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 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 %