Processing Uranium: UO2 to UF6 Conversion

[Sirsh]

8. In the processing of uranium, one of the steps involves converting UO2 to UF6.

UO2(s) + 4HF(g) + F2 (g) --> UF6 (g) + 2H2O (l)

For 7.50kg of UO2, calculate:

a) The mass of hydrogen fluride required.
m(UO2) = 7500g
n(UO2) = 27.78mol


n(UO2) = n(HF)
27.78mol =

m(HF) = 27.78*39.008
= 1083.64g

b) The mass of fluorine required

n(UO2) = 2n(F2)
27.78mol =
27.78*2 =
55.56mol =

m(F2) = 1055.6g or 1.06x10^3g

c) The mass of UF6 produced.

With question (C) could you please do it and explain the mole ratio to me, as when i did it i did a mole ratio of n(UO2) = n(UF6) but the answer makes it seem that it's n(UO2) = 4/3n(UF6).

Thanks, Sirsh.

 

[Borek]

n(UO2) = 27.78mol

OK

n(UO2) = n(HF)

No.

m(F2) = 1055.6g or 1.06x10^3g

OK

when i did it i did a mole ratio of n(UO2) = n(UF6)

That's OK, correct answer is slightly below 10 kg.

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