Question

One of the first steps in the enrichment of uranium for use in nuclear power plants involves a displacement reaction between UO2 and aqueous HF: UO2(s) + HF(aq) → UF4(s) + H2O(l) [unbalanced] How many liters of 1.97 M HF will react with 3.48 kg of UO2?

L?

Answer 1

UO2(s) + 4HF(aq) → UF4(s) + 2H2O(l)

weight of UO2 = 3.48Kg = 3480g

G.M.Wt of UO2   = 270g/mole

no of moles of UO2   = W/G.M.Wt

                                  =3480/270

                                 = 12.89 moles

UO2(s) + 4HF(aq) → UF4(s) + 2H2O(l)

1 mole of UO2 react with 4 moles of HF

12.89 moles of UO2 react with = 4*12.89/1 = 51.56 moles of HF

no of moles of HF = molarity * volume in L

51.56                    = 1.97* volume in L

volume in L             = 51.56/1.97   = 26.17 L

volume of HF = 26.17 L >>>>answer