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.