer & explanation

Answer: 2.8

Answer

The mass of nitrogen in the sample is

2.8\text{ g}

.

The molar mass of

\text{Ca(NO}_3)_2

is calculated as

40 + 2(14 + 3 \times 16) = 164\text{ g/mol}

. A

16.4\text{ g}

sample corresponds to

\frac{16.4}{164} = 0.1\text{ mol}

of

\text{Ca(NO}_3)_2

. Because each formula unit contains 2 nitrogen atoms,

0.1\text{ mol}

of compound yields

0.2\text{ mol}

of nitrogen. Multiplying

0.2\text{ mol}

by the molar mass of atomic nitrogen (

14\text{ g/mol}

) gives

2.8\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of calcium trioxonitrate(V),

\text{Ca(NO}_3)_2

164\text{ g/mol}

Sum the relative atomic masses of all atoms present:

40 + 2(14 + 3 \times 16) = 164\text{ g/mol}

.

2

Calculate the number of moles of

\text{Ca(NO}_3)_2

present in

16.4\text{ g}

0.1\text{ mol}

Use the formula

\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{16.4\text{ g}}{164\text{ g/mol}} = 0.1\text{ mol}

.

3

Determine the number of moles of nitrogen atoms in

0.1\text{ mol}

of

\text{Ca(NO}_3)_2

0.2\text{ mol}

of N atoms

Each formula unit of

\text{Ca(NO}_3)_2

contains 2 nitrogen atoms.

4

Calculate the mass of the nitrogen atoms

2.8\text{ g}

Multiply the moles of nitrogen by its atomic mass:

0.2\text{ mol} \times 14\text{ g/mol} = 2.8\text{ g}

.

Key Concept

Mole Concept and Mass Composition of Compounds

Estimated Time:1m 30s

Question 26Question →

An organic compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. If the vapour density of the compound is $30$ , what is its molecular formula? $[C = 12, H = 1, O = 16]$

CH_2O

C_2H_4O_2

C_3H_6O_3

C_4H_8O_4

Show answer & explanation

Answer: $C_2H_4O_2$

Answer

The molecular formula of the compound is

C_2H_4O_2

.

Dividing mass percentage by relative atomic mass yields a

1:2:1

mole ratio for

\text{C}:\text{H}:\text{O}

, establishing an empirical formula of

CH_2O

(mass

= 30\text{ g/mol}

). Multiplying the vapour density (

30

) by

2

gives a molar mass of

60\text{ g/mol}

. The molecular formula multiplier

n = 60 / 30 = 2

, yielding

C_2H_4O_2

.

Step-by-Step Solution

1

Determine the mole ratio of each constituent element

Moles of

C = \frac{40.0}{12} = 3.33

, Moles of

H = \frac{6.7}{1} = 6.70

, Moles of

O = \frac{53.3}{16} = 3.33

Mass percentages are divided by their respective relative atomic masses to yield mole quantities.

2

Calculate the simplest whole-number ratio to obtain the empirical formula

Ratio

C : H : O = \frac{3.33}{3.33} : \frac{6.70}{3.33} : \frac{3.33}{3.33} = 1 : 2 : 1

. Empirical formula is

CH_2O

.

Dividing all mole values by the smallest value (

3.33

) converts the mole ratio into simple integers.

3

Compute the molar mass from the given vapour density

\text{Molar Mass} = 2 \times \text{Vapour Density} = 2 \times 30 = 60\text{ g/mol}

The molar mass of a volatile substance is equal to twice its vapour density.

4

Find the molecular formula multiplier

n

and determine the molecular formula

Empirical formula mass of

CH_2O = 12 + (2 \times 1) + 16 = 30\text{ g/mol}

.

n = \frac{60}{30} = 2

. Molecular formula =

(CH_2O)_2 = C_2H_4O_2

.

The molecular formula is obtained by multiplying the subscripts of the empirical formula by

n

, where

n = \frac{\text{Molar Mass}}{\text{Empirical Mass}}

.

Key Concept

Deriving Empirical and Molecular Formulae using Vapour Density

Question 27Question →

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 28Question →

A hydrocarbon was analyzed and found to contain $85.7\%$ carbon and $14.3\%$ hydrogen by mass. What is the empirical formula of the hydrocarbon? [Relative atomic masses: $\text{C} = 12$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: CH2; CH_2; CH₂

Answer

CH₂

By dividing the mass percentage of each element by its relative atomic mass (Carbon:

85.7/12 = 7.142

, Hydrogen:

14.3/1 = 14.3

), the mole ratio obtained is

1:2

after dividing by the smallest value (

7.142

). Hence, the empirical formula is

\text{CH}_2

.

Step-by-Step Solution

1

Calculate the mole ratio of each element by dividing their mass percentages by their respective relative atomic masses.

Moles of Carbon =

\frac{85.7}{12} = 7.142\text{ mol}

; Moles of Hydrogen =

\frac{14.3}{1} = 14.3\text{ mol}

Molar mass converts mass percentages into relative molar quantities.

2

Divide each mole value by the smallest number of moles to obtain the simplest whole-number ratio.

Carbon ratio =

\frac{7.142}{7.142} = 1

; Hydrogen ratio =

\frac{14.3}{7.142} = 2

Empirical formula represents the simplest whole-number ratio of atoms in a compound.

3

Write the chemical formula using the calculated whole-number ratio.

Empirical Formula =

\text{CH}_2

Combining the simplest ratio yields 1 atom of C for every 2 atoms of H.

Key Concept

Empirical Formula Calculation from Percentage Composition

Estimated Time:1m 30s

Question 29Question →

A

5.00\text{ g}

sample of impure sodium trioxocarbonate(IV),

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

, reacted completely with excess dilute hydrochloric acid to produce

0.896\text{ dm}^3

of carbon(IV) oxide gas at STP according to the equation:

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

What is the percentage purity of the sodium trioxocarbonate(IV) sample?

[\text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}, \text{Na} = 23, \text{C} = 12, \text{O} = 16]

84.8%

79.1%

66.4%

42.4%

Show answer & explanation

Answer: 84.8%

Answer

The percentage purity of the sodium trioxocarbonate(IV) sample is 84.8%.

The correct answer of 84.8% is obtained by converting the volume of carbon(IV) oxide gas at STP to moles using the molar volume of 22.4 dm³/mol, determining the equivalent mass of pure sodium trioxocarbonate(IV) using its molar mass (106 g/mol), and finding its proportion relative to the 5.00 g sample mass.

Step-by-Step Solution

1

Calculate the number of moles of carbon(IV) oxide gas produced at STP.

n(\text{CO}_2) = \frac{0.896\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.04\text{ mol}

At STP, one mole of any gas occupies a volume of

22.4\text{ dm}^3

.

2

Determine the molar mass and the mass of pure sodium trioxocarbonate(IV).

Molar mass of

\text{Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106\text{ g/mol}

. Mass of pure

\text{Na}_2\text{CO}_3 = 0.04\text{ mol} \times 106\text{ g/mol} = 4.24\text{ g}

.

From the balanced chemical equation,

1\text{ mol}

of

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

yields

1\text{ mol}

of

\text{CO}_2

.

3

Calculate the percentage purity of the sample.

\text{Percentage Purity} = \left(\frac{\text{Mass of pure } \text{Na}_2\text{CO}_3}{\text{Mass of impure sample}}\right) \times 100\% = \left(\frac{4.24\text{ g}}{5.00\text{ g}}\right) \times 100\% = 84.8\%

.

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

Key Concept

Percentage Purity from Gas Stoichiometry

Estimated Time:1m 30s

Question 30Question →

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 31Question →

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 32Question →

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 33Question →

A mixture containing

28\text{ g}

of nitrogen gas (

N_2

) and

12\text{ g}

of hydrogen gas (

H_2

) reacts to completion according to the equation:

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

What mass of the excess reactant remains unreacted at the end of the reaction?

[N = 14,\ H = 1]

6.0\text{ g}

16.0\text{ g}

10.0\text{ g}

3.0\text{ g}

Show answer & explanation

Answer: $6.0\text{ g}$

Answer

The mass of the excess reactant remaining unreacted is

6.0\text{ g}

.

The correct answer is

6.0\text{ g}

.

28\text{ g}

of

N_2

corresponds to

1.0\text{ mol}

, while

12\text{ g}

of

H_2

corresponds to

6.0\text{ mol}

. According to the equation

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

,

1.0\text{ mol}

of

N_2

consumes

3.0\text{ mol}

of

H_2

. Therefore,

N_2

limits the reaction, and

3.0\text{ mol}

of

H_2

(

6.0\text{ g}

) remains unreacted.

Step-by-Step Solution

1

Calculate the initial number of moles for each reactant.

\text{Moles of } N_2 = \frac{28\text{ g}}{28\text{ g/mol}} = 1.0\text{ mol}

;

\text{Moles of } H_2 = \frac{12\text{ g}}{2\text{ g/mol}} = 6.0\text{ mol}

.

Converting given masses to moles is necessary to apply stoichiometric coefficients.

2

Determine the limiting reactant using the mole ratio from the balanced equation.

The reaction ratio is

1\text{ mol } N_2 : 3\text{ mol } H_2

.

1.0\text{ mol } N_2

requires

3.0\text{ mol } H_2

. Since

6.0\text{ mol } H_2

is available,

N_2

is the limiting reactant and

H_2

is in excess.

Comparing available mole ratios against stoichiometric requirements identifies which reactant is completely consumed.

3

Calculate the unreacted moles and mass of the excess reactant (

H_2

).

\text{Unreacted moles of } H_2 = 6.0\text{ mol} - 3.0\text{ mol} = 3.0\text{ mol}

.

\text{Unreacted mass} = 3.0\text{ mol} \times 2\text{ g/mol} = 6.0\text{ g}

.

Multiplying unreacted moles by the molar mass of

H_2

yields the remaining mass.

Key Concept

Limiting and Excess Reactants in Stoichiometry

Estimated Time:1m 30s

Question 34Question →

A naturally occurring sample of boron consists of two stable isotopes, $^{10}\text{B}$ and $^{11}\text{B}$ . If the relative percentage abundance of $^{10}\text{B}$ is $20.0\%$ and that of $^{11}\text{B}$ is $80.0\%$ , what is the relative atomic mass of boron?

Show answer & explanation

Answer: 10.8

Answer

The relative atomic mass of boron is

10.8

.

The relative atomic mass of an element is defined as the weighted average mass of its naturally occurring isotopes relative to

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

the mass of a carbon-12 atom. Applying the formula

\text{RAM} = \frac{\sum (\text{isotopic mass} \times \% \text{ abundance})}{100}

, we get

\frac{(10 \times 20) + (11 \times 80)}{100} = \frac{200 + 880}{100} = 10.8

.

Step-by-Step Solution

1

Determine the mass contribution of the

^{10}\text{B}

isotope

10 \times 0.20 = 2.0

The weighted contribution of an isotope is its mass multiplied by its fractional abundance.

2

Determine the mass contribution of the

^{11}\text{B}

isotope

11 \times 0.80 = 8.8

The weighted contribution of the second isotope is calculated using its percentage abundance.

3

Sum the weighted contributions to find the relative atomic mass

2.0 + 8.8 = 10.8

The relative atomic mass of an element is the weighted average mass of all naturally occurring isotopes relative to carbon-12.

Key Concept

Calculation of Relative Atomic Mass from Isotopic Abundances

Question 35Question →

A sample of pure ammonia gas ( $\text{NH}_3$ ) has a mass of $3.4\text{ g}$ . What is the total number of atoms contained in this sample? $[\text{N} = 14, \text{H} = 1, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

4.816 \times 10^{23}

1.204 \times 10^{23}

3.612 \times 10^{23}

2.408 \times 10^{23}

Show answer & explanation

Answer: $4.816 \times 10^{23}$

Answer

4.816 \times 10^{23}

atoms

The molar mass of ammonia (

\text{NH}_3

) is

17\text{ g/mol}

. A mass of

3.4\text{ g}

corresponds to

0.2\text{ moles}

of

\text{NH}_3

molecules, which contains

1.204 \times 10^{23}

molecules. Because one molecule of

\text{NH}_3

consists of

4

atoms (

1

nitrogen atom and

3

hydrogen atoms), multiplying

1.204 \times 10^{23}

by

4

gives

4.816 \times 10^{23}

total atoms.

Step-by-Step Solution

1

Calculate the molar mass of ammonia (

\text{NH}_3

)

\text{Molar Mass} = 14 + 3(1) = 17\text{ g/mol}

Molar mass is required to convert sample mass into moles.

2

Calculate the number of moles of

\text{NH}_3

\text{Moles} = \frac{3.4\text{ g}}{17\text{ g/mol}} = 0.2\text{ mol}

Dividing given mass by molar mass yields the mole quantity.

3

Determine the number of molecules of

\text{NH}_3

\text{Molecules} = 0.2\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 1.204 \times 10^{23}\text{ molecules}

Multiplying moles by Avogadro's constant gives total molecules.

4

Calculate the total number of individual atoms

\text{Total atoms} = 1.204 \times 10^{23} \times 4 = 4.816 \times 10^{23}\text{ atoms}

Each

\text{NH}_3

molecule contains 1 nitrogen atom and 3 hydrogen atoms (4 atoms in total).

Key Concept

Converting mass to number of constituent atoms using Avogadro's constant and molecular stoichiometry.

Question 36Question →

A $4.00\text{ g}$ sample of a copper oxide is completely reduced by dry hydrogen gas to yield $3.20\text{ g}$ of metallic copper. According to the Law of Definite Proportions, what mass of this same copper oxide, in grams, will be produced when $5.00\text{ g}$ of pure copper is completely oxidized?

Show answer & explanation

Answer: 6.25

Answer

6.25 g

According to the Law of Definite Proportions (or Constant Composition), a chemical compound always contains its component elements in fixed mass ratios. In the first sample, copper makes up

3.20\text{ g} / 4.00\text{ g} = 0.80

or

80\%

of the total mass. Therefore, in any sample of this oxide,

5.00\text{ g}

of copper represents

80\%

of the total mass. Dividing

5.00\text{ g}

by

0.80

gives

6.25\text{ g}

of copper oxide.

Step-by-Step Solution

1

Determine the mass percentage (or mass fraction) of copper in the compound from the first experiment

Mass fraction of Cu = 3.20 / 4.00 = 0.80 (80%)

The first experiment provides quantitative data regarding the mass of copper contained in a known mass of oxide.

2

Apply the Law of Definite Proportions to calculate the required mass of copper oxide for 5.00 g of copper

Mass of Copper Oxide = 5.00 / 0.80 = 6.25 g

The Law of Definite Proportions dictates that the mass composition ratio remains constant regardless of the sample source or size.

Key Concept

Law of Definite Proportions

Question 37Question →

According to Gay-Lussac's Law of Combining Volumes, what volume of ammonia gas ( $NH_3$ ) is produced when $20\text{ cm}^3$ of nitrogen gas ( $N_2$ ) reacts completely with excess hydrogen gas ( $H_2$ ) at constant temperature and pressure?

10\text{ cm}^3

20\text{ cm}^3

40\text{ cm}^3

60\text{ cm}^3

Show answer & explanation

Answer: $40\text{ cm}^3$

Answer

40\text{ cm}^3

of ammonia gas (

NH_3

) is produced.

According to the balanced chemical equation

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

, 1 mole (or volume) of

N_2

reacts to form 2 moles (or volumes) of

NH_3

. Therefore,

20\text{ cm}^3

of

N_2

produces

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

of

NH_3

gas.

Step-by-Step Solution

1

Write the balanced chemical equation for the synthesis of ammonia

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

Stoichiometric coefficients determine the combining volume ratios of reacting and product gases under constant temperature and pressure.

2

Determine the volume ratio between

N_2

and

NH_3

1\text{ volume of } N_2 : 2\text{ volumes of } NH_3

By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number ratios by volume.

3

Calculate the volume of

NH_3

formed from

20\text{ cm}^3

of

N_2

\text{Volume of } NH_3 = 20\text{ cm}^3 \times 2 = 40\text{ cm}^3

Directly scaling the volume of

N_2

by the coefficient ratio (2/1).

Key Concept

Gay-Lussac's Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple whole-number ratio to one another and to the volume of the product if gaseous, provided temperature and pressure remain constant.

Estimated Time:45s

Question 38Question →

Two samples of pure sodium chloride obtained from different sources were analyzed quantitatively. The first sample contained $4.60\text{ g}$ of sodium combined with $7.10\text{ g}$ of chlorine. If the second sample contains $14.20\text{ g}$ of chlorine, what mass of sodium is present in the second sample to satisfy the Law of Definite Proportions?

4.60\text{ g}

9.20\text{ g}

11.70\text{ g}

18.80\text{ g}

Show answer & explanation

Answer: $9.20\text{ g}$

Answer

9.20\text{ g}

According to the Law of Definite Proportions (or Constant Composition), a pure chemical compound always contains its elements combined in a fixed ratio by mass, regardless of its source. Since the chlorine mass in the second sample (

14.20\text{ g}

) is twice that of the first sample (

7.10\text{ g}

), the mass of sodium must also be twice as large, giving

2 \times 4.60\text{ g} = 9.20\text{ g}

.

Step-by-Step Solution

1

Determine the mass ratio of sodium to chlorine in the first sample.

The ratio of mass of sodium to mass of chlorine is

\frac{4.60\text{ g}}{7.10\text{ g}} = \frac{46}{71}

.

The Law of Definite Proportions states that a chemical compound always contains its constituent elements in a fixed ratio by mass.

2

Set up the proportion for the second sample with

14.20\text{ g}

of chlorine.

\frac{m_{\text{Na}}}{14.20\text{ g}} = \frac{4.60\text{ g}}{7.10\text{ g}}

.

The mass ratio must remain constant across all samples of the same compound.

3

Solve for the unknown mass of sodium (

m_{\text{Na}}

).

m_{\text{Na}} = 4.60\text{ g} \times \left(\frac{14.20\text{ g}}{7.10\text{ g}}\right) = 4.60\text{ g} \times 2 = 9.20\text{ g}

.

Since

14.20\text{ g}

is exactly twice

7.10\text{ g}

, the mass of sodium required is twice

4.60\text{ g}

.

Key Concept

Law of Definite Proportions (Law of Constant Composition)

Estimated Time:1m 0s

Question 39Question →

When $10.0\text{ g}$ of pure calcium carbonate ( $\text{CaCO}_3$ ) is strongly heated, it completely decomposes into solid calcium oxide ( $\text{CaO}$ ) and carbon(IV) oxide gas ( $\text{CO}_2$ ). According to the Law of Conservation of Mass, if $5.6\text{ g}$ of calcium oxide remains in the container, what is the mass of carbon(IV) oxide gas released in grams?

Show answer & explanation

Answer: 4.4

Answer

The mass of carbon(IV) oxide gas released is

4.4\text{ g}

.

According to the Law of Conservation of Mass, the total mass of reactants must equal the total mass of products in a chemical change. For the reaction

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

, the initial mass of

10.0\text{ g}

of

\text{CaCO}_3

must equal the combined mass of

\text{CaO}

(

5.6\text{ g}

) and

\text{CO}_2

. Subtracting

5.6\text{ g}

from

10.0\text{ g}

yields

4.4\text{ g}

for the gas produced.

Step-by-Step Solution

1

Apply the Law of Conservation of Mass

Total mass of reactants (

10.0\text{ g}

) = Total mass of products (solid residue + gas)

Mass cannot be created or destroyed in a chemical reaction.

2

Calculate the missing mass of carbon(IV) oxide gas

\text{Mass of CO}_2 = 10.0\text{ g} - 5.6\text{ g} = 4.4\text{ g}

Subtracting the mass of the solid product from the initial mass of reactant gives the mass of the gaseous product evolved.

Key Concept

Law of Conservation of Mass

Estimated Time:45s

Question 40Question →

A sample of sulfur(IV) oxide gas ( $\text{SO}_2$ ) occupies a volume of $1.20\text{ dm}^3$ at room temperature and pressure ( $\text{RTP}$ ). What is the total number of oxygen atoms present in this sample? $[\text{Molar volume of gas at RTP} = 24.0\text{ dm}^3\text{ mol}^{-1}, \text{Avogadro's constant } N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

6.02 \times 10^{22}

6.45 \times 10^{22}

3.01 \times 10^{22}

3.23 \times 10^{22}

Show answer & explanation

Answer: $6.02 \times 10^{22}$

Answer

6.02 \times 10^{22}

oxygen atoms

The correct answer is

6.02 \times 10^{22}

. Dividing the gas volume (

1.20\text{ dm}^3

) by the molar volume at RTP (

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

) gives

0.05\text{ mol}

of

\text{SO}_2

gas. Since each molecule of

\text{SO}_2

contains 2 atoms of oxygen, the sample contains

0.10\text{ mol}

of oxygen atoms. Multiplying

0.10\text{ mol}

by Avogadro's constant (

6.02 \times 10^{23}\text{ mol}^{-1}

) yields

6.02 \times 10^{22}

oxygen atoms.

Step-by-Step Solution

1

Calculate the number of moles of

\text{SO}_2

gas at RTP

n(\text{SO}_2) = \frac{1.20\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.050\text{ mol}

Molar volume at room temperature and pressure is

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

.

2

Determine the number of moles of oxygen atoms present

n(\text{O}) = 2 \times 0.050\text{ mol} = 0.100\text{ mol}

Each molecule of

\text{SO}_2

contains 2 oxygen atoms.

3

Calculate the total number of oxygen atoms using Avogadro's constant

N(\text{O}) = 0.100\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 6.02 \times 10^{22}\text{ atoms}

Multiplying moles of atoms by Avogadro's constant gives the exact atom count.

Key Concept

Molar volume of gases at RTP and stoichometric atom counting within polyatomic molecules.

Chemical Combination and Stoichiometry

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