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20 September, 23:50

Using 25.0cm3 of aqueous lithium hydroxide, concentration 2.48mol/dm3, 2.20g of hydrated lithium sulfate was obtained. Calculate the percentage yield, giving your answer to one decimal place.

2LiOH H2SO4 → Li2SO4 2H2O Li2SO4 H2O → Li2SO4. H2O

The number of moles of LiOH used = ...

The number of moles of Li2SO4. H2O which could be formed = ...

Mass of one mole of Li2SO4. H2O = 128 g

The maximum yield of Li2SO4. H2O = ...

percentage yield = ... % [4]

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Answers (1)
  1. 21 September, 00:05
    0
    Percentage yield = 27.8 %

    Explanation:

    Given dа ta:

    Volume of lithium hydroxide = 25.0 cm³ (0.025 dm³)

    Concentration of lithium hydroxide = M = 2.48 mol/dm³

    Actual yield of hydrated lithium sulfate = 2.20 g

    Percentage yield = ?

    Solution:

    Chemical equation:

    2LiOH + H₂SO₄ → Li₂SO₄.2H₂O

    Moles of lithium hydroxide:

    Molarity = moles / volume of solution

    2.48 mol/dm³ = moles / 0.025 dm³

    moles = 2.48 mol/dm³ * 0.025 dm³

    moles = 0.062 mol

    Now we compare the moles of hydrated lithium sulfate with lithium hydroxide:

    LiOH : Li₂SO₄.2H₂O

    1 : 1

    0.062 : 0.062

    Mass of Li₂SO₄.2H₂O:

    Mass = moles * molar mass

    Mass = 0.062 mol * 128 g/mol

    Mass = 7.9 g

    Percentage yield:

    Percentage yield = actual yield / theoretical yield * 100

    Percentage yield = 2.20 g / 7.9 g * 100

    Percentage yield = 27.8 %
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