All practice questions

Question 1001Question →

A sample of pure potassium trioxonitrate(V), $\text{KNO}_3$ , contains $1.806 \times 10^{23}$ oxygen atoms. What is the mass, in grams, of this sample of $\text{KNO}_3$ ? $[\text{K} = 39, \text{N} = 14, \text{O} = 16, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

Show answer & explanation

Answer: 10.1

Answer

The mass of the potassium trioxonitrate(V) sample is 10.1 g.

First, the number of moles of oxygen atoms is found by dividing

1.806 \times 10^{23}

atoms by Avogadro's number (

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

), yielding

0.3\text{ mol}

of oxygen. Since one mole of

\text{KNO}_3

contains three moles of oxygen atoms, the moles of

\text{KNO}_3

in the sample is

0.3 / 3 = 0.1\text{ mol}

. Multiplying

0.1\text{ mol}

by the molar mass of

\text{KNO}_3

(

101\text{ g/mol}

) gives the correct mass of

10.1\text{ g}

.

Step-by-Step Solution

1

Calculate the moles of oxygen atoms in the sample

n(\text{O}) = 0.3\text{ mol}

Divide the total number of oxygen atoms by Avogadro's constant (

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

).

2

Determine the moles of potassium trioxonitrate(V),

\text{KNO}_3

n(\text{KNO}_3) = 0.1\text{ mol}

Each formula unit of

\text{KNO}_3

contains 3 oxygen atoms, so divide the moles of oxygen atoms by 3.

3

Calculate the molar mass of

\text{KNO}_3

M(\text{KNO}_3) = 101\text{ g/mol}

Sum the relative atomic masses:

39\text{ (K)} + 14\text{ (N)} + 3 \times 16\text{ (O)} = 101\text{ g/mol}

.

4

Multiply moles of

\text{KNO}_3

by its molar mass to get total mass

\text{Mass} = 10.1\text{ g}

\text{Mass} = \text{moles} \times \text{molar mass} = 0.1\text{ mol} \times 101\text{ g/mol} = 10.1\text{ g}

.

Key Concept

Relationship between particle count, mole quantity of constituent atoms, and molar mass

Question 1002Question →

Given the following standard reduction potentials at

25^\circ\text{C}

:

\text{Al}^{3+}_{\text{(aq)}} + 3\text{e}^- \rightarrow \text{Al}_{\text{(s)}} \quad E^\circ = -1.66\text{ V}

\text{Cu}^{2+}_{\text{(aq)}} + 2\text{e}^- \rightarrow \text{Cu}_{\text{(s)}} \quad E^\circ = +0.34\text{ V}

Calculate the standard electromotive force ( $E^\circ_{\text{cell}}$ ), in volts, of the galvanic cell formed by coupling these two half-cells.

Show answer & explanation

Answer: 2

Answer

The standard electromotive force (

E^\circ_{\text{cell}}

) of the galvanic cell is

2.00\text{ V}

.

The standard electromotive force of a galvanic cell is defined as

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

. Since copper has a higher standard reduction potential (

+0.34\text{ V}

) than aluminium (

-1.66\text{ V}

), reduction occurs at the copper electrode (cathode) and oxidation occurs at the aluminium electrode (anode). Evaluating the potential difference yields

E^\circ_{\text{cell}} = 0.34\text{ V} - (-1.66\text{ V}) = 2.00\text{ V}

.

Step-by-Step Solution

1

Determine which half-cell undergoes reduction (cathode) and which undergoes oxidation (anode).

Copper is the cathode (

E^\circ = +0.34\text{ V}

) and aluminium is the anode (

E^\circ = -1.66\text{ V}

).

The half-cell with the higher standard reduction potential spontaneously undergoes reduction at the cathode.

2

Apply the formula for calculating standard electromotive force (

E^\circ_{\text{cell}}

).

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

The cell EMF is the standard potential difference between the reduction half-reaction and the oxidation half-reaction.

3

Perform the subtraction to evaluate

E^\circ_{\text{cell}}

.

E^\circ_{\text{cell}} = 0.34 - (-1.66) = 2.00\text{ V}

Subtracting a negative quantity is mathematically equivalent to adding its positive value.

Key Concept

Calculation of standard cell potential (

E^\circ_{\text{cell}}

) from standard electrode reduction potentials.

Question 1003Question →

The standard reduction potentials for aluminium and nickel half-cells at

25^\circ\text{C}

are given below:

\text{Al}^{3+}(aq) + 3e^- \rightarrow \text{Al}(s) \quad E^\circ = -1.66\text{ V}

\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \quad E^\circ = -0.25\text{ V}

Calculate the standard electromotive force ( $E^\circ_{\text{cell}}$ ) in volts for the spontaneous reaction between these two half-cells.

Show answer & explanation

Answer: 1.41

Answer

The standard electromotive force (

E^\circ_{\text{cell}}

) for the spontaneous galvanic cell reaction is +1.41 V.

For a spontaneous electrochemical reaction, the standard cell electromotive force (

E^\circ_{\text{cell}}

) must be positive. The half-reaction with the more positive standard reduction potential (

\text{Ni}^{2+}/\text{Ni}

at

-0.25\text{ V}

) proceeds as a reduction at the cathode. The half-reaction with the less positive potential (

\text{Al}^{3+}/\text{Al}

at

-1.66\text{ V}

) proceeds as an oxidation at the anode. Calculating

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.25\text{ V} - (-1.66\text{ V}) = +1.41\text{ V}

. Because

E^\circ

is an intensive property, the stoichiometric coefficients used to balance electrons (

2\text{Al} + 3\text{Ni}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Ni}

) do not affect the numerical values of the half-cell potentials.

Step-by-Step Solution

1

Determine which electrode undergoes reduction (cathode) and which undergoes oxidation (anode)

Nickel ion reduction occurs at the cathode (

E^\circ = -0.25\text{ V}

), and aluminium metal oxidation occurs at the anode (

E^\circ = -1.66\text{ V}

).

A galvanic cell operates spontaneously (

E^\circ_{\text{cell}} > 0

) when the half-cell with the more positive standard reduction potential acts as the cathode.

2

State the equation for standard cell potential

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

The cell potential measures the potential difference between the reduction half-reaction and the oxidation half-reaction under standard conditions.

3

Substitute the reduction potential values into the equation

E^\circ_{\text{cell}} = -0.25\text{ V} - (-1.66\text{ V}) = +1.41\text{ V}

Standard electrode potentials are intensive properties; hence, balancing electron stoichiometry does not scale

E^\circ

values.

Key Concept

Calculating standard cell electromotive force (

E^\circ_{\text{cell}}

) and predicting spontaneity from standard reduction potentials

Question 1004Question →

In the auto-reduction stage of copper extraction, copper(I) oxide (

\text{Cu}_2\text{O}

) reacts with copper(I) sulfide (

\text{Cu}_2\text{S}

) according to the following balanced equation:

\text{Cu}_2\text{S (s)} + 2\text{Cu}_2\text{O (s)} \rightarrow 6\text{Cu (s)} + \text{SO}_2\text{ (g)}

If

14.3\text{ g}

of copper(I) oxide reacts completely with excess copper(I) sulfide, what mass of metallic copper in grams is produced?

[\text{Cu} = 63.5, \text{O} = 16.0]

Show answer & explanation

Answer: 19.05

Answer

The mass of metallic copper produced is 19.05 g.

According to the balanced chemical equation, 2 moles of copper(I) oxide react with copper(I) sulfide to yield 6 moles of metallic copper, which simplifies to a 1:3 molar ratio. Given that the molar mass of Cu2O is 143 g/mol, 14.3 g represents 0.1 mol of Cu2O. Based on the 1:3 ratio, this produces 0.3 mol of copper metal, corresponding to 19.05 g of Cu.

Step-by-Step Solution

1

Calculate the molar mass of copper(I) oxide

Molar mass of Cu2O = 143.0 g/mol

Required to convert the given mass of reactant to moles.

2

Determine moles of Cu2O reacted

Moles of Cu2O = 14.3 g / 143.0 g/mol = 0.10 mol

Establishes the quantitative amount of Cu2O in the chemical system.

3

Determine moles of copper metal produced using stoichiometry

Moles of Cu = 0.10 mol * (6 / 2) = 0.30 mol

The balanced chemical equation shows 2 moles of Cu2O yield 6 moles of Cu metal.

4

Calculate the mass of copper metal produced

Mass of Cu = 0.30 mol * 63.5 g/mol = 19.05 g

Converts the stoichiometric amount of product moles into grams.

Key Concept

Auto-reduction in copper extraction and stoichiometric mass calculations

Question 1005Question →

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

Show answer & explanation

Answer: 28

Answer

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

According to the balanced chemical equation

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

, 2 volumes of

SO_2

gas require 1 volume of

O_2

gas for complete oxidation. Therefore,

56\text{ dm}^3

of

SO_2

requires half its volume in oxygen, which equals

28\text{ dm}^3

.

Step-by-Step Solution

1

Determine the stoichiometric ratio between

SO_2

and

O_2

from the balanced equation.

2 volumes of

SO_2(g)

react with 1 volume of

O_2(g)

.

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

2

Compute the required volume of

O_2

gas for

56\text{ dm}^3

of

SO_2

.

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

.

Since the ratio of

SO_2

to

O_2

is

2:1

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

Key Concept

Stoichiometric Volume Calculations in Gas Reactions (Contact Process)

Question 1006Question →

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 1007Question →

The solubility of a salt $X$ is $0.80\text{ mol dm}^{-3}$ at $60^\circ\text{C}$ and $0.30\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . Calculate the mass of salt $X$ (in grams) that will crystallize out of solution when $1.0\text{ dm}^3$ of its saturated solution is cooled from $60^\circ\text{C}$ to $20^\circ\text{C}$ . (Molar mass of salt $X = 100\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 50

Answer

The mass of salt

X

that crystallizes out of the solution is

50\text{ g}

.

At

60^\circ\text{C}

,

1.0\text{ dm}^3

of saturated solution contains

0.80\text{ mol}

of salt

X

. When cooled to

20^\circ\text{C}

, the solution can only hold

0.30\text{ mol}

. The excess amount that crystallizes out is

0.80\text{ mol} - 0.30\text{ mol} = 0.50\text{ mol}

. Converting this amount to mass yields

0.50\text{ mol} \times 100\text{ g mol}^{-1} = 50\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in molar solubility between the two temperatures

0.80\text{ mol dm}^{-3} - 0.30\text{ mol dm}^{-3} = 0.50\text{ mol dm}^{-3}

This difference represents the amount of solute in moles per cubic decimeter that can no longer remain dissolved when cooled to

20^\circ\text{C}

.

2

Convert the precipitated moles into mass in grams for

1.0\text{ dm}^3

of solution

0.50\text{ mol} \times 100\text{ g mol}^{-1} = 50\text{ g}

Multiplying the precipitated amount in moles by the molar mass gives the total mass in grams that crystallizes out.

Key Concept

Solubility and crystallization calculations upon cooling

Question 1008Question →

The solubility of sodium trioxonitrate(V), $\text{NaNO}_3$ , in water is $4.5\text{ mol dm}^{-3}$ at $60^\circ\text{C}$ and $2.0\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . If the molar mass of $\text{NaNO}_3$ is $85\text{ g mol}^{-1}$ , what mass of $\text{NaNO}_3$ in grams will crystallize out when $400\text{ cm}^3$ of a saturated solution of the salt at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ?

Show answer & explanation

Answer: 85

Answer

The mass of

\text{NaNO}_3

that will crystallize out is

85\text{ g}

.

The difference in solubility between

60^\circ\text{C}

and

20^\circ\text{C}

is

4.5 - 2.0 = 2.5\text{ mol dm}^{-3}

. In

400\text{ cm}^3

(

0.4\text{ dm}^3

) of solution, the amount of salt precipitated is

2.5\text{ mol dm}^{-3} \times 0.4\text{ dm}^3 = 1.0\text{ mol}

. Multiplying by the molar mass (

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

) gives

85\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in solubility between

60^\circ\text{C}

and

20^\circ\text{C}

\Delta S = 4.5\text{ mol dm}^{-3} - 2.0\text{ mol dm}^{-3} = 2.5\text{ mol dm}^{-3}

Cooling causes the excess solute to precipitate out based on the difference in saturation concentration.

2

Convert solution volume to

\text{dm}^3

V = \frac{400\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.4\text{ dm}^3

Solubility is given per

\text{dm}^3

, so volume must be in

\text{dm}^3

.

3

Calculate the moles of solute precipitated

n = 2.5\text{ mol dm}^{-3} \times 0.4\text{ dm}^3 = 1.0\text{ mol}

Multiplying the concentration difference by the solution volume yields total precipitated moles.

4

Convert moles to mass in grams

m = 1.0\text{ mol} \times 85\text{ g mol}^{-1} = 85\text{ g}

Mass is found by multiplying moles by molar mass.

Key Concept

Crystallization and solubility change with temperature

Question 1009Question →

When excess dry ammonia gas is passed over

24.0\text{ g}

of heated copper(II) oxide (

CuO

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

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

What is the volume of nitrogen gas, in

\text{dm}^3

, evolved at s.t.p.?
(

Cu = 64.0

,

O = 16.0

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

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

)

Show answer & explanation

Answer: 2.24

Answer

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

2.24\text{ dm}^3

.

According to the balanced chemical equation

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

, 3 moles (

240.0\text{ g}

) of copper(II) oxide produce 1 mole (

22.4\text{ dm}^3

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

24.0\text{ g}

of

CuO

yields

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

of nitrogen gas.

Step-by-Step Solution

1

Calculate the molar mass of

CuO

Molar mass of

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

Required to convert the given mass of reactant to moles.

2

Find the number of moles of

CuO

Moles of

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

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

3

Apply stoichiometry to find moles of

N_2

gas produced

Moles of

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

The mole ratio between

CuO

and

N_2

in the balanced chemical equation is

3:1

.

4

Convert moles of

N_2

to volume at s.t.p.

Volume of

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

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

at s.t.p.

Key Concept

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

Question 1010Question →

A sample containing $2.5\text{ moles}$ of ammonia gas ( $\text{NH}_3$ ) is held in a vessel at $400\text{ K}$ under a pressure of $50.0\text{ atm}$ . Under these conditions, the compressibility factor ( $Z$ ) of ammonia is $0.880$ . What is the actual volume occupied by the gas sample in $\text{dm}^3$ ? (Take $R = 0.0821\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ )

Show answer & explanation

Answer: 1.44

Answer

The actual volume occupied by the ammonia gas sample under the given conditions is

1.44\text{ dm}^3

.

The compressibility factor

Z

is defined as

Z = \frac{P V}{n R T} = \frac{V_{\text{real}}}{V_{\text{ideal}}}

. Substituting

n = 2.5\text{ mol}

,

T = 400\text{ K}

,

P = 50.0\text{ atm}

,

R = 0.0821\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}

, and

Z = 0.880

into

V = \frac{Z \cdot n R T}{P}

gives

1.44\text{ dm}^3

.

Step-by-Step Solution

1

Identify the relationship between real volume and compressibility factor Z

Z = \frac{P V_{\text{actual}}}{n R T}

The compressibility factor quantifies the deviation of a real gas from ideal gas behavior.

2

Rearrange the compressibility formula to solve for actual volume (

V_{\text{actual}}

)

V_{\text{actual}} = \frac{Z \cdot n R T}{P}

Isolating

V_{\text{actual}}

allows direct calculation using the provided parameters.

3

Substitute the given values into the equation and compute the result

V_{\text{actual}} = \frac{0.880 \times 2.5 \times 0.0821 \times 400}{50.0} = 1.44496\text{ dm}^3

Performing the algebraic calculation yields the volume occupied by the real gas.

Key Concept

Compressibility Factor and Real Gas Deviation

Question 1011Question →

Consider the unbalanced redox reaction occurring in acidic solution:

\text{a AsO}_3^{3-}(\text{aq}) + \text{b MnO}_4^-(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d AsO}_4^{3-}(\text{aq}) + \text{e Mn}^{2+}(\text{aq}) + \text{f H}_2\text{O}(\text{l})

When this ionic equation is balanced using the smallest possible whole-number coefficients, what is the value of the coefficient

c

for

\text{H}^+

?

Show answer & explanation

Answer: 6

Answer

The coefficient c for hydrogen ions (H⁺) in the balanced redox equation is 6.

Balancing the oxidation half-reaction shows that each arsenite ion produces 2 electrons and 2 H⁺ ions. The reduction half-reaction shows that each permanganate ion consumes 5 electrons and 8 H⁺ ions. Multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2 balances the total electron transfer at 10 electrons. Combining the equations gives 16 H⁺ on the left and 10 H⁺ on the right, which simplifies to 6 H⁺ on the reactant side.

Step-by-Step Solution

1

Balance the oxidation half-reaction (arsenite to arsenate)

AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻

Arsenic changes oxidation state from +3 to +5, releasing 2 electrons. Oxygen is balanced with H₂O and hydrogen with H⁺.

2

Balance the reduction half-reaction (permanganate to manganese(II))

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Manganese changes oxidation state from +7 to +2, consuming 5 electrons in acidic medium.

3

Equalize the electrons transferred in both half-reactions

5(AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻) and 2(MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O)

The total number of electrons gained and lost must equal 10 electrons.

4

Combine the half-reactions and cancel common terms

5 AsO₃³⁻ + 2 MnO₄⁻ + 6 H⁺ → 5 AsO₄³⁻ + 2 Mn²⁺ + 3 H₂O

Subtracting 10 H⁺ and 5 H₂O from both sides leaves 6 H⁺ on the reactant side.

Key Concept

Balancing Ion-Electron Redox Half-Reactions in Acidic Medium

Question 1012Question →

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 1013Question →

A saturated solution of potassium chlorate ( $\text{KClO}_3$ ) at $20^\circ\text{C}$ contains $7.35\text{ g}$ of solute dissolved in $100\text{ g}$ of distilled water. What is the solubility of $\text{KClO}_3$ at $20^\circ\text{C}$ in $\text{mol/dm}^3$ ? [Molar masses: $\text{K} = 39, \text{Cl} = 35.5, \text{O} = 16$ ; density of water $= 1.00\text{ g/cm}^3$ ]

Show answer & explanation

Answer: 0.6

Answer

The solubility of potassium chlorate at

20^\circ\text{C}

is

0.6\text{ mol/dm}^3

.

To convert mass of salt in a given solvent volume into solubility in

\text{mol/dm}^3

, first calculate the molar mass of

\text{KClO}_3

(

122.5\text{ g/mol}

). The amount of salt in moles is

7.35 / 122.5 = 0.06\text{ mol}

. Since

100\text{ g}

of water equals

0.1\text{ dm}^3

, the concentration of the saturated solution is

0.06\text{ mol} / 0.1\text{ dm}^3 = 0.6\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{KClO}_3

122.5\text{ g/mol}

Sum the relative atomic masses:

39 (\text{K}) + 35.5 (\text{Cl}) + 3 \times 16 (\text{O}) = 122.5\text{ g/mol}

.

2

Calculate the number of moles of solute

0.06\text{ mol}

Divide the given mass by the molar mass:

\frac{7.35\text{ g}}{122.5\text{ g/mol}} = 0.06\text{ mol}

.

3

Convert the mass of solvent to volume in

\text{dm}^3

0.1\text{ dm}^3

Water density is

1.00\text{ g/cm}^3

, so

100\text{ g} = 100\text{ cm}^3 = 0.1\text{ dm}^3

.

4

Determine the molar solubility

0.6\text{ mol/dm}^3

Divide moles of solute by volume of solvent in

\text{dm}^3

:

\frac{0.06\text{ mol}}{0.1\text{ dm}^3} = 0.6\text{ mol/dm}^3

.

Key Concept

Solubility in mol/dm³

Question 1014Question →

A solution containing trivalent ions $M^{3+}$ of an unknown metal $M$ is electrolyzed using inert electrodes. If a steady current of $2.50\text{ A}$ passed for $5790\text{ s}$ deposits $1.35\text{ g}$ of metal $M$ at the cathode, what is the relative atomic mass of metal $M$ ? [ $1\text{ Faraday} = 96500\text{ C/mol}$ ]

Show answer & explanation

Answer: 27

Answer

The relative atomic mass of metal

M

is

27\text{ g/mol}

.

Using Faraday's first law (

Q = I \times t

), the total charge passed is

2.50 \times 5790 = 14475\text{ C}

. Dividing by Faraday's constant (

96500\text{ C/mol}

) gives

0.15\text{ mol}

of electrons. Since the metal ion is trivalent (

M^{3+}

), the half-reaction is

M^{3+} + 3e^- \rightarrow M

, meaning

0.15\text{ mol}

of electrons deposits

0.05\text{ mol}

of metal. Dividing the mass (

1.35\text{ g}

) by the amount in moles (

0.05\text{ mol}

) yields a relative atomic mass of

27\text{ g/mol}

.

Step-by-Step Solution

1

Calculate the total electric charge passed (

Q

).

Q = 2.50\text{ A} \times 5790\text{ s} = 14475\text{ C}

.

Electric charge is the product of current in amperes and time in seconds.

2

Calculate the moles of electrons transferred.

n_e = \frac{14475\text{ C}}{96500\text{ C/mol}} = 0.15\text{ mol } e^-

.

One mole of electrons carries a charge of 1 Faraday (

96500\text{ C}

).

3

Determine the amount (in moles) of metal

M

deposited.

From

M^{3+} + 3e^- \rightarrow M

,

3\text{ moles of } e^-

deposit

1\text{ mole of } M

. Thus,

n_M = \frac{0.15\text{ mol}}{3} = 0.05\text{ mol}

.

The reduction of a trivalent ion requires 3 moles of electrons per mole of metal deposited.

4

Calculate the relative atomic mass (

M_r

).

M_r = \frac{1.35\text{ g}}{0.05\text{ mol}} = 27\text{ g/mol}

.

Molar mass is calculated by dividing the mass of the substance by the amount in moles.

Key Concept

Faraday's Laws of Electrolysis and Quantitative Calculations

Question 1015Question →

What volume of nitrogen(IV) oxide gas ( $\text{NO}_2$ ), measured at s.t.p. in $\text{dm}^3$ , is evolved when $3.2\text{ g}$ of pure copper metal reacts completely with excess concentrated trioxonitrate(V) acid?

(Molar mass of $\text{Cu} = 64\text{ g mol}^{-1}$ ; Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{ mol}^{-1}$ )

Show answer & explanation

Answer: 2.24

Answer

2.24 dm^3 of nitrogen(IV) oxide gas is produced at s.t.p.

When copper reacts with concentrated trioxonitrate(V) acid, the reaction follows the stoichiometry

\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}

. Thus,

1\text{ mole}

of copper metal (

64\text{ g}

) yields

2\text{ moles}

of

\text{NO}_2

gas (

44.8\text{ dm}^3

at s.t.p.). For

3.2\text{ g}

(

0.05\text{ moles}

) of copper, the volume of

\text{NO}_2

formed is

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

.

Step-by-Step Solution

1

Write the balanced equation for the reaction of copper with concentrated trioxonitrate(V) acid

\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}

Concentrated trioxonitrate(V) acid acts as a strong oxidizing agent, converting copper to copper(II) ions and reducing itself to brown nitrogen(IV) oxide gas.

2

Calculate the moles of copper metal present in 3.2 g

3.2 / 64 = 0.05 mol

Number of moles is mass divided by relative molar mass.

3

Determine the amount of nitrogen(IV) oxide gas produced in moles

2 * 0.05 = 0.10 mol

The stoichiometric mole ratio between Cu and NO2 is 1:2.

4

Convert moles of nitrogen(IV) oxide gas into volume at s.t.p.

0.10 * 22.4 = 2.24 dm^3

One mole of any ideal gas occupies 22.4 dm^3 at standard temperature and pressure.

Key Concept

Stoichiometry of copper redox reaction with concentrated trioxonitrate(V) acid producing nitrogen(IV) oxide gas.

Question 1016Question →

For the reversible reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ , the equilibrium concentrations of $\text{PCl}_5$ , $\text{PCl}_3$ , and $\text{Cl}_2$ in a $1.0\text{ dm}^3$ vessel are $0.20\text{ mol dm}^{-3}$ , $0.60\text{ mol dm}^{-3}$ , and $0.30\text{ mol dm}^{-3}$ respectively. What is the numerical value of the equilibrium constant, $K_c$ ?

Show answer & explanation

Answer: 0.9

Answer

The equilibrium constant

K_c

is

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

.

The equilibrium constant

K_c

is calculated by substituting the equilibrium concentrations into

K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.60 \times 0.30}{0.20} = 0.90\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Formulate the equilibrium constant expression

K_c

for the reaction.

K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}

The equilibrium constant expression is defined as the product of the concentrations of the reaction products divided by the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficient.

2

Substitute the equilibrium concentrations into the expression.

K_c = \frac{0.60 \times 0.30}{0.20}

Given equilibrium values:

[\text{PCl}_5] = 0.20\text{ mol dm}^{-3}

,

[\text{PCl}_3] = 0.60\text{ mol dm}^{-3}

, and

[\text{Cl}_2] = 0.30\text{ mol dm}^{-3}

.

3

Calculate the arithmetic result.

K_c = \frac{0.18}{0.20} = 0.90\text{ mol dm}^{-3}

Multiplying

0.60 \times 0.30

gives

0.18

, and dividing by

0.20

yields

0.90

.

Key Concept

Direct calculation of equilibrium constant

K_c

from equilibrium concentrations

Question 1017Question →

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 1018Question →

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 1019Question →

In the Contact Process for the manufacture of tetraoxosulfate(VI) acid, $44.8\text{ dm}^3$ of sulfur(IV) oxide ( $SO_2$ ) gas measured at standard temperature and pressure (STP) is reacted with excess oxygen over a vanadium(V) oxide ( $V_2O_5$ ) catalyst. If the catalytic conversion efficiency of $SO_2$ to sulfur(VI) oxide ( $SO_3$ ) is $85\%$ , and all produced $SO_3$ is subsequently absorbed in concentrated $H_2SO_4$ and hydrated to form pure $H_2SO_4$ , what mass of pure $H_2SO_4$ in grams is produced? (Molar mass of $H_2SO_4 = 98\text{ g mol}^{-1}$ , Molar volume of gas at STP $= 22.4\text{ dm}^3\text{ mol}^{-1}$ )

Show answer & explanation

Answer: 166.6

Answer

The mass of pure H2SO4 produced is 166.6 g.

The correct answer of 166.6 g is derived by converting 44.8 dm³ of SO2 at STP to 2.0 moles, taking 85% of that value to find the actual 1.70 moles of SO3 produced, and multiplying by the molar mass of H2SO4 (98 g/mol).

Step-by-Step Solution

1

Calculate the moles of SO2 gas supplied at STP

Moles of SO2 = 2.0 mol

Dividing the gas volume at STP (44.8 dm³) by the molar gas volume (22.4 dm³/mol) gives the molar quantity.

2

Determine theoretical yield of SO3

Theoretical moles of SO3 = 2.0 mol

From the stoichiometric mole ratio in 2SO2 + O2 -> 2SO3, 2 moles of SO2 produce 2 moles of SO3.

3

Calculate actual moles of SO3 produced considering catalytic efficiency

Actual moles of SO3 = 1.70 mol

Multiplying the theoretical yield (2.0 mol) by the 85% conversion efficiency gives the actual yield of 1.70 mol.

4

Calculate mass of H2SO4 produced from the actual SO3 formed

Mass of H2SO4 = 166.6 g

Overall absorption and hydration converts SO3 to H2SO4 in a 1:1 mole ratio (SO3 + H2O -> H2SO4). Multiplying 1.70 mol by molar mass 98 g/mol yields 166.6 g.

Key Concept

Stoichiometry of the Contact Process involving molar volume at STP and percentage conversion efficiency

Question 1020Question →

Excess solid carbon,

\text{C}(s)

, is allowed to react with

1.0\text{ mol}

of carbon dioxide gas,

\text{CO}_2(g)

, in a sealed

2.0\text{ dm}^3

reaction vessel at a constant high temperature according to the equation:

\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)

If

1.0\text{ mol}

of carbon monoxide gas,

\text{CO}(g)

, is present at equilibrium, calculate the numerical value of the equilibrium constant,

K_c

, in

\text{mol dm}^{-3}

.

Show answer & explanation

Answer: 1

Answer

1.0

To find

K_c

, convert initial and equilibrium amounts to molar concentrations by dividing by the volume (

2.0\text{ dm}^3

). Initial concentration of

\text{CO}_2

is

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

. At equilibrium,

[\text{CO}] = \frac{1.0}{2.0} = 0.50\text{ mol dm}^{-3}

. From the

1:2

stoichiometry,

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

of

\text{CO}_2

reacted, leaving

[\text{CO}_2] = 0.25\text{ mol dm}^{-3}

. For a heterogeneous system, solid carbon is excluded from the equilibrium constant expression. Thus,

K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]} = \frac{(0.50)^2}{0.25} = 1.0\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Calculate initial concentrations from the given moles and volume

Initial

[\text{CO}_2] = 0.50\text{ mol dm}^{-3}

, initial

[\text{CO}] = 0.0\text{ mol dm}^{-3}

Concentration is calculated using

C = \frac{n}{V}

where

V = 2.0\text{ dm}^3

.

2

Determine equilibrium concentrations using stoichiometric mole ratios

Equilibrium

[\text{CO}] = 0.50\text{ mol dm}^{-3}

and

[\text{CO}_2] = 0.25\text{ mol dm}^{-3}

Producing

1.0\text{ mol}

of

\text{CO}

consumes

0.50\text{ mol}

of

\text{CO}_2

. Remaining moles of

\text{CO}_2 = 1.0 - 0.50 = 0.50\text{ mol}

.

3

Write the equilibrium constant expression for the heterogeneous system

K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]}

Pure solids like

\text{C}(s)

have constant activity and are omitted from the

K_c

expression.

4

Substitute concentrations into the

K_c

expression and solve

K_c = \frac{(0.50)^2}{0.25} = 1.0\text{ mol dm}^{-3}

Calculates the numerical value of the equilibrium constant.

Key Concept

Heterogeneous equilibrium constant expression and ICE calculations

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