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A) i-iv. Calculate the amount in moles of dichromate (VI) used and Fe2+ present in 25 cm³ of solution.

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Here are the calculations for question 1)a) i-iv.

1)a) i) The amount in mole of dichromate (VI) used during the reaction

  1. Convert the volume of dichromate solution to dm3\text{dm}^3.

    Volume of Cr2O72−\text{Cr}_2\text{O}_7^{2-} solution = 27.30 cm3=27.30×10−3 dm3=0.02730 dm327.30 \text{ cm}^3 = 27.30 \times 10^{-3} \text{ dm}^3 = 0.02730 \text{ dm}^3.

  2. Calculate the moles of dichromate (VI) used.

    Concentration of Cr2O72−\text{Cr}_2\text{O}_7^{2-} solution = 0.020 mol/dm30.020 \text{ mol/dm}^3.

    Moles of Cr2O72−=Concentration×Volume\text{Moles of } \text{Cr}_2\text{O}_7^{2-} = \text{Concentration} \times \text{Volume} Moles of Cr2O72−=0.020 mol/dm3×0.02730 dm3\text{Moles of } \text{Cr}_2\text{O}_7^{2-} = 0.020 \text{ mol/dm}^3 \times 0.02730 \text{ dm}^3 Moles of Cr2O72−=0.000546 mol\text{Moles of } \text{Cr}_2\text{O}_7^{2-} = 0.000546 \text{ mol} Amount of dichromate (VI) used=5.46×10−4 mol\boxed{\text{Amount of dichromate (VI) used} = 5.46 \times 10^{-4} \text{ mol}}

    1)a) ii) The amount in mole of Fe2+\text{Fe}^{2+} present in 25 cm³ of solution.

  3. Use the mole ratio from the balanced chemical equation.

    The balanced equation is:

    Cr2O72−(aq)+14H+(aq)+6Fe2+(aq)⟶2Cr3+(aq)+6Fe3+(aq)+7H2O(l)\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 14\text{H}^+\text{(aq)} + 6\text{Fe}^{2+}\text{(aq)} \longrightarrow 2\text{Cr}^{3+}\text{(aq)} + 6\text{Fe}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}

    From the equation, 1 mole of Cr2O72−1 \text{ mole of } \text{Cr}_2\text{O}_7^{2-} reacts with 6 moles of Fe2+6 \text{ moles of } \text{Fe}^{2+}.

  4. Calculate the moles of Fe2+\text{Fe}^{2+} in 25 cm325 \text{ cm}^3 of solution.

    Moles of Fe2+=Moles of Cr2O72−×6\text{Moles of } \text{Fe}^{2+} = \text{Moles of } \text{Cr}_2\text{O}_7^{2-} \times 6 Moles of Fe2+=0.000546 mol×6\text{Moles of } \text{Fe}^{2+} = 0.000546 \text{ mol} \times 6 Moles of Fe2+=0.003276 mol\text{Moles of } \text{Fe}^{2+} = 0.003276 \text{ mol} Amount of Fe2+ in 25 cm3 solution=3.28×10−3 mol\boxed{\text{Amount of Fe}^{2+} \text{ in } 25 \text{ cm}^3 \text{ solution} = 3.28 \times 10^{-3} \text{ mol}}

    1)a) iii) The amount in mole of Fe2+\text{Fe}^{2+} present in the ore sample.

  5. Determine the dilution factor.

    The 25 cm325 \text{ cm}^3 sample was taken from a 250 cm3250 \text{ cm}^3 solution.

    Dilution factor=Total volumeAliquot volume=250 cm325 cm3=10\text{Dilution factor} = \frac{\text{Total volume}}{\text{Aliquot volume}} = \frac{250 \text{ cm}^3}{25 \text{ cm}^3} = 10

  6. Calculate the total moles of Fe2+\text{Fe}^{2+} in the 250 cm3250 \text{ cm}^3 solution (which came from the ore sample).

    Total moles of Fe2+=Moles of Fe2+ in 25 cm3×Dilution factor\text{Total moles of } \text{Fe}^{2+} = \text{Moles of } \text{Fe}^{2+} \text{ in } 25 \text{ cm}^3 \times \text{Dilution factor} Total moles of Fe2+=0.003276 mol×10\text{Total moles of } \text{Fe}^{2+} = 0.003276 \text{ mol} \times 10 Total moles of Fe2+=0.03276 mol\text{Total moles of } \text{Fe}^{2+} = 0.03276 \text{ mol} Amount of Fe2+ in the ore sample=0.0328 mol\boxed{\text{Amount of Fe}^{2+} \text{ in the ore sample} = 0.0328 \text{ mol}}

    1)a) iv) The percentage of iron (II) carbonate in the sample.

    To calculate the percentage of iron (II) carbonate (FeCO3\text{FeCO}_3) in the sample, the mass of the original ore sample is required. This information is not provided in the image.

    Assuming that all the Fe2+\text{Fe}^{2+} in the ore came from FeCO3\text{FeCO}_3:

  7. Calculate the molar mass of FeCO3\text{FeCO}_3.

    • Molar mass of Fe = 55.85 g/mol55.85 \text{ g/mol}
    • Molar mass of C = 12.01 g/mol12.01 \text{ g/mol}
    • Molar mass of O = 16.00 g/mol16.00 \text{ g/mol}
    Molar mass of FeCO3=55.85+12.01+(3×16.00)=115.86 g/mol\text{Molar mass of } \text{FeCO}_3 = 55.85 + 12.01 + (3 \times 16.00) = 115.86 \text{ g/mol}

  8. Calculate the mass of FeCO3\text{FeCO}_3 in the ore sample.

    Mass of FeCO3=Total moles of Fe2+×Molar mass of FeCO3\text{Mass of } \text{FeCO}_3 = \text{Total moles of } \text{Fe}^{2+} \times \text{Molar mass of } \text{FeCO}_3 Mass of FeCO3=0.03276 mol×115.86 g/mol\text{Mass of } \text{FeCO}_3 = 0.03276 \text{ mol} \times 115.86 \text{ g/mol} Mass of FeCO3=3.795 g\text{Mass of } \text{FeCO}_3 = 3.795 \text{ g}

  9. Calculate the percentage of FeCO3\text{FeCO}_3.

    Percentage of FeCO3=Mass of FeCO3Mass of ore sample×100%\text{Percentage of } \text{FeCO}_3 = \frac{\text{Mass of } \text{FeCO}_3}{\text{Mass of ore sample}} \times 100\%

    Since the mass of the ore sample is not given, the percentage cannot be calculated.

    Please provide the mass of the original ore sample if you would like to complete this calculation.

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