Day: December 25, 2024

Chapter 4 Stoichiometry SLO based solved Exercise 9th class new syllabus

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Get comprehensive solutions for Chapter 4 Stoichiometry from the 9th Class New Syllabus. This SLO-based solved exercise is designed to help students master the concepts of stoichiometry effectively.

(i) How many atoms are present in one gram of H2O?

1 mole of H2O = 18 g and contains 6.022×1023 molecules.
1 molecule of H2O = 3 atoms (2 H + 1 O).

Thus,
Atoms in 1 gram of H2O: 6.022×1023/18×3≈10.02×1022

Correct answer: (a) 10.02×1022 atoms

(ii) Which is the correct formula of calcium phosphide?

The formula of calcium phosphide is based on the charges of Ca2+ and P3−. To balance charges, the formula is Ca3P2

Correct answer: (c) Ca3P2

(iii) How many atomic mass units (amu) are there in one gram?

1 amu = 1.66×10−24g.

Number of amu in 1 g=1/1.66×10−24≈6.022×1023

Correct answer: (b) 6.022×1023

(iv) Structural formula of 2-hexene is CH3−CH=CH−(CH2)2−CH3. What is its empirical formula?

The molecular formula of 2-hexene is C6H12
Empirical formula is obtained by dividing subscripts by their greatest common divisor (6). Empirical formula=CH2

Correct answer: (b) CH2

(v) How many moles are there in 25 g of H2SO4?

Molar mass of H2SO4 = 2+32+(4×16)=98 g/mol

Moles=Mass/Molar mass=25/98≈0.255 moles.

Correct answer: (c) 0.255 moles

(vi) A necklace has 6 g of diamonds in it. What are the number of carbon atoms in it?

Diamonds are made of carbon.
Molar mass of carbon = 12 g/mol.
Number of moles of carbon: Moles=6/12=0.5 moles

Number of carbon atoms: 0.5×6.022×1023≈3.01×1023

Correct answer: (d) 3.01×1023

(vii) What is the mass of Al in 204 g of aluminum oxide (Al2O3)?

Molar mass of Al2O3 = 2(27)+3(16)=102 g/mol
Mass fraction of Al: 2(27)/102=54/102=0.5294

Mass of Al in 204 g of Al2O3: 0.5294×204≈108 g

Correct answer: (d) 108 g

(viii) Which one of the following compounds will have the highest percentage by mass of nitrogen?

  • For CO(NH2)2(urea):
    Molar mass = 12+16+2(14)+4(1)=60
    Nitrogen mass = 28
    Percentage of N = 28/60×100=46.67%
  • For NH4NO3:
    Molar mass = 14+4+14+3(16)=80
    Nitrogen mass = 28.
    Percentage of N = 28/80×100=35%

Correct answer: (a) CO(NH2)2

(ix) When one mole of each of the following compounds is reacted with oxygen, which will produce the maximum amount of CO2?

The number of moles of CO2 depends on the number of carbon atoms in the compound. Among the given options:

  • Methane (CH4): Produces 1 mole of CO2.
  • Ethane (C2H6): Produces 2 moles of CO2.
  • Diamond (C): Produces 1 mole of CO2

Correct answer: (b) Ethane

(x) What mass of 95% Ca(OH)2 will be required to neutralize 50 cm3 of 0.5 M H2SO4?

  • Reaction: Ca(OH)2+H2SO4→CaSO4+2H2O
    1 mole of H2SO4 reacts with 1 mole of Ca(OH)2.
    Moles of H2SO4: 0.5×0.05=0.025 moles
    Mass of pure Ca(OH)2: 0.025×(40+2(16+1))=0.025×74=1.85 g.
  • Mass of 95% pure Ca(OH)2: 1.850.95≈1.95 g

Correct answer: (d) 1.95 g

2. Questions for Short Answers

i. Write down the chemical formula of barium nitride.
The chemical formula of barium nitride is Ba₃N₂.
Explanation: Barium (Ba) is an alkaline earth metal with a +2 oxidation state, and nitride (N) is a nonmetal with a -3 charge. To balance the charges, three barium ions combine with two nitride ions, forming Ba₃N₂.

ii. Find out the molecular formula of a compound whose empirical formula is CH₂O and its molar mass is 180.
Step 1: Calculate the molar mass of the empirical formula (CH₂O).
Molar mass of CH₂O = 12 (C) + 2(1) (H) + 16 (O) = 30 g/mol.

Step 2: Determine the ratio of the molecular formula to the empirical formula.
Ratio=Molar mass of compound / Molar mass of empirical formula=180/30=6.

Step 3: Multiply the empirical formula by the ratio.
Molecular formula = (CH2O)6=C6H12O6 = C₆H₁₂O₆.
The molecular formula of the compound is C₆H₁₂O₆ (glucose).

iii. How many molecules are present in 1.5 g H₂O?
Step 1: Calculate the molar mass of water (H₂O).
Molar mass = 2(1) + 16 = 18 g/mol.

Step 2: Find the number of moles in 1.5 g of H₂O.
Moles = Mass/Molar mass=1.5/18=0.0833 mol

Step 3: Calculate the number of molecules using Avogadro’s number (6.022×1023 molecules/mol)
Number of molecules = 0.0833×6.022×1023=5.02×1022
Thus, there are approximately 5.02×1022

iv. What is the difference between a mole and Avogadro’s number?

  • Mole: A mole is a unit in chemistry that represents a quantity of substance. One mole of any substance contains the same number of particles (atoms, molecules, or ions) as there are in 12 g of carbon-12.
  • Avogadro’s Number: Avogadro’s number is a constant that defines the number of particles in one mole of a substance. It is equal to 6.022×1023 particles/mol

Difference: The mole is a quantity (like a dozen), while Avogadro’s number provides the specific count of particles in one mole.

v. Write down the chemical equation for the given reaction.
Copper + Sulphuric acid → Copper sulphate + Sulphur dioxide + Water
Balanced chemical equation:
Cu+2H₂SO₄→CuSO₄+SO₂+2H₂O

3. Constructed Response Questions

i. Different compounds will never have the same molecular formula but can have the same empirical formula. Explain.

  • The empirical formula represents the simplest whole-number ratio of elements in a compound, whereas the molecular formula shows the exact number of atoms of each element in a molecule.
  • Example: Acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆) have the same empirical formula (CH₂O), but their molecular formulas are different.
  • Explanation: Compounds with different molecular structures can share the same empirical formula because the molecular formula is a multiple of the empirical formula.

ii. Write down the chemical formulas of the following compounds.

  1. Calcium phosphate: Ca₃(PO₄)₂
  2. Aluminium nitride: AlN
  3. Sodium acetate: CH₃COONa
  4. Ammonium carbonate: (NH₄)₂CO₃
  5. Bismuth sulphate: Bi₂(SO₄)₃

iii. Why does Avogadro’s number have immense importance in chemistry?
Avogadro’s number (6.022×1023 is essential in chemistry because:

  • It establishes a connection between the microscopic (atoms/molecules) and macroscopic (grams/moles) scales.
  • It allows chemists to calculate the number of particles in a given mass of substance.
  • It is crucial for determining stoichiometric relationships in chemical reactions, enabling precise calculations for reactants and products.

iv. When 8.657 g of a compound were converted into elements, it gave 5.217 g of carbon, 0.962 g of hydrogen, and 2.478 g of oxygen. Calculate the percentage of each element present in this compound.
Step 1: Calculate the total mass of the compound.
Total mass = 5.217+0.962+2.478=8.657 g

Step 2: Calculate the percentage of each element.

  • Carbon: 5.2178.657×100=60.3%
  • Hydrogen: 0.9628.657×100=11.1%
  • Oxygen: 2.4788.657×100=28.6%

Thus, the percentages are:

  • Carbon: 60.3%
  • Hydrogen: 11.1%
  • Oxygen: 28.6%.

4. Descriptive Questions

i. Which conditions must be fulfilled before writing a chemical equation for a reaction?
To write a chemical equation, the following conditions must be fulfilled:

  1. Correct identification of reactants and products: Ensure the substances involved in the reaction are accurately identified.
  2. Balancing the equation: The equation must follow the law of conservation of mass, where the number of atoms for each element is the same on both sides.
  3. Physical states of substances: Indicate the physical states of reactants and products (solid (s), liquid (l), gas (g), or aqueous (aq)).
  4. Reaction conditions: Specify the conditions such as temperature, pressure, or the use of catalysts if required for the reaction.

ii. Explain the concepts of Avogadro’s number and mole.

  • Avogadro’s Number: Avogadro’s number (6.022×1023 is the number of particles (atoms, molecules, or ions) in one mole of a substance. It bridges the gap between the microscopic scale and the macroscopic scale.
  • Mole: A mole is a unit of measurement used in chemistry to represent 6.022×1023 particles of a substance. It is the amount of a substance that contains as many entities as there are atoms in 12 grams of carbon-12.

iii. How many grams of CO₂ will be produced when we react 10 g of CH₄ with excess of O₂ according to the following equation?
CH4+2O2→CO2+2H2OCH₄ + 2O₂ → CO₂ + 2H₂O

  1. Molar masses:
    • CH₄ = 16 g/mol
    • CO₂ = 44 g/mol
  2. Moles of CH₄:
    Moles = Mass/Molar mass=10/16=0.625 mol
  3. Moles of CO₂:
    From the equation, 1 mole of CH₄ produces 1 mole of CO₂.
    Moles of CO₂ = 0.625 mol.
  4. Mass of CO₂:
    Mass = Moles × Molar mass = 0.625×44=27.5 g

Answer: 27.5 g of CO₂ will be produced.

iv. How many moles of coal are needed to produce 10 moles of CO according to the following equation?
3C+O2+H2O→H2+3CO3C + O₂ + H₂O → H₂ + 3CO

From the equation, 3 moles of carbon (C) produce 3 moles of CO.
Thus, to produce 10 moles of CO: Moles of coal=103=3.33 moles.\text{Moles of coal} = \frac{10}{3} = 3.33 \, \text{moles}.

Answer: 3.33 moles of coal are needed.

v. How much SO₂ is needed in grams to produce 10 moles of sulphur?
2H2S+SO2→2H2O+3S

From the equation, 1 mole of SO₂ produces 1.5 moles of sulphur.
To produce 10 moles of sulphur: Moles of SO₂=10/1.5=6.67 moles

Molar mass of SO₂ = 32 (S) + 2(16) = 64 g/mol.
Mass = Moles × Molar mass = 6.67×64=426.88 g

Answer: 426.88 g of SO₂ are needed.

vi. How much ammonia is needed in grams to produce 1 kg of urea fertilizer?
2NH3+CO2→CO(NH2)2+H2O

  1. Molar masses:
    • NH₃ = 17 g/mol
    • CO(NH₂)₂ (urea) = 60 g/mol
  2. Moles of urea:
    Moles = Mass/Molar mass=1000/60=16.67 mol
  3. Moles of NH₃:
    From the equation, 2 moles of NH₃ produce 1 mole of urea.
    Moles of NH₃ = 16.67×2=33.34 mol.
  4. Mass of NH₃:
    Mass = Moles × Molar mass = 33.34×17=566.78 g

Answer: 566.78 g of ammonia is needed.

vii. Calculate the number of atoms in the following:

(a) 3 g of H₂:
Moles of H₂ = Mass/Molar mass=3/2=1.5 mol
Number of molecules = 1.5×6.022×1023=9.033×1023
Each molecule of H₂ contains 2 atoms, so:
Number of atoms = 9.033×1023×2=1.8066×1024
Number of molecules = 3.4×6.022×1023=2.0475×1024
Each molecule of N₂ contains 2 atoms, so:
Number of atoms = 2.0475×1024×2=4.095×1024
Molar mass of C₆H₁₂O₆ = 6(12)+12(1)+6(16)=180 g/mol
Number of molecules = 0.0556×6.022×1023=3.348×1022
Each molecule contains 6+12+6=246 + 12 + 6 = 24 atoms.
Number of atoms = 3.348×1022×24=8.035×1023

5. Investigative Questions

i. How many moles of water are needed for a single adult?
Volume of water per glass = 400 cm3
Density of water = 1 g/cm3
Mass of water = 400×8=3200 g

Molar mass of water = 18 g/mol.
Moles of water = Mass/Molar mass=3200/18=177.78 mol

Answer: 177.78 moles of water are needed.

ii. How many moles of SiO₂ are present in the glass?
Assuming the mass of glass is provided, moles = Mass of glass/Molar mass of SiO₂

where Molar mass of SiO₂=28+2(16)=60 g/mol

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