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]

Percentage yield = 27.8 %

Explanation:

Given data:

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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