Chemical Combination and Stoichiometry Alıştırma Soruları — JAMB UTME — Sayfa 3

Soru 41Soru →

A gaseous mixture containing carbon monoxide ( $\text{CO}$ ) and carbon dioxide ( $\text{CO}_2$ ) has a total mass of $10.0\text{ g}$ . If the mixture contains a total of $2.408 \times 10^{23}$ oxygen atoms, calculate the mass, in grams, of carbon dioxide ( $\text{CO}_2$ ) present in the mixture. $[\text{C} = 12.0, \text{O} = 16.0, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

Cevabı ve açıklamayı göster

Cevap: 4.4

Cevap

The mass of carbon dioxide (

\text{CO}_2

) present in the mixture is

4.4\text{ g}

.

The correct calculation yields 4.4 g by converting the oxygen atom count to 0.40 moles of O atoms, formulating the system of equations for total mass (28x + 44y = 10.0) and total oxygen moles (x + 2y = 0.40), and solving for the mass of CO₂.

Adım Adım Çözüm

1

Determine the molar masses of carbon monoxide and carbon dioxide.

\text{Molar mass of CO} = 12.0 + 16.0 = 28.0\text{ g mol}^{-1}

;

\text{Molar mass of CO}_2 = 12.0 + 2(16.0) = 44.0\text{ g mol}^{-1}

.

Molar masses are required to relate the mass of each component to its molar quantity.

2

Calculate the total number of moles of oxygen atoms in the mixture using Avogadro's constant.

n_{\text{O}} = \frac{2.408 \times 10^{23}}{6.02 \times 10^{23}\text{ mol}^{-1}} = 0.40\text{ mol of O atoms}

.

Avogadro's constant converts particle count to mole quantity.

3

Set up a system of linear equations representing the total mass and total moles of oxygen atoms.

Let

x = \text{moles of CO}

and

y = \text{moles of CO}_2

.
Equation 1 (Mass):

28x + 44y = 10.0

Equation 2 (Oxygen atoms):

x + 2y = 0.40

CO contains 1 O atom per molecule and CO₂ contains 2 O atoms per molecule.

4

Solve the system of linear equations for

y

(moles of

\text{CO}_2

).

From Equation 2,

x = 0.40 - 2y

. Substituting into Equation 1 gives

28(0.40 - 2y) + 44y = 10.0 \implies 11.2 - 56y + 44y = 10.0 \implies 12y = 1.2 \implies y = 0.10\text{ mol}

.

Algebraic substitution yields the mole quantity of carbon dioxide.

5

Calculate the mass of

\text{CO}_2

present in the sample.

\text{Mass of CO}_2 = y \times \text{Molar mass} = 0.10\text{ mol} \times 44.0\text{ g mol}^{-1} = 4.4\text{ g}

.

Multiplying moles of CO₂ by its molar mass gives the required mass in grams.

Anahtar Kavram

Mole Concept, Avogadro's Constant, and Gas Mixture Stoichiometry

Soru 42Soru →

An element $Z$ exists naturally as two isotopes, $^{24}Z$ and $^{26}Z$ . On the carbon-12 scale, one atomic mass unit ( $1\text{ amu}$ ) is defined as $\frac{1}{12}$ of the mass of a single $^{12}\text{C}$ atom, where the mass of one $^{12}\text{C}$ atom is $1.992 \times 10^{-23}\text{ g}$ . If the average absolute mass of one atom of element $Z$ is $4.0504 \times 10^{-23}\text{ g}$ , what is the percentage abundance of the heavier isotope, $^{26}Z$ ?

20\%

40\%

50\%

80\%

Cevabı ve açıklamayı göster

Cevap: $20\%$

Cevap

The percentage abundance of the heavier isotope

^{26}Z

is

20\%

.

One atomic mass unit (

1\text{ amu}

) on the carbon-12 scale equals

\frac{1.992 \times 10^{-23}\text{ g}}{12} = 1.66 \times 10^{-24}\text{ g}

. Dividing the absolute mass of one atom of

Z

(

4.0504 \times 10^{-23}\text{ g}

) by

1.66 \times 10^{-24}\text{ g}

yields a relative atomic mass of

24.4

. Setting up the weighted average formula

24.4 = 24(1 - p) + 26p

, where

p

is the fractional abundance of

^{26}Z

, gives

2p = 0.4

, so

p = 0.20

or

20\%

.

Adım Adım Çözüm

1

Calculate the value of

1\text{ amu}

in grams using the carbon-12 reference standard

1\text{ amu} = \frac{1.992 \times 10^{-23}\text{ g}}{12} = 1.66 \times 10^{-24}\text{ g}

Relative atomic mass is based on the carbon-12 scale, where

1\text{ amu}

equals one-twelfth the mass of a carbon-12 atom.

2

Determine the relative atomic mass (RAM) of element

Z

\text{RAM of } Z = \frac{4.0504 \times 10^{-23}\text{ g}}{1.66 \times 10^{-24}\text{ g}} = 24.4

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

1\text{ amu}

.

3

Set up and solve the isotopic abundance equation for percentage

x

of

^{26}Z

24.4 = \frac{24(100 - x) + 26x}{100} \implies 2440 = 2400 - 24x + 26x \implies 2x = 40 \implies x = 20\%

The relative atomic mass of an element is the weighted average of the mass numbers of its naturally occurring isotopes.

Anahtar Kavram

Relative Atomic Mass calculation on the Carbon-12 scale and isotopic abundance determination

Tahmini Süre:2m 30s

Soru 43Soru →

What is the relative molecular mass of hydrated sodium trioxocarbonate(IV) crystals, $Na_2CO_3 \cdot 10H_2O$ ? $[Na = 23, C = 12, O = 16, H = 1]$

286

106

124

268

Cevabı ve açıklamayı göster

Cevap: 286

Cevap

The relative molecular mass of hydrated sodium trioxocarbonate(IV),

Na_2CO_3 \cdot 10H_2O

, is 286.

The relative molecular mass of a hydrated compound is calculated by taking the sum of the relative atomic masses of all constituent elements in both the anhydrous salt and the water molecules of crystallization. For

Na_2CO_3 \cdot 10H_2O

,

(23 \times 2) + 12 + (16 \times 3) + 10 \times [(1 \times 2) + 16] = 46 + 12 + 48 + 180 = 286

.

Adım Adım Çözüm

1

Calculate the relative formula mass of anhydrous

Na_2CO_3

(23 \times 2) + 12 + (16 \times 3) = 46 + 12 + 48 = 106

Sum the relative atomic masses of all atoms present in the anhydrous salt portion.

2

Calculate the relative mass of the water of crystallization (

10H_2O

)

10 \times [(1 \times 2) + 16] = 10 \times 18 = 180

Multiply the relative molecular mass of one water molecule (18) by the coefficient 10.

3

Add the mass of the anhydrous salt to the mass of the water of crystallization

106 + 180 = 286

Combine both components to obtain the total relative molecular mass of the hydrated crystal.

Anahtar Kavram

Relative Molecular Mass of Hydrated Salts

Soru 44Soru →

An oxide of nitrogen contains $30.4\%$ nitrogen and $69.6\%$ oxygen by mass. If its relative molar mass is $92\text{ g/mol}$ , what is the molecular formula of the compound? $[N = 14, O = 16]$

Cevabı ve açıklamayı göster

Cevap: N2O4; N₂O₄; $N_2O_4$

Cevap

The molecular formula of the oxide is

N_2O_4

.

First, the empirical formula is derived by dividing the percentage composition by relative atomic masses (

30.4/14 = 2.17

for N and

69.6/16 = 4.35

for O), which gives a simple whole number ratio of

1:2

(

NO_2

). The empirical formula mass of

NO_2

is

46\text{ g/mol}

. Dividing the molar mass (

92\text{ g/mol}

) by the empirical mass (

46\text{ g/mol}

) yields an integer factor of

2

. Multiplying the empirical formula by

2

gives the molecular formula

N_2O_4

.

Adım Adım Çözüm

1

Calculate the relative number of moles of each element.

Moles of

N = \frac{30.4}{14} = 2.17

; Moles of

O = \frac{69.6}{16} = 4.35

.

Dividing mass percentage by atomic mass gives the mole ratio.

2

Determine the simplest whole number mole ratio.

Ratio

N : O = \frac{2.17}{2.17} : \frac{4.35}{2.17} = 1 : 2

. Empirical formula =

NO_2

.

Dividing by the smallest mole value yields the empirical formula subscripts.

3

Calculate the empirical formula mass and the multiplier integer

n

.

Empirical formula mass =

14 + 2(16) = 46\text{ g/mol}

.

n = \frac{92}{46} = 2

.

The integer multiplier

n

is the ratio of molar mass to empirical formula mass.

4

Multiply empirical formula subscripts by

n

.

Molecular formula =

(NO_2)_2 = N_2O_4

.

Applying the integer multiplier gives the actual number of atoms in the molecule.

Anahtar Kavram

Determining empirical and molecular formulas from elemental percentage composition and relative molar mass.

Tahmini Süre:1m 30s

Soru 45Soru →

A sample of hydrated magnesium tetraoxosulfate(VI) has the formula $MgSO_4 \cdot xH_2O$ . If the relative molecular mass of the hydrated salt is $246$ , what is the value of $x$ ? $[Mg = 24, S = 32, O = 16, H = 1]$

5

6

7

10

Cevabı ve açıklamayı göster

Cevap: $7$

Cevap

The value of

x

is

7

.

The relative formula mass of anhydrous

MgSO_4

is calculated as

24 + 32 + (4 \times 16) = 120

. Subtracting this from the total relative molecular mass of

246

leaves

126

for the water of crystallization (

xH_2O

). Since each

H_2O

molecule has a relative mass of

(2 \times 1) + 16 = 18

, dividing

126

by

18

gives

x = 7

.

Adım Adım Çözüm

1

Calculate the relative formula mass of the anhydrous salt

MgSO_4

.

24 + 32 + (4 \times 16) = 120

Summing the relative atomic masses of magnesium, sulfur, and four oxygen atoms gives the mass of the anhydrous portion.

2

Calculate the relative molecular mass of a single water molecule

H_2O

.

(2 \times 1) + 16 = 18

Two hydrogen atoms and one oxygen atom combine to give a relative molecular mass of 18.

3

Set up the equation for the relative molecular mass of the hydrated salt and solve for

x

.

120 + 18x = 246 \implies 18x = 126 \implies x = 7

Subtracting the anhydrous mass of 120 from the total mass of 246 gives 126 for the water content, and dividing by 18 yields 7 molecules of water of crystallization.

Anahtar Kavram

Relative Molecular Mass of Hydrated Salts

Tahmini Süre:1m 30s

Soru 46Soru →

A compound has an empirical formula of $CH_2$ and a relative molecular mass of $42$ . What is the molecular formula of the compound? $[C = 12, H = 1]$

C_3H_6

C_2H_4

C_4H_8

CH_2

Cevabı ve açıklamayı göster

Cevap: $C_3H_6$

Cevap

The molecular formula of the compound is

C_3H_6

.

The empirical formula mass of

CH_2

is

14\text{ g/mol}

. Dividing the given relative molecular mass (

42

) by

14

gives an integer factor

n = 3

. Multiplying the empirical formula indices by

3

gives

C_3H_6

.

Adım Adım Çözüm

1

Calculate the empirical formula mass of

CH_2

.

Empirical formula mass

= 12 + 2(1) = 14\text{ g/mol}

.

The empirical formula mass is required to find the integer ratio multiplier

n

.

2

Determine the integer multiplier

n

by dividing the relative molecular mass by the empirical formula mass.

n = \frac{42}{14} = 3

.

The integer multiplier indicates how many empirical units constitute the molecular formula.

3

Multiply the subscripts of the empirical formula

CH_2

by

n = 3

.

Molecular formula

= (CH_2)_3 = C_3H_6

.

Applying the integer multiplier yields the true formula of the compound.

Anahtar Kavram

Molecular Formula Derivation from Empirical Formula and Molar Mass

Soru 47Soru →

Calculate the relative molecular mass of hydrated magnesium tetraoxosulfate(VI), $MgSO_4 \cdot 7H_2O$ . $[Mg = 24, S = 32, O = 16, H = 1]$

Cevabı ve açıklamayı göster

Cevap: 246

Cevap

246

The relative molecular mass of

MgSO_4 \cdot 7H_2O

is calculated by summing the atomic masses of all constituent atoms. The anhydrous part

MgSO_4

contributes

24 + 32 + (4 \times 16) = 120

. The water of crystallization

7H_2O

contributes

7 \times (2 + 16) = 126

. The total relative molecular mass is

120 + 126 = 246

.

Adım Adım Çözüm

1

Calculate the mass of the anhydrous salt component,

MgSO_4

24 + 32 + (4 \times 16) = 120

Sum the relative atomic masses of one magnesium atom, one sulfur atom, and four oxygen atoms.

2

Calculate the mass of the water of crystallization component,

7H_2O

7 \times (2 + 16) = 126

Multiply the molecular mass of one water molecule (18) by the coefficient 7.

3

Combine the mass of

MgSO_4

and

7H_2O

120 + 126 = 246

The total relative molecular mass of a hydrated salt is the sum of the anhydrous salt mass and the water of crystallization mass.

Anahtar Kavram

Relative Molecular Mass of Hydrated Salts

Soru 48Soru →

An organic compound containing carbon, hydrogen, and nitrogen was analyzed and found to contain $61.02\%$ carbon and $15.25\%$ hydrogen by mass, with the remainder being nitrogen. Given that the relative molecular mass of the compound is $118\text{ g/mol}$ , what is the value of the integer multiplier $n$ that relates its empirical formula to its molecular formula? $[C = 12, H = 1, N = 14]$

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of the integer multiplier n is 2.

Subtracting the carbon (

61.02\%

) and hydrogen (

15.25\%

) percentages from

100\%

gives a nitrogen content of

23.73\%

. Converting these mass percentages to mole ratios by dividing by relative atomic masses (

C=12, H=1, N=14

) gives

5.085\text{ mol}

of C,

15.25\text{ mol}

of H, and

1.695\text{ mol}

of N. Dividing by the smallest value (

1.695

) produces the mole ratio

3:9:1

, establishing the empirical formula as

C_3H_9N

. The empirical mass is

(3 \times 12) + (9 \times 1) + 14 = 59\text{ g/mol}

. Dividing the relative molecular mass (

118\text{ g/mol}

) by the empirical mass (

59\text{ g/mol}

) yields

n = 2

.

Adım Adım Çözüm

1

Calculate percentage composition of nitrogen

23.73%

The sum of percentages of all constituent elements in a compound must equal 100%.

2

Calculate relative number of moles for each element

C = 5.085 mol, H = 15.25 mol, N = 1.695 mol

Dividing the mass percentage of each element by its relative atomic mass gives its relative mole quantity.

3

Determine simplest whole number atomic ratio

C : H : N = 3 : 9 : 1

Dividing all mole amounts by the smallest value (1.695) gives the empirical mole ratio.

4

Determine the empirical formula mass

59 g/mol

Summing the atomic masses of elements in C3H9N yields (3 x 12) + (9 x 1) + 14 = 59 g/mol.

5

Calculate the integer multiplier n

2

Dividing relative molecular mass (118 g/mol) by empirical formula mass (59 g/mol) gives n = 2.

Anahtar Kavram

Calculation of Empirical Formula and Molecular Formula Multiplier

Soru 49Soru →

A sample of pure hydrated aluminum nitrate, $\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}$ , has a mass of $75.0\text{ g}$ . What is the mass, in grams, of oxygen contained in this sample? $[\text{Al} = 27, \text{N} = 14, \text{O} = 16, \text{H} = 1]$

Cevabı ve açıklamayı göster

Cevap: 57.6

Cevap

57.6

The molar mass of

\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}

is

375\text{ g/mol}

. A

75.0\text{ g}

sample equals

0.20\text{ mol}

of the hydrated salt. Because each formula unit contains

18

oxygen atoms (

9

from the nitrate groups and

9

from the water molecules),

0.20\text{ mol}

of the salt contains

3.60\text{ mol}

of oxygen atoms. Multiplying

3.60\text{ mol}

by the atomic mass of oxygen (

16\text{ g/mol}

) yields

57.6\text{ g}

.

Adım Adım Çözüm

1

Calculate the molar mass of hydrated aluminum nitrate,

\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}

.

375\text{ g/mol}

Sum the atomic masses of all constituent atoms:

\text{Al} = 27

,

3 \times \text{NO}_3 = 3 \times 62 = 186

,

9 \times \text{H}_2\text{O} = 9 \times 18 = 162

. Total

= 27 + 186 + 162 = 375\text{ g/mol}

.

2

Determine the amount (in moles) of the compound in the

75.0\text{ g}

sample.

0.20\text{ mol}

Moles of compound

= \frac{\text{mass}}{\text{molar mass}} = \frac{75.0\text{ g}}{375\text{ g/mol}} = 0.20\text{ mol}

.

3

Determine the total moles of oxygen atoms per mole of the hydrated compound.

18\text{ mol of O}

Each formula unit contains

9

oxygen atoms in the nitrate groups (

3 \times 3

) and

9

oxygen atoms in the water molecules (

9 \times 1

), giving a total of

18

oxygen atoms.

4

Calculate the total mass of oxygen atoms in the sample.

57.6\text{ g}

Mass of oxygen

= \text{moles of compound} \times 18 \times \text{molar mass of O} = 0.20 \times 18 \times 16 = 3.60 \times 16 = 57.6\text{ g}

.

Anahtar Kavram

Mole Concept and Stoichiometric Mass Composition in Hydrated Compounds

Soru 50Soru →

A $2.016\text{ g}$ sample of a hydrated dicarboxylic acid, $(\text{COOH})_2 \cdot n\text{H}_2\text{O}$ , was dissolved in distilled water and made up to $250.0\text{ cm}^3$ in a volumetric flask. A $25.0\text{ cm}^3$ portion of this solution required exactly $20.0\text{ cm}^3$ of $0.160\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the value of the integer $n$ in the formula of the hydrated acid? $[\text{H} = 1, \text{C} = 12, \text{O} = 16]$

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of the integer n is 2.

Titration of 20.0 cm³ of 0.160 mol dm⁻³ NaOH consumes 0.0032 mol of NaOH. Because the dicarboxylic acid is diprotic ((COOH)₂), it reacts in a 1:2 ratio with NaOH, giving 0.0016 mol of acid in the 25.0 cm³ aliquot. Scaling to the full 250.0 cm³ flask yields 0.0160 mol of hydrated acid. The molar mass of the hydrated acid is 2.016 g / 0.0160 mol = 126 g/mol. Since the anhydrous formula mass of (COOH)₂ is 90 g/mol, the water of crystallization contributes 126 - 90 = 36 g/mol. Dividing 36 by 18 (the molar mass of H₂O) gives n = 2.

Adım Adım Çözüm

1

Calculate the amount in moles of NaOH used in the titration.

Moles of NaOH = 0.0032 mol

Moles = concentration × volume = 0.160 mol dm⁻³ × (20.0 / 1000) dm³ = 0.0032 mol.

2

Determine the moles of acid present in the 25.0 cm³ titration sample using the stoichiometric mole ratio.

Moles of acid in 25.0 cm³ = 0.0016 mol

Ethanoic/oxalic acid is diprotic ((COOH)₂ + 2NaOH → (COONa)₂ + 2H₂O), requiring 2 moles of NaOH per mole of acid. Moles of acid = 0.0032 / 2 = 0.0016 mol.

3

Scale up to find the total moles of acid in the original 250.0 cm³ solution.

Total moles of acid = 0.0160 mol

Total moles = 0.0016 mol × (250.0 cm³ / 25.0 cm³) = 0.0160 mol.

4

Calculate the molar mass of the hydrated dicarboxylic acid.

Molar mass = 126 g mol⁻¹

Molar mass M = sample mass / total moles = 2.016 g / 0.0160 mol = 126 g mol⁻¹.

5

Calculate the value of integer n by comparing the molar mass to the anhydrous acid mass.

n = 2

Formula mass of anhydrous (COOH)₂ = 2(12) + 2(1) + 4(16) = 90 g mol⁻¹. The water component mass is 18n = 126 - 90 = 36 g mol⁻¹, giving n = 36 / 18 = 2.

Anahtar Kavram

Empirical and Molecular Formula Calculations with Water of Crystallization and Volumetric Stoichiometry

Soru 51Soru →

A

3.25\text{ g}

sample of impure zinc granules reacts completely with excess dilute tetraoxosulfate(VI) acid according to the equation:

\text{Zn(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{H}_2\text{(g)}

If

896\text{ cm}^3

of hydrogen gas is collected at STP, calculate the percentage purity of the zinc sample.

[\text{Zn} = 65\text{; Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

80%

The volume of hydrogen gas produced at STP (

896\text{ cm}^3 = 0.896\text{ dm}^3

) corresponds to

0.04\text{ mol}

of

\text{H}_2

. Based on the

1:1

stoichiometric reaction between zinc and tetraoxosulfate(VI) acid,

0.04\text{ mol}

of pure zinc was present in the sample. The mass of pure zinc is

0.04\text{ mol} \times 65\text{ g/mol} = 2.60\text{ g}

. Dividing this pure mass by the total sample mass of

3.25\text{ g}

and multiplying by

100\%

yields a percentage purity of

80\%

.

Adım Adım Çözüm

1

Convert the collected volume of hydrogen gas from cm³ to dm³

\text{Volume of H}_2 = \frac{896\text{ cm}^3}{1000\text{ cm}^3/\text{dm}^3} = 0.896\text{ dm}^3

Molar gas volume is given in dm³/mol, so gas volume must be expressed in dm³.

2

Calculate the amount in moles of hydrogen gas produced at STP

\text{Moles of H}_2 = \frac{0.896\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.04\text{ mol}

Number of moles of gas at STP equals volume divided by molar volume.

3

Determine the moles and mass of pure zinc that reacted

\text{Moles of pure Zn} = 0.04\text{ mol}

;

\text{Mass of pure Zn} = 0.04\text{ mol} \times 65\text{ g/mol} = 2.60\text{ g}

The reaction stoichiometry shows a 1:1 mole ratio between Zn and H₂.

4

Calculate the percentage purity of the zinc sample

\text{Percentage Purity} = \left( \frac{2.60\text{ g}}{3.25\text{ g}} \right) \times 100\% = 80\%

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

Anahtar Kavram

Percentage purity determination via gas volume stoichiometry at STP

Soru 52Soru →

A $5.00\text{ g}$ sample of impure zinc granules reacts completely with excess dilute tetraoxosulfate(VI) acid to produce $1.12\text{ dm}^3$ of hydrogen gas at s.t.p. What is the percentage purity of the zinc sample? $[\text{Zn} = 65, \text{Molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]$

$30.0\%

$60.7\%

$65.0\%

$35.0\%

Cevabı ve açıklamayı göster

Cevap: $65.0\%

Cevap

65.0\%

The balanced chemical equation

\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2

shows a

1:1

mole ratio between zinc and hydrogen gas. At s.t.p.,

1.12\text{ dm}^3

of

\text{H}_2

represents

\frac{1.12}{22.4} = 0.05\text{ mol}

. Thus,

0.05\text{ mol}

of pure zinc was present, corresponding to

0.05 \times 65 = 3.25\text{ g}

. Dividing the pure zinc mass by the total sample mass (

5.00\text{ g}

) gives

\frac{3.25}{5.00} \times 100\% = 65.0\%

.

Adım Adım Çözüm

1

Calculate the amount of hydrogen gas produced in moles at s.t.p.

\text{Moles of H}_2 = \frac{1.12\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.05\text{ mol}

The reaction produces

1\text{ mol}

of

\text{H}_2

for every

1\text{ mol}

of pure zinc reacted:

\text{Zn (s)} + \text{H}_2\text{SO}_4\text{ (aq)} \rightarrow \text{ZnSO}_4\text{ (aq)} + \text{H}_2\text{ (g)}

.

2

Determine the mass of pure zinc that reacted.

\text{Mass of pure Zn} = 0.05\text{ mol} \times 65\text{ g mol}^{-1} = 3.25\text{ g}

Molar mass converts the stoichiometric mole amount to pure mass of the element.

3

Calculate the percentage purity of the zinc sample.

\text{Percentage Purity} = \left( \frac{3.25\text{ g}}{5.00\text{ g}} \right) \times 100\% = 65.0\%

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

Anahtar Kavram

Calculating percentage purity from stoichiometric gas yields at s.t.p.

Tahmini Süre:1m 30s

Soru 53Soru →

A $5.00\text{ g}$ sample of an impure hydrated calcium tetraoxosulfate(VI) salt, $\text{CaSO}_4 \cdot x\text{H}_2\text{O}$ , containing $14\%$ non-volatile, inert impurities by mass, is heated strongly to constant mass to remove all water of crystallization. If the mass of the remaining dry residue is $4.10\text{ g}$ , what is the value of $x$ ? (Relative atomic masses: $\text{Ca} = 40$ , $\text{S} = 32$ , $\text{O} = 16$ , $\text{H} = 1$ )

1

2

3

5

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of

x

is 2.

The original

5.00\text{ g}

sample contains

14\%

non-volatile impurity (

0.70\text{ g}

). Upon heating, only water evaporates, leaving

4.10\text{ g}

of residue. Subtracting the

0.70\text{ g}

of impurity gives

3.40\text{ g}

of pure anhydrous

\text{CaSO}_4

(

0.025\text{ mol}

). The mass of evaporated water is

5.00\text{ g} - 4.10\text{ g} = 0.90\text{ g}

(

0.050\text{ mol}

). Dividing moles of water by moles of anhydrous salt yields

x = 2

.

Adım Adım Çözüm

1

Calculate the mass of the inert, non-volatile impurity in the original sample.

\text{Mass of impurity} = 14\% \times 5.00\text{ g} = 0.70\text{ g}

.

The sample consists of pure hydrated salt and

14\%

inert non-volatile impurity.

2

Determine the mass of pure anhydrous

\text{CaSO}_4

in the dry residue.

\text{Mass of pure } \text{CaSO}_4 = 4.10\text{ g} - 0.70\text{ g} = 3.40\text{ g}

.

Since the impurity is non-volatile, it remains in the

4.10\text{ g}

dry residue along with the anhydrous salt.

3

Calculate the mass of water of crystallization driven off.

\text{Mass of } \text{H}_2\text{O} = 5.00\text{ g} - 4.10\text{ g} = 0.90\text{ g}

.

Heating to constant mass removes only the volatile water of crystallization.

4

Find the molar amounts of pure

\text{CaSO}_4

and

\text{H}_2\text{O}

.

\text{Moles of } \text{CaSO}_4 = \frac{3.40\text{ g}}{136\text{ g/mol}} = 0.025\text{ mol}

;

\text{Moles of } \text{H}_2\text{O} = \frac{0.90\text{ g}}{18\text{ g/mol}} = 0.050\text{ mol}

.

Molar masses are

136\text{ g/mol}

for

\text{CaSO}_4

and

18\text{ g/mol}

for

\text{H}_2\text{O}

.

5

Determine the mole ratio

x = \frac{n(\text{H}_2\text{O})}{n(\text{CaSO}_4)}

.

x = \frac{0.050}{0.025} = 2

.

The formula ratio requires the relative mole ratio of water to anhydrous salt.

Anahtar Kavram

Accounting for non-volatile impurities in hydrated salt analysis to determine water of crystallization.

Soru 54Soru →

A

3.50\text{ g}

sample of impure potassium trioxochlorate(V),

\text{KClO}_3

, was completely decomposed by heating in the presence of a manganese(IV) oxide catalyst according to the equation:

2\text{KClO}_3(\text{s}) \rightarrow 2\text{KCl}(\text{s}) + 3\text{O}_2(\text{g})

If

0.672\text{ dm}^3

of oxygen gas was collected at STP, what is the percentage purity of the

\text{KClO}_3

sample?

(\text{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})

70.0\%

105.0\%

65.3\%

35.0\%

Cevabı ve açıklamayı göster

Cevap: $70.0\%$

Cevap

The percentage purity of the potassium trioxochlorate(V) sample is 70.0%.

The option stating 70.0% is correct. Converting

0.672\text{ dm}^3

of

\text{O}_2

gas at STP yields

0.030\text{ mol}

of

\text{O}_2

. Using the balanced stoichiometric mole ratio (

2\text{ KClO}_3 : 3\text{ O}_2

), this corresponds to

0.020\text{ mol}

of pure

\text{KClO}_3

. Multiplying by its molar mass (

122.5\text{ g/mol}

) gives

2.45\text{ g}

of pure

\text{KClO}_3

. Dividing

2.45\text{ g}

by the total sample mass of

3.50\text{ g}

and multiplying by

100\%

gives

70.0\%

.

Adım Adım Çözüm

1

Calculate the moles of oxygen gas collected at STP

\text{Moles of O}_2 = \frac{0.672\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.030\text{ mol}

At STP, one mole of any ideal gas occupies 22.4 dm³.

2

Determine the moles of pure potassium trioxochlorate(V) reacted using stoichiometric coefficients

\text{Moles of KClO}_3 = 0.030\text{ mol O}_2 \times \frac{2\text{ mol KClO}_3}{3\text{ mol O}_2} = 0.020\text{ mol}

From the balanced chemical equation, 2 moles of KClO3 decompose to yield 3 moles of O2.

3

Calculate the mass of pure potassium trioxochlorate(V)

\text{Molar mass of KClO}_3 = 39.0 + 35.5 + 3(16.0) = 122.5\text{ g/mol}

\text{Mass of pure KClO}_3 = 0.020\text{ mol} \times 122.5\text{ g/mol} = 2.45\text{ g}

Mass is obtained by multiplying moles by molar mass.

4

Calculate the percentage purity of the original sample

\text{Percentage purity} = \left(\frac{2.45\text{ g}}{3.50\text{ g}}\right) \times 100\% = 70.0\%

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

Anahtar Kavram

Percentage purity calculation via gas stoichiometry at STP

Soru 55Soru →

A

10.0\text{ g}

sample of impure limestone containing calcium trioxocarbonate(IV),

\text{CaCO}_3

, was strongly heated until decomposition was complete according to the equation:

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

If

1.792\text{ dm}^3

of carbon(IV) oxide gas was evolved at s.t.p., what is the percentage purity of the limestone 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}]

44.8\%

74.7\%

80.0\%

89.6\%

Cevabı ve açıklamayı göster

Cevap: $80.0\%$

Cevap

The percentage purity of the limestone sample is 80.0%.

From the balanced chemical equation, 1 mole of calcium trioxocarbonate(IV) produces 1 mole of carbon(IV) oxide gas. At s.t.p.,

1.792\text{ dm}^3

of

\text{CO}_2

equals

\frac{1.792}{22.4} = 0.08\text{ mol}

. Consequently,

0.08\text{ mol}

of pure

\text{CaCO}_3

was present. With a molar mass of

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

, the mass of pure

\text{CaCO}_3

is

8.00\text{ g}

. Dividing this pure mass by the

10.0\text{ g}

total sample mass yields a purity of

80.0\%

.

Adım Adım Çözüm

1

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

\text{Moles of CO}_2 = \frac{\text{Volume at s.t.p.}}{\text{Molar volume at s.t.p.}} = \frac{1.792\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.08\text{ mol}

Gas volume at standard temperature and pressure directly determines mole quantity using standard molar gas volume.

2

Determine the molar mass of pure calcium trioxocarbonate(IV),

\text{CaCO}_3

.

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

Required to convert moles of pure reactant into mass.

3

Use the mole ratio from the balanced chemical equation to find the mass of pure

\text{CaCO}_3

.

Mole ratio of

\text{CaCO}_3 : \text{CO}_2 = 1 : 1

. Moles of pure

\text{CaCO}_3 = 0.08\text{ mol}

. Mass of pure

\text{CaCO}_3 = 0.08\text{ mol} \times 100\text{ g mol}^{-1} = 8.00\text{ g}

.

Stoichiometry dictates that 1 mole of

\text{CaCO}_3

produces 1 mole of

\text{CO}_2

upon complete decomposition.

4

Calculate the percentage purity of the limestone sample.

\text{Percentage purity} = \left( \frac{\text{Mass of pure CaCO}_3}{\text{Mass of impure sample}} \right) \times 100 = \left( \frac{8.00\text{ g}}{10.0\text{ g}} \right) \times 100 = 80.0\%

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

Anahtar Kavram

Percentage Purity and Stoichiometry from Gas Volumes

Tahmini Süre:1m 30s

Soru 56Soru →

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

Aşağıdaki boşlukları doldurun

When

11.2\text{ g}

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

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

the volume of chlorine gas consumed at STP is

\text{dm}^3

.
[Relative atomic mass:

\text{Fe} = 56

; Molar gas volume at STP

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

]

Cevabı ve açıklamayı göster

Cevap

The volume of chlorine gas consumed at STP is

6.72\text{ dm}^3

.

To find the volume of

Cl_2

gas consumed at STP, first determine the amount of

Fe

in moles:

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

. From the balanced equation

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

,

2\text{ moles of } Fe

react with

3\text{ moles of } Cl_2

. Therefore,

0.20\text{ mol of } Fe

requires

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

. At STP,

1\text{ mole of gas}

occupies

22.4\text{ dm}^3

, so the volume of

Cl_2

is

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

.

Adım Adım Çözüm

1

Calculate the number of moles of iron reacted

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

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

2

Determine the moles of chlorine gas (

Cl_2

) required using the stoichiometric mole ratio

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

The balanced chemical equation shows that

2\text{ moles of } Fe

react with

3\text{ moles of } Cl_2

(a

2:3

mole ratio).

3

Calculate the volume of

Cl_2

gas consumed at STP

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

Multiply the number of moles of

Cl_2

by the molar volume of a gas at STP.

Anahtar Kavram

Mass-Volume Stoichiometric Calculations at STP

Tahmini Süre:2m 0s

Soru 57Soru →

Hydrogen gas reacts with oxygen gas to form water according to the balanced chemical equation:

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

If

4.0\text{ moles}

of hydrogen gas (

\text{H}_2

) are mixed with

3.0\text{ moles}

of oxygen gas (

\text{O}_2

) and allowed to react completely, which of the following correctly identifies the limiting reactant and the amount of excess reactant remaining?

\text{H}_2

is the limiting reactant, and

1.0\text{ mole}

of

\text{O}_2

remains unreacted.

\text{O}_2

is the limiting reactant, and

1.0\text{ mole}

of

\text{H}_2

remains unreacted.

\text{H}_2

is the limiting reactant, and

2.0\text{ moles}

of

\text{O}_2

remain unreacted.

\text{O}_2

is the limiting reactant, and

2.0\text{ moles}

of

\text{H}_2

remain unreacted.

Cevabı ve açıklamayı göster

Cevap: $\text{H}_2$ is the limiting reactant, and $1.0\text{ mole}$ of $\text{O}_2$ remains unreacted.

Cevap

\text{H}_2

is the limiting reactant, and

1.0\text{ mole}

of

\text{O}_2

remains unreacted.

The balanced chemical equation indicates a

2:1

mole ratio between

\text{H}_2

and

\text{O}_2

. Reacting

4.0\text{ moles}

of

\text{H}_2

requires

2.0\text{ moles}

of

\text{O}_2

. Since

3.0\text{ moles}

of

\text{O}_2

are present,

\text{H}_2

is completely consumed first (limiting reactant), leaving

1.0\text{ mole}

of

\text{O}_2

unreacted as excess.

Adım Adım Çözüm

1

Determine the mole ratio from the balanced equation

The mole ratio of

\text{H}_2

to

\text{O}_2

is

2 : 1

.

The stoichiometric coefficients in

2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}

dictate that

2\text{ moles}

of

\text{H}_2

consume

1\text{ mole}

of

\text{O}_2

.

2

Calculate the required moles of oxygen for the given hydrogen

\text{Required moles of } \text{O}_2 = 4.0\text{ moles } \text{H}_2 \times \frac{1\text{ mole } \text{O}_2}{2\text{ moles } \text{H}_2} = 2.0\text{ moles } \text{O}_2

.

This determines how much oxygen is actually needed to completely react with all

4.0\text{ moles}

of hydrogen.

3

Identify the limiting reactant and excess amount

Available

\text{O}_2 = 3.0\text{ moles}

, which is greater than the required

2.0\text{ moles}

. Therefore,

\text{H}_2

is the limiting reactant and excess

\text{O}_2 = 3.0 - 2.0 = 1.0\text{ mole}

.

The reactant that runs out first (

\text{H}_2

) limits the reaction, leaving unreacted excess oxygen.

Anahtar Kavram

Limiting Reactant Determination

Soru 58Soru →

What is the total number of nitrogen atoms contained in a $5.60\text{ dm}^3$ sample of dinitrogen monoxide gas ( $\text{N}_2\text{O}$ ) measured at standard temperature and pressure ( $\text{STP}$ )? $[N_A = 6.02 \times 10^{23}\text{ mol}^{-1}\text{; Molar volume of gas at STP } = 22.4\text{ dm}^3\text{mol}^{-1}]$

3.01 \times 10^{23}

2.81 \times 10^{23}

1.51 \times 10^{23}

4.52 \times 10^{23}

Cevabı ve açıklamayı göster

Cevap: $3.01 \times 10^{23}$

Cevap

3.01 \times 10^{23}

nitrogen atoms

The correct answer of

3.01 \times 10^{23}

is calculated by determining the amount in moles of dinitrogen monoxide gas (

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

), multiplying by 2 because there are two nitrogen atoms per

\text{N}_2\text{O}

molecule (

0.50\text{ mol}

of N atoms), and multiplying by Avogadro's constant (

0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}

).

Adım Adım Çözüm

1

Calculate the moles of dinitrogen monoxide gas at STP

\text{Moles of N}_2\text{O} = \frac{5.60\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.25\text{ mol}

At STP, one mole of any ideal gas occupies a molar volume of 22.4 dm³.

2

Determine the number of moles of nitrogen atoms

\text{Moles of N atoms} = 0.25\text{ mol} \times 2 = 0.50\text{ mol}

Each formula unit of dinitrogen monoxide (N₂O) contains 2 atoms of nitrogen.

3

Convert moles of nitrogen atoms to the total number of atoms using Avogadro's constant

\text{Number of N atoms} = 0.50\text{ mol} \times 6.02 \times 10^{23}\text{ atoms/mol} = 3.01 \times 10^{23}\text{ atoms}

Avogadro's constant provides the number of particles per mole of substance.

Anahtar Kavram

Mole Concept and Avogadro's Constant applied to Gas Molar Volume

Soru 59Soru →

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 $\text{KNO}_3$ ? $[\text{K} = 39, \text{N} = 14, \text{O} = 16, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

Cevabı ve açıklamayı göster

Cevap: 10.1

Cevap

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}

.

Adım Adım Çözüm

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}

.

Anahtar Kavram

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

Soru 60Soru →

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

Aşağıdaki boşlukları doldurun

When

16.0\text{ g}

of iron(III) oxide (

Fe_2O_3

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

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

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

[Relative atomic masses:

\text{Fe} = 56

,

\text{O} = 16

; Molar volume of gas at STP

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

]

Cevabı ve açıklamayı göster

Cevap

The volume of carbon monoxide consumed at STP is

6.72\text{ dm}^3

and the mass of iron metal produced is

11.2\text{ g}

.

Based on the stoichiometry of the reaction

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

,

1\text{ mol}

(

160\text{ g}

) of

Fe_2O_3

reacts with

3\text{ mol}

(

0.30\text{ mol}

for

16.0\text{ g}

) of

CO

gas, yielding a volume of

6.72\text{ dm}^3

at STP, and produces

2\text{ mol}

(

0.20\text{ mol}

for

16.0\text{ g}

) of

Fe

, corresponding to

11.2\text{ g}

.

Adım Adım Çözüm

1

Calculate the molar mass of

Fe_2O_3

and determine the number of moles of

Fe_2O_3

present.

Molar mass of

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

. Moles of

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

.

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

2

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

CO(g)

required at STP.

From the balanced equation,

1\text{ mol } Fe_2O_3

reacts with

3\text{ mol } CO

.
Moles of

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

.
Volume of

CO

at STP

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

.

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

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

).

3

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

Fe

) produced.

From the equation,

1\text{ mol } Fe_2O_3

produces

2\text{ mol } Fe

.
Moles of

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

.
Mass of

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

.

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

Anahtar Kavram

Mass-Mass and Mass-Volume Stoichiometric Calculations

Chemical Combination and Stoichiometry

Cevap

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