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1.50 g of ethyl ethanoate was hydrolysed with 50.0 cm3 of 0.5 M NaOH solution and the mixture was made up to 250 cm3 with distilled water in a volumetric flask. Calculate: A: The number of moles of et

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Here are the solutions for question 1(b):

1. (b) 1.50 g of ethyl ethanoate was hydrolysed with 50.0 cm3^3 of 0.5 M NaOH solution and the mixture was made up to 250 cm3^3 with distilled water in a volumetric flask.

A: The number of moles of ethyl ethanoate in 1.50 g. (RMM: CH3_3COOC2_2H5_5 = 88)

  1. Use the formula for moles.

    Moles=MassRelative Molecular Mass (RMM)\text{Moles} = \frac{\text{Mass}}{\text{Relative Molecular Mass (RMM)}}

  2. Substitute the given values.

    Moles of ethyl ethanoate=1.50 g88 g/mol\text{Moles of ethyl ethanoate} = \frac{1.50 \text{ g}}{88 \text{ g/mol}} Moles of ethyl ethanoate=0.017045... mol\text{Moles of ethyl ethanoate} = 0.017045... \text{ mol}

  3. Round to an appropriate number of significant figures (e.g., 3 s.f.).

    Moles of ethyl ethanoate≈0.0170 mol\text{Moles of ethyl ethanoate} \approx 0.0170 \text{ mol}

    The number of moles of ethyl ethanoate is 0.0170 mol\boxed{\text{0.0170 mol}}.

    B: The number of moles of NaOH in 50 cm3^3 of 0.5 M solution.

  4. Convert the volume from cm3^3 to dm3^3.

    Volume in dm3=50.0 cm31000 cm3/dm3=0.050 dm3\text{Volume in dm}^3 = \frac{50.0 \text{ cm}^3}{1000 \text{ cm}^3/\text{dm}^3} = 0.050 \text{ dm}^3

  5. Use the formula for moles in solution.

    Moles=Concentration×Volume (in dm3)\text{Moles} = \text{Concentration} \times \text{Volume (in dm}^3\text{)}

  6. Substitute the given values.

    Moles of NaOH=0.5 mol/dm3×0.050 dm3\text{Moles of NaOH} = 0.5 \text{ mol/dm}^3 \times 0.050 \text{ dm}^3 Moles of NaOH=0.025 mol\text{Moles of NaOH} = 0.025 \text{ mol}

    The number of moles of NaOH is 0.025 mol\boxed{\text{0.025 mol}}.

    C: The percentage yield when 1.50 g of ethyl ethanoate reacts completely with sodium hydroxide to give 0.63 g of ethanol. (RAM: C=12.0, H=1.0, O=16.0, Na=23.0)

  7. Write the balanced chemical equation for the hydrolysis of ethyl ethanoate.

    CH3COOC2H5(aq)+NaOH (aq)→CH3COONa (aq)+C2H5OH (aq)\text{CH}_3\text{COOC}_2\text{H}_5 \text{(aq)} + \text{NaOH (aq)} \rightarrow \text{CH}_3\text{COONa (aq)} + \text{C}_2\text{H}_5\text{OH (aq)}

    From the equation, 1 mole of ethyl ethanoate produces 1 mole of ethanol.

  8. Calculate the molar mass of ethanol (C2_2H5_5OH).

    RMM of C2H5OH=(2×12.0)+(6×1.0)+(1×16.0)\text{RMM of C}_2\text{H}_5\text{OH} = (2 \times 12.0) + (6 \times 1.0) + (1 \times 16.0) RMM of C2H5OH=24.0+6.0+16.0=46.0 g/mol\text{RMM of C}_2\text{H}_5\text{OH} = 24.0 + 6.0 + 16.0 = 46.0 \text{ g/mol}

  9. Determine the theoretical moles of ethanol produced from 1.50 g of ethyl ethanoate.

    From part A, moles of ethyl ethanoate = 0.017045 mol.
    Since the mole ratio is 1:1, theoretical moles of ethanol = 0.017045 mol.

  10. Calculate the theoretical mass of ethanol.

    Theoretical mass of ethanol=Theoretical moles×RMM of ethanol\text{Theoretical mass of ethanol} = \text{Theoretical moles} \times \text{RMM of ethanol} Theoretical mass of ethanol=0.017045 mol×46.0 g/mol\text{Theoretical mass of ethanol} = 0.017045 \text{ mol} \times 46.0 \text{ g/mol} Theoretical mass of ethanol=0.78407 g\text{Theoretical mass of ethanol} = 0.78407 \text{ g}

  11. Calculate the percentage yield.

    Percentage yield=Actual yieldTheoretical yield×100%\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%

    Given actual yield of ethanol = 0.63 g.

    Percentage yield=0.63 g0.78407 g×100%\text{Percentage yield} = \frac{0.63 \text{ g}}{0.78407 \text{ g}} \times 100\% Percentage yield=80.35%\text{Percentage yield} = 80.35\%

  12. Round to an appropriate number of significant figures (e.g., 2 s.f.

    based on 0.63 g).

    Percentage yield≈80%\text{Percentage yield} \approx 80\%

    The percentage yield is \boxed{\text{80%}}.

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