This is a question that i picked up and shared with a student that i am coaching at the moment. This will be a question that tests on the different sub-sections that is covered in Mole Concepts and Chemical Calculations.
Example:
Hydrogen Fluoride attacks glass and is used to draw on glass. It is prepared by adding concentrated sulphuric acid to calcium fluoride. The reaction takes place at r.t.p: CaF2 (s) + H2SO4 (l)?–> 2HF (g) + CaSO4 (s)1) When 15.6g of calcium fluoride was reacted with 12.5cm^3 of 20.0 mol/dm^3 sulphuric acid, 7.68dm^3 of hydrogen fluoride was formed.
i) What is the limiting reactant in this reaction?
ii) Calculate the theoretical volume of hydrogen fluoride that should be formed.2) Calculate the mass of the calcium fluoride in the impure calcium fluoride.
3) What is the percentage purity of the calcium fluoride?
Please work out your answer on your own first on a piece of paper, before you click on “Read More” or the Headline itself for the suggested answer.
Suggested Answer:
1 i)
Mr of CaF2 = 40 + (19 x 2) = 78
No. of moles of CaF2 used = 15.6 / 78 = 0.20mol
No. of moles of H2SO4 used = 20.0 x (12.5/1000) = 0.25mol
Since 1 mol of H2S04 reacts with 1 mol of CaF2, 0.25 mol of H2S04 will react with 0.25 mol of CaF2
Hence, CaF2 is the Limiting Reactant.ii)
Based on the equation and 0.20 mol of CaF2 used,
No. of moles of HF formed = 2 x 0.20 = 0.40 mol
Theoretical volume of HF formed = 0.40 x 24 dm^3 = 9.6dm^32)
Moles of HF formed = 7.68/24 = 0.32mol
From the equation, 1 mol of CaF2 produces 2 mol of HF.
Moles of pure CaF2 = 0.32/2 = 0.16 mol
Mass of calcium fluoride = 0.16 x 78 = 12.48g3)
Percentage purity of CaF2 = (Mass of pure CaF2 / Mass of CaF2 Sample) x 100%
= (12.48/15.6) x 100% = 80%
Note:
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