- #1
Sirsh
- 267
- 10
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.
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.