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Calculate the moles of nitric acid (HNO₃) used.

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4. Calculate the mass D of Potassium hydroxide.

  1. Calculate the moles of nitric acid (HNO3\text{HNO}_3) used.

    Moles of HNO3=Concentration×Volume\text{Moles of HNO}_3 = \text{Concentration} \times \text{Volume} Moles of HNO3=2.0 mol/L×501000 L=0.1 mol\text{Moles of HNO}_3 = 2.0 \text{ mol/L} \times \frac{50}{1000} \text{ L} = 0.1 \text{ mol}

  2. Determine the moles of potassium hydroxide (KOH\text{KOH}) in 50 cm3\text{cm}^3 of solution.

    The balanced equation is KOH(aq)+HNO3(aq)→KNO3(aq)+H2O(l)\text{KOH}_{(aq)} + \text{HNO}_{3(aq)} \rightarrow \text{KNO}_{3(aq)} + \text{H}_2\text{O}_{(l)}.
    From the equation, the mole ratio of KOH:HNO3\text{KOH} : \text{HNO}_3 is 1:11:1.
    Therefore, moles of KOH\text{KOH} in 50 cm3\text{cm}^3 solution = moles of HNO3\text{HNO}_3 = 0.1 mol.

  3. Calculate the moles of KOH\text{KOH} in 100 cm3\text{cm}^3 of solution.

    Since 50 cm3\text{cm}^3 of solution contains 0.1 mol of KOH\text{KOH}, then 100 cm3\text{cm}^3 of solution contains:

    Moles of KOH in 100 cm3=0.1 mol×100 cm350 cm3=0.2 mol\text{Moles of KOH in 100 cm}^3 = 0.1 \text{ mol} \times \frac{100 \text{ cm}^3}{50 \text{ cm}^3} = 0.2 \text{ mol}

  4. Calculate the mass D of KOH\text{KOH}.

    Relative formula mass (RFM) of KOH=56 g/mol\text{KOH} = 56 \text{ g/mol}.

    Mass D=Moles×RFM\text{Mass D} = \text{Moles} \times \text{RFM} Mass D=0.2 mol×56 g/mol=11.2 g\text{Mass D} = 0.2 \text{ mol} \times 56 \text{ g/mol} = 11.2 \text{ g}

    The mass D of Potassium hydroxide is 11.2 g\boxed{11.2 \text{ g}}.

    5. Calculate the mass of sodium sulphite that was used.

  5. Write the balanced chemical equation for the reaction.

    Na2SO3(s)+2HCl(aq)→2NaCl(aq)+H2O(l)+SO2(g)\text{Na}_2\text{SO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{SO}_2(g)

  6. Calculate the moles of sulphur dioxide (SO2\text{SO}_2) gas produced.

    Molar gas volume = 24000 cm3\text{cm}^3/mol.

    Moles of SO2=Volume of SO2Molar gas volume\text{Moles of SO}_2 = \frac{\text{Volume of SO}_2}{\text{Molar gas volume}} Moles of SO2=960 cm324000 cm3/mol=0.04 mol\text{Moles of SO}_2 = \frac{960 \text{ cm}^3}{24000 \text{ cm}^3/\text{mol}} = 0.04 \text{ mol}

  7. Determine the moles of sodium sulphite (Na2SO3\text{Na}_2\text{SO}_3) used.

    From the balanced equation, the mole ratio of Na2SO3:SO2\text{Na}_2\text{SO}_3 : \text{SO}_2 is 1:11:1.
    Therefore, moles of Na2SO3\text{Na}_2\text{SO}_3 used = 0.04 mol.

  8. Calculate the mass of sodium sulphite used.

    Molar mass of sodium sulphite = 126 g/mol.

    Mass of Na2SO3=Moles×Molar mass\text{Mass of Na}_2\text{SO}_3 = \text{Moles} \times \text{Molar mass} Mass of Na2SO3=0.04 mol×126 g/mol=5.04 g\text{Mass of Na}_2\text{SO}_3 = 0.04 \text{ mol} \times 126 \text{ g/mol} = 5.04 \text{ g}

    The mass of sodium sulphite used is 5.04 g\boxed{5.04 \text{ g}}.

    6. Calculate the volume of chlorine which will react with iron to form 0.5g of Iron (III) chloride.

  9. Calculate the molar mass of Iron (III) chloride (FeCl3\text{FeCl}_3).

    Atomic mass of Fe = 56
    Atomic mass of Cl = 35.5

    Molar mass of FeCl3=56+(3×35.5)=56+106.5=162.5 g/mol\text{Molar mass of FeCl}_3 = 56 + (3 \times 35.5) = 56 + 106.5 = 162.5 \text{ g/mol}

  10. Calculate the moles of FeCl3\text{FeCl}_3 formed.

    Moles of FeCl3=Mass of FeCl3Molar mass of FeCl3\text{Moles of FeCl}_3 = \frac{\text{Mass of FeCl}_3}{\text{Molar mass of FeCl}_3} Moles of FeCl3=0.5 g162.5 g/mol≈0.0030769 mol\text{Moles of FeCl}_3 = \frac{0.5 \text{ g}}{162.5 \text{ g/mol}} \approx 0.0030769 \text{ mol}

  11. Determine the moles of chlorine (Cl2\text{Cl}_2) required.

    The balanced equation is 2Fe(s)+3Cl2(g)→2FeCl3(s)2\text{Fe}_{(s)} + 3\text{Cl}_{2(g)} \rightarrow 2\text{FeCl}_{3(s)}.
    From the equation, the mole ratio of Cl2:FeCl3\text{Cl}_2 : \text{FeCl}_3 is 3:23:2.

    Moles of Cl2=32×Moles of FeCl3\text{Moles of Cl}_2 = \frac{3}{2} \times \text{Moles of FeCl}_3 Moles of Cl2=32×0.0030769 mol=0.00461535 mol\text{Moles of Cl}_2 = \frac{3}{2} \times 0.0030769 \text{ mol} = 0.00461535 \text{ mol}

  12. Calculate the volume of chlorine gas.

    Molar gas volume at 298K = 24 dm3\text{dm}^3/mol.

    Volume of Cl2=Moles of Cl2×Molar gas volume\text{Volume of Cl}_2 = \text{Moles of Cl}_2 \times \text{Molar gas volume} Volume of Cl2=0.00461535 mol×24 dm3/mol=0.1107684 dm3\text{Volume of Cl}_2 = 0.00461535 \text{ mol} \times 24 \text{ dm}^3/\text{mol} = 0.1107684 \text{ dm}^3

    Converting to cm3\text{cm}^3:

    0.1107684 dm3×1000 cm3/dm3=110.7684 cm30.1107684 \text{ dm}^3 \times 1000 \text{ cm}^3/\text{dm}^3 = 110.7684 \text{ cm}^3

    The volume of chlorine required is approximately 110.8 cm3\boxed{110.8 \text{ cm}^3}.

    7. Calculate the concentration of the solution in moles per litre.

  13. Calculate the mass of ethanoic acid (CH3COOH\text{CH}_3\text{COOH}) used.

    Density of ethanoic acid = 1.05 g/cm3\text{g/cm}^3.
    Volume of ethanoic acid = 15.0 cm3\text{cm}^3.

    Mass of CH3COOH=Density×Volume\text{Mass of CH}_3\text{COOH} = \text{Density} \times \text{Volume} Mass of CH3COOH=1.05 g/cm3×15.0 cm3=15.75 g\text{Mass of CH}_3\text{COOH} = 1.05 \text{ g/cm}^3 \times 15.0 \text{ cm}^3 = 15.75 \text{ g}

  14. Calculate the molar mass of ethanoic acid (CH3COOH\text{CH}_3\text{COOH}).

    Atomic masses: C=12, H=1, O=16.

    Molar mass of CH3COOH=(2×12)+(4×1)+(2×16)\text{Molar mass of CH}_3\text{COOH} = (2 \times 12) + (4 \times 1) + (2 \times 16) Molar mass of CH3COOH=24+4+32=60 g/mol\text{Molar mass of CH}_3\text{COOH} = 24 + 4 + 32 = 60 \text{ g/mol}

  15. Calculate the moles of ethanoic acid.

    Moles of CH3COOH=Mass of CH3COOHMolar mass of CH3COOH\text{Moles of CH}_3\text{COOH} = \frac{\text{Mass of CH}_3\text{COOH}}{\text{Molar mass of CH}_3\text{COOH}} Moles of CH3COOH=15.75 g60 g/mol=0.2625 mol\text{Moles of CH}_3\text{COOH} = \frac{15.75 \text{ g}}{60 \text{ g/mol}} = 0.2625 \text{ mol}

  16. Calculate the concentration of the solution in moles per litre.

    Total volume of solution = 500 cm3=0.5 L\text{cm}^3 = 0.5 \text{ L}.

    Concentration=Moles of CH3COOHVolume of solution (L)\text{Concentration} = \frac{\text{Moles of CH}_3\text{COOH}}{\text{Volume of solution (L)}} Concentration=0.2625 mol0.5 L=0.525 mol/L\text{Concentration} = \frac{0.2625 \text{ mol}}{0.5 \text{ L}} = 0.525 \text{ mol/L}

    The concentration of the solution is 0.525 mol/L\boxed{0.525 \text{ mol/L}}.

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