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Calculate the concentration of HCl.

Chemistry

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Answer

0.210 mol/dm3\text{0.210 mol/dm}^3

90 cm3\text{90 cm}^3

Step-by-step solution

10 steps
  1. Calculate the concentration of HCl.

    3)

  2. Write the balanced chemical equation for the neutralization reaction.

    NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)\text{NaOH(aq)} + \text{HCl(aq)} \to \text{NaCl(aq)} + \text{H}_2\text{O(l)}

    From the equation, the mole ratio of NaOH\text{NaOH} to HCl\text{HCl} is 1:11:1.

  3. Calculate the moles of NaOH\text{NaOH} used.

    Given: Volume of NaOH=26 cm3=0.026 dm3\text{NaOH} = 26 \text{ cm}^3 = 0.026 \text{ dm}^3
    Concentration of NaOH=0.1 M\text{NaOH} = 0.1 \text{ M}

    Moles of NaOH=Concentration×Volume\text{Moles of NaOH} = \text{Concentration} \times \text{Volume} Moles of NaOH=0.1 mol/dm3×0.026 dm3\text{Moles of NaOH} = 0.1 \text{ mol/dm}^3 \times 0.026 \text{ dm}^3 =0.0026 mol= 0.0026 \text{ mol}
  4. Determine the moles of HCl\text{HCl} reacted.

    Since the mole ratio of NaOH\text{NaOH} to HCl\text{HCl} is 1:11:1,

    Moles of HCl=Moles of NaOH=0.0026 mol\text{Moles of HCl} = \text{Moles of NaOH} = 0.0026 \text{ mol}
  5. Calculate the concentration of HCl\text{HCl}.

    Given: Volume of HCl=12.4 cm3=0.0124 dm3\text{HCl} = 12.4 \text{ cm}^3 = 0.0124 \text{ dm}^3

    Concentration of HCl=Moles of HClVolume of HCl\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}} Concentration of HCl=0.0026 mol0.0124 dm3\text{Concentration of HCl} = \frac{0.0026 \text{ mol}}{0.0124 \text{ dm}^3} ≈0.209677 mol/dm3\approx 0.209677 \text{ mol/dm}^3

    Rounding to three significant figures:

    Concentration of HCl≈0.210 mol/dm3\text{Concentration of HCl} \approx 0.210 \text{ mol/dm}^3

    The concentration of HCl is 0.210 mol/dm3\boxed{\text{0.210 mol/dm}^3}.

    1)

  6. Write the balanced chemical equation for the combustion of carbon(II) oxide.

    2CO(g)+O2(g)→2CO2(g)\text{2CO(g)} + \text{O}_2\text{(g)} \to \text{2CO}_2\text{(g)}
  7. Determine the limiting reactant using the given volumes and the mole ratio from the balanced equation.

    According to the equation, 2 volumes of CO2 \text{ volumes of CO} react with 1 volume of O21 \text{ volume of O}_2.
    Given: Volume of CO=36 cm3\text{CO} = 36 \text{ cm}^3
    Volume of O2=72 cm3\text{O}_2 = 72 \text{ cm}^3

    For 36 cm336 \text{ cm}^3 of CO\text{CO}, the required volume of O2\text{O}_2 is:

    36 cm3 CO×1 volume O22 volumes CO=18 cm3 O236 \text{ cm}^3 \text{ CO} \times \frac{1 \text{ volume O}_2}{2 \text{ volumes CO}} = 18 \text{ cm}^3 \text{ O}_2

    Since 18 cm318 \text{ cm}^3 of O2\text{O}_2 is required and 72 cm372 \text{ cm}^3 of O2\text{O}_2 is available, O2\text{O}_2 is in excess. CO\text{CO} is the limiting reactant.

  8. Calculate the volume of CO2\text{CO}_2 produced.

    From the equation, 2 volumes of CO2 \text{ volumes of CO} produce 2 volumes of CO22 \text{ volumes of CO}_2.

    Volume of CO2=36 cm3 CO×2 volumes CO22 volumes CO=36 cm3 CO2\text{Volume of CO}_2 = 36 \text{ cm}^3 \text{ CO} \times \frac{2 \text{ volumes CO}_2}{2 \text{ volumes CO}} = 36 \text{ cm}^3 \text{ CO}_2
  9. Calculate the volume of excess O2\text{O}_2 remaining.

    Excess O2=Initial O2−O2 reacted\text{Excess O}_2 = \text{Initial O}_2 - \text{O}_2 \text{ reacted} Excess O2=72 cm3−18 cm3=54 cm3\text{Excess O}_2 = 72 \text{ cm}^3 - 18 \text{ cm}^3 = 54 \text{ cm}^3
  10. Determine the total volume of the resulting gaseous mixture.

    The gaseous mixture consists of the product (CO2\text{CO}_2) and the unreacted excess reactant (O2\text{O}_2).

    Total volume=Volume of CO2+Volume of excess O2\text{Total volume} = \text{Volume of CO}_2 + \text{Volume of excess O}_2 Total volume=36 cm3+54 cm3=90 cm3\text{Total volume} = 36 \text{ cm}^3 + 54 \text{ cm}^3 = 90 \text{ cm}^3

    The total volume of the resulting gaseous mixture is 90 cm3\boxed{\text{90 cm}^3}.

    4)
    To prepare crystals of copper(II) chloride from copper metal:

    1. Heat the copper metal in air to form copper(II) oxide. This is because copper does not react with dilute hydrochloric acid. 2Cu(s)+O2(g)→heat2CuO(s)\text{2Cu(s)} + \text{O}_2\text{(g)} \xrightarrow{\text{heat}} \text{2CuO(s)}
    2. Add the black copper(II) oxide to a beaker containing dilute hydrochloric acid. Stir the mixture gently. The copper(II) oxide will react with the acid to form a blue solution of copper(II) chloride. CuO(s)+2HCl(aq)→CuCl2(aq)+H2O(l)\text{CuO(s)} + \text{2HCl(aq)} \to \text{CuCl}_2\text{(aq)} + \text{H}_2\text{O(l)}
    3. Heat the solution gently to evaporate some of the water, concentrating the solution until it is saturated. This can be tested by dipping a clean glass rod into the solution and observing if crystals form on cooling.
    4. Allow the concentrated solution to cool slowly at room temperature. Copper(II) chloride crystals will form.
    5. Filter the crystals from the remaining solution (mother liquor) and wash them with a small amount of cold distilled water to remove impurities.
    6. Dry the crystals between filter papers or in a desiccator.

    5)
    An isotope refers to atoms of the same element that have the same number of protons (atomic number) but different numbers of neutrons, resulting in different mass numbers. They have identical chemical properties but differ in their physical properties.

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