Solid calcium fluorite (CaF_2) reacts with sulfuric acid to form solid calcium s

Question

Solid calcium fluorite (CaF_2) reacts with sulfuric acid to form solid calcium sulfate and gaseous hydrogen fluorite (HF). CaF_2(s) + H_2SO_4(l) rightarrow CaSO_4(s) + 2HF(g) The HF is then dissolved in water to form hydrofluoric acid. A source of calcium fluorite is fluorite ore containing 96wt% CaF_2 and 4 % SiO_2. In a typical hydrochloric acid manufacturing process, fluorite ore is reacted with 93wt% aqueous sulfuric acid, supplied 15% in excess to the stoichiometric amount. Ninety-five percent of the ore dissolves in the acid. Some of the HF formed reacts with the dissolved silica In the reaction. 6HF + SiO_2(aq) rightarrow H_2SIF_6(s) +2H_20(I) The hydrogen fluoride exiting from the reactor is subsequently dissolved in enough water to produce 60wt% hydrofluoric acid. Calculate the quantity of fluorite ore needed to produce a metric ton of aqueous hydrofluoric acid Some of the given data are not needed to solve the problem. Given that the molecular weight of: Ca = 40, F = 19, Si = 28, H =1, 0 = 16

Explanation / Answer

the molecular weight of water (H2O) =18g/mol (H=1; O=16)

the molecular weight of hydrofluoric acid (HF)= 20g/mol (H=1; F=19)

> if we have a tonne of 60% HF by weight, we need 600kg of HF, which is 600,000 grammes, which is 600,000/20 moles; this is 30,000moles of HF

we require 30,000 moles of the fluoride ion to produce one tonne of 60% w/w fluoric acid



the molecular weight of calcium fluoride (CaF2) = 78 (Ca=40; F=19) –

so fluoride is 38/78 of this by weight

the molecular weight of silica (SiO2) = 60 (Si=28; O=16)

one tonne of ore, 96% by weight, or 960kg, will be CaF2, and of those 960kg, fluoride is 960x(38/78) = 468kg

however, we are told that only 95% of the fluoride reacts, so the fluoride available for the process is 0.95x468kg = 444kg

the atomic weight of fluoride is 19g/mol, so we have 444,000/19 = 23,400moles of fluoride available from one tonne of ore

one tonne of ore yields 40kg of silica (4% by weight)
the molecular weight of silica is 60g/mol
> therefore

we have 40,000/60 = 667moles of silica from one tonne of ore

from the stoichiometry of the situation (see the equation you are given), we know that we lose 6 moles of fluoride ion to every mole of silica in the mix
> therefore,

we have 667x6 = 4,000moles of fluoride wastage per tonne of ore submitted to the process


fluoride wastage to by-product) from Step 2 (fluoride availability),

we obtain 23,400-4,000 = 19,400moles of fluoride (and therefore 19,400moles of HF) from one tonne of ore

30,000moles of HF for one tonne of the product

hence

the amount of ore required for one tonne of 60% w/w hydrofluoric acid is

30,000moles/19,400moles-per-tonne, which is 1.533 tonnes of fluorite ore, =1533 kg ore

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