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Calculate the moles of ethanoic acid and the number of water molecules formed in the reaction between ethanoic acid and potassium hydroxide.

Chemistry

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Answer

2

Number of water molecules=8.6688×1020 molecules\text{Number of water molecules} = 8.6688 \times 10^{20} \text{ molecules}

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Part 1· 2 steps

Amount in moles of ethanoic acid in the average volume of A used.

  1. Convert the average volume from cm3\text{cm}^3 to dm3\text{dm}^3.

    Average volume of A=14.40 cm3=14.40×10−3 dm3=0.01440 dm3\text{Average volume of A} = 14.40 \text{ cm}^3 = 14.40 \times 10^{-3} \text{ dm}^3 = 0.01440 \text{ dm}^3

    This conversion is correct.

  2. Calculate the moles of CH3COOH\text{CH}_3\text{COOH} using the concentration and volume.

    Moles of CH3COOH=Concentration×Volume\text{Moles of CH}_3\text{COOH} = \text{Concentration} \times \text{Volume} Moles of CH3COOH=0.100 mol dm−3×0.01440 dm3\text{Moles of CH}_3\text{COOH} = 0.100 \text{ mol dm}^{-3} \times 0.01440 \text{ dm}^3 Moles of CH3COOH=0.00144 mol\text{Moles of CH}_3\text{COOH} = 0.00144 \text{ mol} Moles of CH3COOH=1.44×10−3 mol\text{Moles of CH}_3\text{COOH} = 1.44 \times 10^{-3} \text{ mol}

    This calculation is correct.

Part 2

Number of water molecules=8.6688×1020 molecules\text{Number of water molecules} = 8.6688 \times 10^{20} \text{ molecules}

Number of water molecules formed in the reaction.

  1. Write the balanced chemical equation and determine the mole ratio.

    The reaction is between ethanoic acid (CH3COOH\text{CH}_3\text{COOH}) and potassium hydroxide (KOH\text{KOH}).

    CH3COOH+KOH→CH3COOK+H2O\text{CH}_3\text{COOH} + \text{KOH} \rightarrow \text{CH}_3\text{COOK} + \text{H}_2\text{O}

    From the balanced equation, the mole ratio of CH3COOH:H2O\text{CH}_3\text{COOH} : \text{H}_2\text{O} is 1:11:1.
    This is correct.

  2. Calculate the moles of H2O\text{H}_2\text{O} formed.

    Since the mole ratio is 1:11:1, the moles of H2O\text{H}_2\text{O} formed are equal to the moles of CH3COOH\text{CH}_3\text{COOH} used.

    Moles of H2O formed=1.44×10−3 mol\text{Moles of H}_2\text{O formed} = 1.44 \times 10^{-3} \text{ mol}

    This is correct.

  3. Calculate the number of water molecules using Avogadro's number.

    Number of water molecules=Moles×Avogadro’s number(NA)\text{Number of water molecules} = \text{Moles} \times \text{Avogadro's number} (\text{N}_{\text{A}})

    Using NA=6.02×1023 mol−1\text{N}_{\text{A}} = 6.02 \times 10^{23} \text{ mol}^{-1}:

    Number of water molecules=1.44×10−3 mol×6.02×1023 mol−1\text{Number of water molecules} = 1.44 \times 10^{-3} \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} Number of water molecules=(1.44×6.02)×(10−3×1023)\text{Number of water molecules} = (1.44 \times 6.02) \times (10^{-3} \times 10^{23}) Number of water molecules=8.6688×10(−3+23)\text{Number of water molecules} = 8.6688 \times 10^{(-3+23)} Number of water molecules=8.6688×1020 molecules\text{Number of water molecules} = 8.6688 \times 10^{20} \text{ molecules}

    Your calculation for the numerical part (1.44×6.02=8.66881.44 \times 6.02 = 8.6688) is correct, but there is an error in the exponent. The exponent should be 102010^{20}, not 102310^{23}.

    Therefore, the final answer for the number of water molecules is:

    Number of water molecules=8.6688×1020 molecules\boxed{\text{Number of water molecules} = 8.6688 \times 10^{20} \text{ molecules}}

    The first part of your answer is correct, but the second part has an error in the exponent for the number of water molecules.

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