Question

8. One of the first steps in the enrichment of uranium for use in nuclear power plans involves a
displacement reaction between UO2 and aqueous HF to produce the green salt, UF4, and water.

a. Write the balance reaction.
b. In a reaction of 4.25 kg of UO2 and 7.30 L of 6.50 M aqueous HF, what mass of UF4 is produced?
c. If the reaction in part b resulted in 3481 g of UF4, what is the percent yield of that reaction?

Answer 1

a)

UO2 + 4HF -------------------> UF4 + 2H2O

b).

mass of UO2 = 4.25 kg = 4.25 x 10^3 g

molar mass of UO2 = 270.03 g/mol

moles of UO2 = mass / molar mass = 4.25 x 10^3 / 270.03

= 15.74

Molarity = moles solute/ volume of solution

6.50 = moles / 7.30

moles HF = 47.75 moles HF.

UO2 + 4HF -------------------> UF4 + 2H2O

1 4 1

15.74 47.75 ?

here limiting reagent is HF. so UF4 produced based on that.

4 mol HF -----------------> 1 mol UF4

47.75 mol HF ---------------> ??

moles of UF4 = 47.75 x 1 / 4

= 11.94

molar mass of UF4 = 314 g/mol

mass of UF4 = moles x molar mass

= 11.94 x 314

= 3748 g

mass of UF4 = 3748 g

c)

% yield = (actual / theoretical ) x 100

= (3481 / 3748 ) x 100

= 92.87%