Chemical Combination and Stoichiometry Alıştırma Soruları — JAMB UTME

Soru 1Soru →

A metal $X$ forms two distinct chlorides. Quantitative analysis shows that in Chloride 1, $5.40\text{ g}$ of $X$ combines with $10.65\text{ g}$ of chlorine. In Chloride 2, $3.60\text{ g}$ of $X$ combines with $10.65\text{ g}$ of chlorine. If the empirical formula of Chloride 1 is $XCl_2$ , calculate the subscript value $y$ in the empirical formula $XCl_y$ of Chloride 2.

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The value of the subscript y is 3.

Applying the Law of Multiple Proportions, when a fixed mass of chlorine (

10.65\text{ g}

) reacts with different masses of metal

X

(

5.40\text{ g}

and

3.60\text{ g}

), the mass ratio of

X

is

5.40 : 3.60 = 3 : 2

. This implies that for a fixed amount of metal

X

, the ratio of chlorine atoms in Chloride 1 to Chloride 2 is

2 : 3

. Since Chloride 1 is

XCl_2

, Chloride 2 must be

XCl_3

, yielding

y = 3

.

Adım Adım Çözüm

1

Determine the mass of chlorine per gram of metal X in Chloride 1.

\frac{10.65\text{ g}}{5.40\text{ g}} = 1.9722\text{ g } Cl / \text{g } X

This establishes the baseline quantitative relationship for the formula

XCl_2

.

2

Determine the mass of chlorine per gram of metal X in Chloride 2.

\frac{10.65\text{ g}}{3.60\text{ g}} = 2.9583\text{ g } Cl / \text{g } X

This determines the mass of chlorine per unit mass of metal in the second compound.

3

Calculate the simple multiple proportion ratio between the two compounds.

\frac{2.9583}{1.9722} = 1.5

According to the Law of Multiple Proportions, the masses of chlorine combining with a fixed mass of X stand in a simple whole-number ratio.

4

Multiply the subscript of chlorine in the first compound by the calculated ratio.

y = 2 \times 1.5 = 3

Since Chloride 1 has 2 chlorine atoms (

XCl_2

), Chloride 2 must have

2 \times 1.5 = 3

chlorine atoms (

XCl_3

).

Anahtar Kavram

Law of Multiple Proportions

Soru 2Soru →

A $10.00\text{ g}$ sample of impure hydrated iron(II) tetraoxosulfate(VI), $\text{FeSO}_4 \cdot 7\text{H}_2\text{O}$ , was dissolved in acidic medium and titrated against $0.050\text{ mol dm}^{-3}$ potassium tetraoxomanganate(VII) solution. If exactly $40.00\text{ cm}^3$ of the $\text{KMnO}_4$ solution was required for complete oxidation of the $\text{Fe}^{2+}$ ions, what is the percentage purity of the hydrated salt sample? ( $[\text{H}=1,\text{ O}=16,\text{ S}=32,\text{ Fe}=56]$ )

$27.80\%

$15.20\%

$5.56\%

$5.60\%

Cevabı ve açıklamayı göster

Cevap: $27.80\%

Cevap

The percentage purity of the hydrated salt sample is

27.80\%

.

The correct answer is

27.80\%

. Calculating the moles of potassium tetraoxomanganate(VII) used (

0.050 \times 0.04000 = 0.0020\text{ mol}

) and applying the redox stoichiometry ratio (

5\text{ Fe}^{2+} : 1\text{ MnO}_4^-

) gives

0.010\text{ mol}

of pure

\text{FeSO}_4 \cdot 7\text{H}_2\text{O}

. Multiplying by its molar mass (

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

) gives

2.78\text{ g}

of pure salt, which represents

27.80\%

of the

10.00\text{ g}

impure sample.

Adım Adım Çözüm

1

Calculate the moles of

\text{KMnO}_4

used in the titration.

\text{Moles of } \text{KMnO}_4 = \text{Molarity} \times \text{Volume in dm}^3 = 0.050\text{ mol dm}^{-3} \times \frac{40.00}{1000}\text{ dm}^3 = 0.0020\text{ mol}

.

Molarity and volume yield the amount of oxidizing agent delivered at the endpoint.

2

Determine the moles of

\text{Fe}^{2+}

(and thus pure

\text{FeSO}_4 \cdot 7\text{H}_2\text{O}

) present.

From the redox ionic equation

5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}

, the mole ratio is

5\text{ mol Fe}^{2+} : 1\text{ mol MnO}_4^-

. Therefore,

\text{moles of Fe}^{2+} = 5 \times 0.0020\text{ mol} = 0.010\text{ mol}

.

Stoichiometry of the redox reaction establishes the mole relationship between reactant species.

3

Calculate the molar mass of hydrated iron(II) tetraoxosulfate(VI),

\text{FeSO}_4 \cdot 7\text{H}_2\text{O}

.

\text{Molar mass} = 56 + 32 + (4 \times 16) + 7 \times (2 \times 1 + 16) = 56 + 32 + 64 + 126 = 278\text{ g mol}^{-1}

.

The entire hydrated formula mass must be used to convert moles of pure salt to mass.

4

Calculate mass of pure hydrated salt and the percentage purity.

\text{Mass of pure salt} = 0.010\text{ mol} \times 278\text{ g mol}^{-1} = 2.78\text{ g}

.

\text{Percentage Purity} = \left( \frac{\text{Mass of Pure Salt}}{\text{Total Impure Sample Mass}} \right) \times 100\% = \left( \frac{2.78\text{ g}}{10.00\text{ g}} \right) \times 100\% = 27.80\%

.

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

Anahtar Kavram

Redox Titration Stoichiometry and Percentage Purity

Tahmini Süre:2m 30s

Soru 3Soru →

A

4.00\text{ g}

sample of an impure copper(II) oxide ore is heated in a stream of dry hydrogen gas until reduction is complete according to the equation:

\text{CuO}_{(s)} + \text{H}_{2(g)} \rightarrow \text{Cu}_{(s)} + \text{H}_2\text{O}_{(g)}

If

2.54\text{ g}

of pure copper metal is obtained, what is the percentage purity of the copper(II) oxide in the ore?

[\text{Relative atomic masses: Cu} = 63.5, \text{O} = 16.0]

79.5\%

63.5\%

36.5\%

20.5\%

Cevabı ve açıklamayı göster

Cevap: $79.5\%$

Cevap

The percentage purity of the copper(II) oxide in the ore is

79.5\%

.

The option stating

79.5\%

is correct because

2.54\text{ g}

of copper corresponds to

0.04\text{ mol}

of

\text{Cu}

. According to the chemical equation,

0.04\text{ mol}

of

\text{Cu}

requires

0.04\text{ mol}

of pure

\text{CuO}

, which weighs

0.04 \times 79.5\text{ g mol}^{-1} = 3.18\text{ g}

. Dividing

3.18\text{ g}

of pure

\text{CuO}

by the total sample mass of

4.00\text{ g}

yields

79.5\%

.

Adım Adım Çözüm

1

Calculate the molar mass of copper(II) oxide (CuO) and the moles of copper metal produced.

\text{Molar mass of CuO} = 63.5 + 16.0 = 79.5\text{ g mol}^{-1}

.

\text{Moles of Cu} = \frac{2.54\text{ g}}{63.5\text{ g mol}^{-1}} = 0.04\text{ mol}

.

Converting the given mass of product into moles allows stoichiometric ratio calculations.

2

Determine the moles and mass of pure copper(II) oxide in the sample.

From the equation,

1\text{ mol CuO} \rightarrow 1\text{ mol Cu}

. Moles of pure

\text{CuO} = 0.04\text{ mol}

. Mass of pure

\text{CuO} = 0.04\text{ mol} \times 79.5\text{ g mol}^{-1} = 3.18\text{ g}

.

Stoichiometry dictates that 1 mole of CuO produces 1 mole of Cu upon complete reduction.

3

Calculate the percentage purity of the copper(II) oxide sample.

\text{Percentage purity} = \frac{\text{Mass of pure CuO}}{\text{Total mass of sample}} \times 100\% = \frac{3.18\text{ g}}{4.00\text{ g}} \times 100\% = 79.5\%

.

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

Anahtar Kavram

Determining percentage purity using stoichiometric reduction yields

Soru 4Soru →

A naturally occurring element $X$ consists of two stable isotopes, $^{63}X$ and $^{65}X$ . If the relative atomic mass of element $X$ is $63.6$ , what is the percentage abundance of the heavier isotope $^{65}X$ ?

30\%

70\%

60\%

50\%

Cevabı ve açıklamayı göster

Cevap: $30\%$

Cevap

The percentage abundance of the heavier isotope

^{65}X

is

30\%

.

The relative atomic mass of an element with isotopes is calculated using the weighted average formula:

\text{RAM} = \sum (\text{mass} \times \text{fractional abundance})

. Setting

p

as the fraction of

^{65}X

gives

63.6 = 63(1 - p) + 65p = 63 + 2p

. Solving yields

2p = 0.6

, so

p = 0.30

, which corresponds to

30\%

.

Adım Adım Çözüm

1

Set up the relative atomic mass weighted average equation

63.6 = \frac{(m_1 \times x_1) + (m_2 \times x_2)}{100}

Relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes based on their fractional abundances.

2

Express isotopic abundances in terms of a single variable

p

Let

p

be the fraction of

^{65}X

, so the fraction of

^{63}X

is

1 - p

.

The sum of the fractional abundances of all isotopes of an element must equal 1 (or

100\%

).

3

Substitute known values and solve for

p

63.6 = 63(1 - p) + 65p \implies 63.6 = 63 + 2p \implies 2p = 0.6 \implies p = 0.30

Algebraic expansion isolates the variable corresponding to the heavier isotope abundance.

4

Convert the fractional abundance to percentage

0.30 \times 100\% = 30\%

Multiplying the decimal fraction by 100 yields the percentage abundance.

Anahtar Kavram

Calculation of Relative Atomic Mass from Isotopic Abundance

Tahmini Süre:1m 30s

Soru 5Soru →

A metallic element $M$ forms two distinct oxides. Quantitative analysis shows that the first oxide contains $20.0\%$ oxygen by mass and has the empirical formula $MO$ . The second oxide contains $11.1\%$ oxygen by mass. Based on the Law of Multiple Proportions, what is the empirical formula of the second oxide?

M_2O

MO_2

M_2O_3

M_3O_4

Cevabı ve açıklamayı göster

Cevap: $M_2O$

Cevap

M_2O

According to the Law of Multiple Proportions, when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. In the first oxide (

MO

),

20.0\text{ g}

of oxygen combines with

80.0\text{ g}

of metal

M

, giving a ratio of

4.0\text{ g } M

per

1.0\text{ g } O

. In the second oxide,

11.1\text{ g}

of oxygen combines with

88.9\text{ g}

of

M

, giving a ratio of

8.0\text{ g } M

per

1.0\text{ g } O

. Comparing the mass of metal combining with

1.0\text{ g}

of oxygen gives

4.0 : 8.0 = 1 : 2

. Since the first oxide has

1

atom of

M

per atom of

O

, the second oxide contains

2

atoms of

M

per atom of

O

, yielding the formula

M_2O

.

Adım Adım Çözüm

1

Calculate the mass ratio of metal

M

to oxygen

O

in the first oxide

In

100\text{ g}

of the first oxide, mass of

O = 20.0\text{ g}

and mass of

M = 80.0\text{ g}

. Mass ratio

M : O = \frac{80.0\text{ g } M}{20.0\text{ g } O} = 4.0\text{ g } M / \text{g } O

.

Determining the mass of metal that combines with a fixed unit mass (

1.0\text{ g}

) of oxygen in the first compound.

2

Calculate the mass ratio of metal

M

to oxygen

O

in the second oxide

In

100\text{ g}

of the second oxide, mass of

O = 11.1\text{ g}

and mass of

M = 88.9\text{ g}

. Mass ratio

M : O = \frac{88.9\text{ g } M}{11.1\text{ g } O} = 8.01\text{ g } M / \text{g } O \approx 8.0\text{ g } M / \text{g } O

.

Determining the mass of metal that combines with a fixed unit mass (

1.0\text{ g}

) of oxygen in the second compound.

3

Apply the Law of Multiple Proportions to compare the masses of metal

M

combining with fixed oxygen

\text{Ratio of masses of } M = 4.0 : 8.0 = 1 : 2.

The Law of Multiple Proportions states that the masses of one element combining with a fixed mass of another are in simple whole-number ratios.

4

Deduce the empirical formula of the second oxide

Since the first oxide (

MO

) has

1

atom of

M

per atom of

O

, an oxide with twice the mass of

M

per atom of

O

must have

2

atoms of

M

per atom of

O

, giving the empirical formula

M_2O

.

Relating the atomic ratio of the second oxide directly to the known formula of the first oxide.

Anahtar Kavram

Law of Multiple Proportions

Soru 6Soru →

A $5.00\text{ g}$ sample of impure limestone ( $\text{CaCO}_3$ ) is strongly heated until decomposition is complete. If the loss in mass due to the escape of carbon dioxide ( $\text{CO}_2$ ) gas is $1.76\text{ g}$ , what is the percentage purity of the limestone sample? $[\text{Relative atomic masses: } \text{Ca} = 40, \text{C} = 12, \text{O} = 16]$

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The percentage purity of the limestone sample is 80%.

Thermal decomposition of calcium carbonate yields calcium oxide and carbon dioxide. The mass loss of 1.76 g corresponds to the evolved CO2. From the molar masses (CaCO3 = 100 g/mol, CO2 = 44 g/mol), 44 g of CO2 is released by 100 g of pure CaCO3. Thus, 1.76 g of CO2 is released by 4.00 g of pure CaCO3. Dividing the pure mass (4.00 g) by the original sample mass (5.00 g) and multiplying by 100 yields a percentage purity of 80%.

Adım Adım Çözüm

1

Write the balanced equation for the decomposition reaction.

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

The decrease in mass is entirely due to the evolved carbon dioxide gas.

2

Calculate the relative formula mass of calcium carbonate and carbon dioxide.

\text{Molar mass of } \text{CaCO}_3 = 100\text{ g/mol}, \quad \text{Molar mass of } \text{CO}_2 = 44\text{ g/mol}

Required to relate the mass of evolved gas to the mass of reacting calcium carbonate.

3

Calculate the mass of pure calcium carbonate in the sample.

\text{Mass of pure } \text{CaCO}_3 = \left(\frac{100}{44}\right) \times 1.76\text{ g} = 4.00\text{ g}

Direct stoichiometric ratio derived from 1 mol CaCO3 producing 1 mol CO2.

4

Calculate percentage purity.

\text{Percentage purity} = \left(\frac{4.00\text{ g}}{5.00\text{ g}}\right) \times 100 = 80\%

Ratio of pure reactant mass to total sample mass expressed as a percentage.

Anahtar Kavram

Calculating percentage purity using stoichiometry and gravimetric decomposition data.

Soru 7Soru →

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\%

Cevabı ve açıklamayı göster

Cevap: $75.0\%$

Cevap

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\%

.

Adım Adım Çözüm

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.

Anahtar Kavram

Mass-volume stoichiometric calculations and percentage purity determination

Tahmini Süre:2m 0s

Soru 8Soru →

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

Cevabı ve açıklamayı göster

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

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Tahmini Süre:1m 0s

Soru 9Soru →

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

Cevabı ve açıklamayı göster

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

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Mass-Volume Stoichiometric Calculations at STP

Soru 10Soru →

Complete the statement by calculating the required gas volume under constant temperature and pressure conditions.

Aşağıdaki boşlukları doldurun

According to Gay-Lussac's law of combining volumes, in the reaction

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

,

40\text{ cm}^3

of carbon(II) oxide gas requires exactly

\text{cm}^3

of oxygen gas for complete reaction.

Cevabı ve açıklamayı göster

Cevap

20

According to Gay-Lussac's law of combining volumes, gases react in volumes that bear a simple whole-number ratio to one another at constant temperature and pressure. In the balanced equation

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

, the volume combining ratio of

CO

to

O_2

is

2:1

. Therefore,

40\text{ cm}^3

of

CO

requires

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

of

O_2

for complete combustion.

Adım Adım Çözüm

1

Determine the stoichiometric volume ratio from the balanced chemical equation.

2 volumes of

CO_{(g)}

react with 1 volume of

O_{2(g)}

.

By Gay-Lussac's law of combining volumes, the combining volumes of reacting gases at constant temperature and pressure are proportional to their stoichiometric coefficients.

2

Calculate the volume of oxygen needed to react with

40\text{ cm}^3

of carbon(II) oxide.

Volume of

O_{2(g)} = 40\text{ cm}^3 \times \frac{1}{2} = 20\text{ cm}^3

.

Since the ratio of

CO

to

O_2

is

2:1

, the required volume of

O_2

is half the volume of

CO

.

Anahtar Kavram

Gay-Lussac's Law of Combining Volumes

Soru 11Soru →

Calculate the percentage by mass of calcium in a pure sample of calcium carbonate ( $\text{CaCO}_3$ ). $[\text{Ca} = 40, \text{C} = 12, \text{O} = 16]$

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The percentage by mass of calcium in calcium carbonate is 40%.

The molar mass of calcium carbonate (

\text{CaCO}_3

) is

100\text{ g/mol}

. Since one formula unit contains

40\text{ g}

of calcium, the percentage composition by mass of calcium is

\frac{40}{100} \times 100\% = 40\%

.

Adım Adım Çözüm

1

Calculate the molar mass of calcium carbonate (

\text{CaCO}_3

).

Molar mass =

40 + 12 + (3 \times 16) = 100\text{ g/mol}

.

Sum the relative atomic masses of all constituent atoms in one formula unit of

\text{CaCO}_3

.

2

Determine the mass contribution of calcium in one mole of

\text{CaCO}_3

.

Mass of Ca =

40\text{ g}

.

Each formula unit of

\text{CaCO}_3

contains 1 atom of calcium.

3

Calculate the percentage composition of calcium.

Percentage of Ca =

\frac{40}{100} \times 100\% = 40\%

.

Divide the mass of calcium by the total molar mass of the compound and multiply by 100.

Anahtar Kavram

Percentage composition of an element in a compound

Soru 12Soru →

At constant temperature and pressure,

30\text{ cm}^3

of hydrogen gas reacts completely with

15\text{ cm}^3

of oxygen gas to form steam according to the balanced equation:

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

What is the volume of steam produced, and which law of chemical combination governs this gas volume relationship?

30\text{ cm}^3

; Gay-Lussac's Law of Combining Volumes

45\text{ cm}^3

; Law of Conservation of Mass

15\text{ cm}^3

; Law of Definite Proportions

22.4\text{ cm}^3

; Gay-Lussac's Law of Combining Volumes

Cevabı ve açıklamayı göster

Cevap: $30\text{ cm}^3$ ; Gay-Lussac's Law of Combining Volumes

Cevap

The volume of steam produced is

30\text{ cm}^3

, governed by Gay-Lussac's Law of Combining Volumes.

According to Gay-Lussac's Law of Combining Volumes, gases combine in simple numerical ratios by volume at constant temperature and pressure. In the balanced reaction

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

, 2 volumes of hydrogen combine with 1 volume of oxygen to yield 2 volumes of steam. Thus,

30\text{ cm}^3

of hydrogen produces

30\text{ cm}^3

of steam.

Adım Adım Çözüm

1

Identify the mole/volume ratio from the balanced chemical equation.

The coefficient ratio for

\text{H}_2(g) : \text{O}_2(g) : \text{H}_2\text{O}(g)

is

2 : 1 : 2

.

By Gay-Lussac's Law of Combining Volumes, when gases react at constant temperature and pressure, their reacting volumes and the volume of gaseous products are in simple numerical ratios equal to their stoichiometric coefficients.

2

Calculate the volume of steam produced from the volume of reacting hydrogen gas.

\text{Volume of } \text{H}_2\text{O}(g) = 30\text{ cm}^3 \times \frac{2}{2} = 30\text{ cm}^3

Since the ratio of hydrogen to steam is

2 : 2

(or

1 : 1

),

30\text{ cm}^3

of hydrogen produces an equal volume (

30\text{ cm}^3

) of steam.

3

Identify the law of chemical combination illustrated.

Gay-Lussac's Law of Combining Volumes.

This specific law deals with the simple integer ratios between reacting gas volumes under identical conditions of temperature and pressure.

Anahtar Kavram

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 any gaseous product, provided temperature and pressure remain constant.

Tahmini Süre:1m 15s

Soru 13Soru →

A sample of dinitrogen tetraoxide ( $\text{N}_2\text{O}_4$ ) gas occupies a volume of $5.6\text{ dm}^3$ at standard temperature and pressure (STP). What is the total number of oxygen atoms contained in this sample? $[\text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

6.02 \times 10^{23}

5.62 \times 10^{23}

1.51 \times 10^{23}

9.03 \times 10^{23}

Cevabı ve açıklamayı göster

Cevap: $6.02 \times 10^{23}$

Cevap

6.02 \times 10^{23}

oxygen atoms

The sample contains

0.25\text{ mol}

of

\text{N}_2\text{O}_4

gas because

5.6\text{ dm}^3 / 22.4\text{ dm}^3\text{mol}^{-1} = 0.25\text{ mol}

. Since each molecule of

\text{N}_2\text{O}_4

contains 4 oxygen atoms, the total amount of oxygen atoms is

0.25 \times 4 = 1.0\text{ mol}

. Multiplying

1.0\text{ mol}

by Avogadro's constant (

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

) yields

6.02 \times 10^{23}

oxygen atoms.

Adım Adım Çözüm

1

Calculate the number of moles of dinitrogen tetraoxide (

\text{N}_2\text{O}_4

) gas at STP

n(\text{N}_2\text{O}_4) = \frac{\text{Volume}}{\text{Molar Volume}} = \frac{5.6\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.25\text{ mol}

At STP, one mole of any ideal gas occupies

22.4\text{ dm}^3

.

2

Determine the moles of oxygen atoms present in

0.25\text{ mol}

of

\text{N}_2\text{O}_4

n(\text{O}) = 0.25\text{ mol} \times 4 = 1.0\text{ mol}

of oxygen atoms

Each molecule of

\text{N}_2\text{O}_4

contains 4 oxygen atoms.

3

Compute the absolute number of oxygen atoms using Avogadro's constant

N(\text{O}) = n(\text{O}) \times N_A = 1.0\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 6.02 \times 10^{23}

atoms

One mole of any substance contains

N_A

particles.

Anahtar Kavram

Molar gas volume at STP and Avogadro's constant stoichiometric conversions

Tahmini Süre:2m 0s

Soru 14Soru →

A $6.60\text{ g}$ sample of impure ammonium tetraoxosulfate(VI), $(\text{NH}_4)_2\text{SO}_4$ , is heated with excess sodium hydroxide solution. The evolved ammonia gas, $\text{NH}_3$ , is absorbed completely in $100.0\text{ cm}^3$ of $0.50\text{ mol dm}^{-3}$ tetraoxosulfate(VI) acid solution, $\text{H}_2\text{SO}_4$ . The unreacted acid requires $40.0\text{ cm}^3$ of $0.50\text{ mol dm}^{-3}$ sodium hydroxide solution for complete neutralization. What is the percentage purity of the ammonium tetraoxosulfate(VI) sample? $[\text{N} = 14, \text{H} = 1, \text{S} = 32, \text{O} = 16]$

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

80%

The correct answer is 80.0%. Through back-titration analysis, 0.020 mol of NaOH neutralizes 0.010 mol of unreacted excess H₂SO₄ out of the initial 0.050 mol, leaving 0.040 mol of H₂SO₄ to react with 0.080 mol of evolved NH₃ gas. Since 1 mole of pure ammonium tetraoxosulfate(VI) produces 2 moles of NH₃ gas, the sample contained 0.040 mol of pure (NH₄)₂SO₄. Multiplying by its molar mass (132 g/mol) yields 5.28 g of pure compound. Dividing 5.28 g by the total sample mass of 6.60 g and multiplying by 100 gives exactly 80.0%.

Adım Adım Çözüm

1

Calculate the initial moles of H₂SO₄ acid solution used for absorbing ammonia.

0.050 mol H₂SO₄

Total acid available = Volume (in dm³) × Concentration (in mol dm⁻³).

2

Calculate the unreacted moles of H₂SO₄ from the titration with NaOH.

0.010 mol excess H₂SO₄

1 mole of H₂SO₄ reacts with 2 moles of NaOH, so excess H₂SO₄ = 0.5 × moles of NaOH used.

3

Calculate moles of H₂SO₄ neutralized by evolved NH₃ gas.

0.040 mol H₂SO₄ reacted

Reacted acid = Initial total acid - Excess unreacted acid.

4

Calculate the moles of NH₃ evolved from the sample.

0.080 mol NH₃

2 moles of NH₃ react with 1 mole of H₂SO₄.

5

Determine the mass of pure (NH₄)₂SO₄ present in the original sample.

5.28 g of pure (NH₄)₂SO₄

1 mole of (NH₄)₂SO₄ yields 2 moles of NH₃. Mass = moles (0.040 mol) × molar mass (132 g/mol).

6

Compute the percentage purity of the sample.

80%

Percentage Purity = (Mass of pure substance / Total mass of impure sample) × 100%.

Anahtar Kavram

Back-titration quantitative analysis for determining percentage purity

Tahmini Süre:3m 0s

Soru 15Soru →

A sample of pure methane ( $\text{CH}_4$ ) contains $3.00\text{ g}$ of carbon. When this sample undergoes complete combustion in excess oxygen gas, all the hydrogen present is converted into water vapor ( $\text{H}_2\text{O}$ ). Based on the Law of Definite Proportions, what is the total mass (in grams) of water vapor produced? [Atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ ]

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

The total mass of water vapor produced is

9.00\text{ g}

.

According to the Law of Definite Proportions, a chemical compound always contains its component elements in a fixed ratio by mass. In methane (

\text{CH}_4

), the ratio of mass of carbon to hydrogen is

12 : 4

(

3 : 1

). Therefore,

3.00\text{ g}

of carbon is combined with

1.00\text{ g}

of hydrogen. When methane undergoes complete combustion, all

1.00\text{ g}

of hydrogen is converted into water (

\text{H}_2\text{O}

). Since hydrogen makes up

\frac{2}{18}

of the mass of water,

1.00\text{ g}

of hydrogen yields

1.00 \times \frac{18}{2} = 9.00\text{ g}

of water.

Adım Adım Çözüm

1

Calculate the mass of hydrogen present in the methane sample using the Law of Definite Proportions.

The mass of hydrogen in the sample is

1.00\text{ g}

.

In

\text{CH}_4

, the mass ratio of carbon to hydrogen is

12 : 4 = 3 : 1

. Given

3.00\text{ g}

of carbon, the mass of hydrogen is

\frac{3.00\text{ g}}{3} = 1.00\text{ g}

.

2

Determine the mass fraction of hydrogen in water (

\text{H}_2\text{O}

).

Hydrogen accounts for

\frac{2}{18}

of the total mass of water.

The molar mass of

\text{H}_2\text{O}

is

2(1) + 16 = 18\text{ g/mol}

, of which

2\text{ g}

is hydrogen.

3

Calculate the total mass of water vapor formed from the hydrogen.

The total mass of water produced is

9.00\text{ g}

.

All

1.00\text{ g}

of hydrogen from methane is converted into water. Mass of

\text{H}_2\text{O} = 1.00\text{ g} \times \frac{18}{2} = 9.00\text{ g}

.

Anahtar Kavram

Law of Definite Proportions (Constant Composition)

Soru 16Soru →

A sample of hydrated copper(II) sulfate ( $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ ) has a mass of $24.95\text{ g}$ . What is the total number of moles of oxygen atoms contained in this sample? $[\text{Relative atomic masses: } \text{Cu} = 63.5, \text{S} = 32.0, \text{O} = 16.0, \text{H} = 1.0]$

Cevabı ve açıklamayı göster

Cevap: 0.9

Cevap

The total number of moles of oxygen atoms contained in the sample is

0.90\text{ mol}

.

The molar mass of hydrated copper(II) sulfate (

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

) is

249.5\text{ g/mol}

. Dividing

24.95\text{ g}

by

249.5\text{ g/mol}

gives

0.10\text{ mol}

of the compound. Since each mole of

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

contains

9\text{ moles}

of oxygen atoms (

4

from

\text{CuSO}_4

and

5

from

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

), the total quantity of oxygen atoms is

0.10 \times 9 = 0.90\text{ mol}

.

Adım Adım Çözüm

1

Calculate the molar mass of

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

Molar mass =

249.5\text{ g/mol}

Sum the relative atomic masses of all atoms present in one formula unit of the hydrated compound.

2

Calculate the moles of the hydrated salt

Moles of compound =

0.10\text{ mol}

Divide the mass of the sample (

24.95\text{ g}

) by its molar mass (

249.5\text{ g/mol}

).

3

Determine the stoichiometric multiplier for oxygen atoms

9 moles of oxygen atoms per mole of compound

Each formula unit contains 4 oxygen atoms in the sulfate group and 5 oxygen atoms in the water of crystallization.

4

Calculate total moles of oxygen atoms

Moles of oxygen atoms =

0.90\text{ mol}

Multiply the moles of compound (

0.10\text{ mol}

) by the 9 moles of oxygen atoms per mole of compound.

Anahtar Kavram

Stoichiometric relationship of constituent atoms in a hydrated compound

Soru 17Soru →

A mixture of $30\text{ cm}^3$ of carbon(II) oxide and $25\text{ cm}^3$ of oxygen was sparked at constant temperature and pressure to form carbon(IV) oxide. What is the total volume of the resulting gaseous mixture?

30\text{ cm}^3

40\text{ cm}^3

55\text{ cm}^3

10\text{ cm}^3

Cevabı ve açıklamayı göster

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

Cevap

The total volume of the resulting gaseous mixture is

40\text{ cm}^3

.

According to the balanced equation

2\text{CO}_{(g)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)}

, two volumes of carbon(II) oxide react with one volume of oxygen to yield two volumes of carbon(IV) oxide.

30\text{ cm}^3

of carbon(II) oxide consumes

15\text{ cm}^3

of oxygen, leaving

10\text{ cm}^3

of excess oxygen unreacted. The reaction produces

30\text{ cm}^3

of carbon(IV) oxide gas. Adding the volume of carbon(IV) oxide produced to the remaining unreacted oxygen gives a total residual gaseous volume of

30\text{ cm}^3 + 10\text{ cm}^3 = 40\text{ cm}^3

.

Adım Adım Çözüm

1

Write and balance the stoichiometric chemical equation for the reaction.

2\text{CO}_{(g)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)}

To establish the mole and volume combining ratios according to Gay-Lussac's Law.

2

Determine combining volume ratios and identify the limiting reactant and unreacted gas.

By volume ratio,

2\text{ cm}^3

of

\text{CO}

reacts with

1\text{ cm}^3

of

\text{O}_2

. Therefore,

30\text{ cm}^3

of

\text{CO}

requires

\frac{1}{2} \times 30\text{ cm}^3 = 15\text{ cm}^3

of

\text{O}_2

. Unreacted

\text{O}_2 = 25\text{ cm}^3 - 15\text{ cm}^3 = 10\text{ cm}^3

.

Carbon(II) oxide is completely consumed first, making it the limiting reactant.

3

Calculate the volume of gaseous product formed.

Since

2\text{ cm}^3

of

\text{CO}

produces

2\text{ cm}^3

of

\text{CO}_2

(a

1:1

ratio),

30\text{ cm}^3

of

\text{CO}

produces

30\text{ cm}^3

of

\text{CO}_2

.

Gay-Lussac's Law applies directly to gaseous reactants and products under uniform conditions.

4

Sum the volumes of all gases present after the reaction completes.

Total residual volume = Unreacted

\text{O}_2

+ Produced

\text{CO}_2 = 10\text{ cm}^3 + 30\text{ cm}^3 = 40\text{ cm}^3

.

The final mixture contains both the newly formed gaseous product and the remaining excess reactant.

Anahtar Kavram

Gay-Lussac's Law of Combining Volumes states that when gases react under constant temperature and pressure, their combining volumes and the volumes of any gaseous products bear a simple whole-number ratio to one another.

Soru 18Soru →

When $12.0\text{ g}$ of carbon reacts completely with $32.0\text{ g}$ of oxygen gas, carbon(IV) oxide ( $\text{CO}_2$ ) is formed. According to the Law of Conservation of Mass, what is the total mass of carbon(IV) oxide produced?

44.0\text{ g}

20.0\text{ g}

22.4\text{ g}

16.0\text{ g}

Cevabı ve açıklamayı göster

Cevap: $44.0\text{ g}$

Cevap

44.0\text{ g}

of carbon(IV) oxide is produced.

The Law of Conservation of Mass dictates that mass is neither created nor destroyed in a chemical reaction. Therefore, the total mass of carbon(IV) oxide formed is equal to the sum of the mass of carbon (

12.0\text{ g}

) and oxygen (

32.0\text{ g}

), which gives

44.0\text{ g}

.

Adım Adım Çözüm

1

Identify the relevant law of chemical combination.

Law of Conservation of Mass states that total mass of reactants equals total mass of products.

Matter cannot be created or destroyed during a chemical reaction.

2

Sum the masses of all reactants.

Mass of

\text{CO}_2 = 12.0\text{ g (carbon)} + 32.0\text{ g (oxygen)} = 44.0\text{ g}

.

Both carbon and oxygen combine completely to yield carbon(IV) oxide.

Anahtar Kavram

Law of Conservation of Mass

Soru 19Soru →

A mixture of $25\text{ cm}^3$ of methane ( $CH_4$ ) and $60\text{ cm}^3$ of oxygen ( $O_2$ ) is sparked at constant temperature and pressure. Assuming water vapor condenses to liquid upon cooling to room temperature, what is the total volume of residual gas remaining in $\text{cm}^3$ ?

Cevabı ve açıklamayı göster

Cevap: 35

Cevap

The total volume of residual gas remaining after cooling to room temperature is

35\text{ cm}^3

.

According to Gay-Lussac's law of combining volumes, gases react in simple numerical ratios equal to their stoichiometric coefficients at constant temperature and pressure. For the equation

CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)

,

25\text{ cm}^3

of methane reacts completely with

50\text{ cm}^3

of oxygen to produce

25\text{ cm}^3

of

CO_2

gas. Since

60\text{ cm}^3

of oxygen was initially present,

10\text{ cm}^3

of oxygen remains unreacted. Liquid water occupies negligible volume compared to gases. Therefore, the total residual gaseous volume is the sum of unreacted oxygen and produced carbon(IV) oxide:

10\text{ cm}^3 + 25\text{ cm}^3 = 35\text{ cm}^3

.

Adım Adım Çözüm

1

Write the balanced chemical equation for the combustion of methane gas.

CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)

Gay-Lussac's law of combining volumes applies directly to the stoichiometric mole ratios of gaseous reactants and products.

2

Determine the volume of oxygen consumed and identify the excess reactant.

25\text{ cm}^3

of

CH_4

consumes

50\text{ cm}^3

of

O_2

, leaving

10\text{ cm}^3

of unreacted

O_2

.

The reaction stoichiometry requires 2 volumes of oxygen per 1 volume of methane.

3

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

Volume of

CO_2(g) = 25\text{ cm}^3

.

1 volume of methane produces 1 volume of carbon(IV) oxide gas.

4

Sum the volumes of all remaining gaseous species.

Total residual volume =

10\text{ cm}^3\text{ (excess } O_2) + 25\text{ cm}^3\text{ (formed } CO_2) = 35\text{ cm}^3

.

Water formed is liquid at room temperature and contributes negligibly to gaseous volume.

Anahtar Kavram

Gay-Lussac's Law of Combining Volumes and Stoichiometry of Gas Reactions

Soru 20Soru →

A sample of pure ammonium trioxocarbonate(IV), $(\text{NH}_4)_2\text{CO}_3$ , is determined to contain $3.6 \times 10^{24}$ hydrogen atoms. What is the total mass of oxygen present in this sample?

[Relative atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{N} = 14$ , $\text{O} = 16$ ; Avogadro's constant $N_A = 6.0 \times 10^{23} \text{ mol}^{-1}$ ]

12\text{ g}

36\text{ g}

72\text{ g}

2.25\text{ g}

Cevabı ve açıklamayı göster

Cevap: $36\text{ g}$

Cevap

36\text{ g}

The correct answer is

36\text{ g}

. Converting

3.6 \times 10^{24}

hydrogen atoms using Avogadro's constant gives

6.0\text{ moles}

of hydrogen atoms. Because each formula unit of

(\text{NH}_4)_2\text{CO}_3

contains

8

hydrogen atoms and

3

oxygen atoms, the mole ratio of O to H is

3:8

, giving

2.25\text{ moles}

of oxygen atoms. Multiplying

2.25\text{ moles}

by the molar mass of oxygen (

16\text{ g/mol}

) yields

36\text{ g}

.

Adım Adım Çözüm

1

Calculate the total number of moles of hydrogen atoms from the given particle count.

\text{Moles of H atoms} = \frac{3.6 \times 10^{24}}{6.0 \times 10^{23} \text{ mol}^{-1}} = 6.0\text{ moles}

.

Dividing particle count by Avogadro's constant gives the amount in moles.

2

Determine the number of hydrogen atoms and oxygen atoms in one formula unit of

(\text{NH}_4)_2\text{CO}_3

.

One formula unit contains

2 \times 4 = 8

hydrogen atoms and

3

oxygen atoms.

The subscript 2 outside the ammonium group

(\text{NH}_4)

multiplies both N and H inside.

3

Calculate the moles of oxygen atoms present in the sample.

\text{Moles of O atoms} = 6.0\text{ moles of H} \times \frac{3\text{ mol O}}{8\text{ mol H}} = 2.25\text{ moles of O}

.

The mole ratio of O to H in the chemical formula is

3 : 8

.

4

Convert the moles of oxygen atoms to mass.

\text{Mass of O} = 2.25\text{ mol} \times 16\text{ g/mol} = 36\text{ g}

.

Mass is obtained by multiplying the number of moles by the molar mass of oxygen.

Anahtar Kavram

Stoichiometric mole relationships between constituent elements in a chemical compound using Avogadro's constant and molar mass.

Chemical Combination and Stoichiometry

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