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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[Blueshift5]

To calculate the mass of UF6 produced, we first need to determine the number of moles of UF6 produced using the mole ratio of the balanced chemical equation. From the equation, we can see that for every 1 mole of UO2, 1 mole of UF6 is produced. However, we also need to take into account the fact that there are 4 moles of HF and 1 mole of F2 involved in the reaction.

So, the mole ratio between UO2 and UF6 is actually 1:4/3, or 3 moles of UF6 for every 4 moles of UO2. Therefore, the number of moles of UF6 produced from 27.78 moles of UO2 would be:

n(UF6) = (27.78 mol UO2) * (3 mol UF6 / 4 mol UO2) = 20.84 mol UF6

Now, we can calculate the mass of UF6 produced by multiplying the number of moles by the molar mass of UF6:

m(UF6) = 20.84 mol * 352.02 g/mol = 7,337.68 g or 7.34 kg

In summary, for 7.50 kg of UO2, we would need 1.08 kg of HF and 1.06 kg of F2 to produce 7.34 kg of UF6. It is important to understand the mole ratios in chemical reactions to accurately calculate the amounts of reactants and products involved.