Question

Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction

CaF₂ + H₂SO₄ → CaSO₄ + 2HF

In one process 6.00 kg of CaF₂ are treated with an excess of H₂SO₄ and yield 2.86 kg of HF. Calculate the percent yield of HF.

Answer 1

Sol :-

As, number of moles = Given mass in gram / Gram molar mass

So,

Number of moles of CaF₂ = Given mass of CaF₂/Gram molar mass of CaF₂

= 6.00 x 10³ g / 78.07 g/mol

= 76.85 mol

Now,

Given reaction is :

CaF₂ + H₂SO₄ → CaSO₄ + 2HF

In the balance reaction :

1 mole of CaF₂ will gives = 2 moles of of HF

So,

76.85 moles of CaF₂ will gives = 2 mol x 76.85 mol

= 153.7 mol

Now, convert 153.7 moles of HF into mass

Mass of HF = Number of moles of HF x Gram molar mass of HF

= 153.7 mol x 20.01 g/mol

= 3075.537 g

= 3.075537 Kg

So, Theoretical yield of HF = 3.075537 Kg

Now,

The percent yield of HF = Actual yield x 100% / Theoretical yield

= 2.86 Kg x 100 % / 3.075537 Kg

= 93 %

Hence, percent yield of HF = 93 %