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Calculate the mass of calcium oxide that would be obtained when 10 g of calcium oxide reacts with excess oxygen.

Chemistry

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Answer

a

8.4 g\text{8.4 g}

Step-by-step solution

5 steps

Here are the answers to the questions:

Part a· 5 steps

Calculate the mass of calcium oxide that would be obtained when 15 g15 \text{ g} of calcium carbonate is strongly heated in excess oxygen.

(Ca = 40, C = 12, O = 16)

  1. Write the balanced chemical equation for the decomposition of calcium carbonate.

    CaCO3(s)⟶CaO(s)+CO2(g)\text{CaCO}_3(s) \longrightarrow \text{CaO}(s) + \text{CO}_2(g)

  2. Calculate the molar masses of CaCO3\text{CaCO}_3 and CaO\text{CaO}.

    Molar mass of CaCO3=40(Ca)+12(C)+3×16(O)=40+12+48=100 g/mol\text{CaCO}_3 = 40 (\text{Ca}) + 12 (\text{C}) + 3 \times 16 (\text{O}) = 40 + 12 + 48 = 100 \text{ g/mol}
    Molar mass of CaO=40(Ca)+16(O)=56 g/mol\text{CaO} = 40 (\text{Ca}) + 16 (\text{O}) = 56 \text{ g/mol}

  3. Determine the moles of CaCO3\text{CaCO}_3 given.

    Moles of CaCO3=massmolar mass=15 g100 g/mol=0.15 mol\text{CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{15 \text{ g}}{100 \text{ g/mol}} = 0.15 \text{ mol}

  4. Use the mole ratio from the balanced equation to find moles of CaO\text{CaO}.

    From the equation, 1 mol1 \text{ mol} of CaCO3\text{CaCO}_3 produces 1 mol1 \text{ mol} of CaO\text{CaO}.
    So, moles of CaO\text{CaO} produced =0.15 mol= 0.15 \text{ mol}.

  5. Calculate the mass of CaO\text{CaO} produced.

    Mass of CaO=moles×molar mass=0.15 mol×56 g/mol=8.4 g\text{CaO} = \text{moles} \times \text{molar mass} = 0.15 \text{ mol} \times 56 \text{ g/mol} = 8.4 \text{ g}
    The mass of calcium oxide obtained is 8.4 g\boxed{\text{8.4 g}}.

Part b

State three applications of standard electrode potentials.

  • To predict the feasibility or spontaneity of a redox reaction.
  • To calculate the electromotive force (EMF) of an electrochemical cell.
  • To determine the relative strengths of oxidizing and reducing agents.

Part c(i)

What is meant by the term allotropy?

Allotropy is the property of some chemical elements to exist in two or more different physical forms (allotropes) in the same physical state. These different forms have distinct physical properties but similar chemical properties.

Part c(ii)

Apart from carbon, state one element that exhibits allotropy.

Sulfur (e.g., rhombic sulfur, monoclinic sulfur) or Phosphorus (e.g., white phosphorus, red phosphorus).

Part d

Give two characteristics of a homologous series.

  • All members can be represented by a general formula.
  • Successive members differ by a −CH2−-\text{CH}_2- group.
  • They show a gradual change in physical properties (e.g., boiling point) as molecular mass increases.
  • They have similar chemical properties due to the same functional group.

Part e(i)

heavy chemical

Give one example each of a

Sulfuric acid (H2SO4\text{H}_2\text{SO}_4) or Ammonia (NH3\text{NH}_3).

Part e(ii)

fine chemical

Aspirin or a specific laboratory reagent.

Part f

State two factors which influence the rate of a chemical reaction.

  • Concentration of reactants.
  • Temperature of the reaction mixture.
  • Surface area of solid reactants.
  • Presence of a catalyst.
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