Question

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]​

Answer 1

Percentage yield = 27.8 %

Step-by-step 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 %