Quiz 22 OCT 08

• A process developed to treat minerals containing 4000 ppm of tetravalent uranium (UO2) is based on the following conditions

• Consumption of H2SO4: 25 kg/t

• Consumption of NaClO3 = 2 kg/t

• Volume of attacking solution/ mineral mass = 1 m3/t

• This treatment is used to solubilize 96% of uranium content.

• You are asked to• Write the oxidation equation of U(IV) by

ClO3- ions

• Determine the efficiency of the oxidizing treatment

• Determine the composition in uranium and H2SO4, of the solutions after attack (by making the following hypothesis that the variation of [H+] is negligible during the reaction period)

molesaddedofNumber

molesClOnecessaryofNumberE

...

.... 3

• The purification of uranium is performed by chromatography on anion exchange resins columns ( with an exchange capacity for one monovalent ion = 3 eq/kg)

• Determine, by using Figure 1, the distribution coefficient of U(VI) between the attacking solution and the resin ( consider that an excess of resin is used per unitary volume of solution)

• Determine the volume of uraniferous solution that you can transfer on a column composed of 100 kg of resins where we want to saturate to 70% maximum the exchange sites, knowing that the extraction reaction is

• The extraction of 1 U(VI) is saturating 4 sites of the resin

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Dis

trib

utio

n R

atio

n K

D

Concentration (NH4)2SO4 or H2SO4 in Mol/L

CORRECTION

• Write the oxidation equation of U(IV) by ClO3

- ions

OHClUOClOHUO

OHCleClOH

eUOUO

22232

23

222

3363

366

2

OXIDATION

REDUCTION

Determine the efficiency of the oxidizing treatment (1)

• Number of moles of U solubilized4000 ppm of U is equal to 4000 g/t of UKnowing that the molar mass M of U is 238 g /

mole, 4000g of U represents 4000/238 = 16.8067 moles/t. Since 96% of U is solubilized, we will have 0.96*16.8067 = 16.13 moles/t of U solubilized.

• Number of moles of NaClO3 used• M of NaClO3 is 106.5 g/mole• Consumption is 2 kg/t, = 2000g/t = 2000/106.5 =

18.78 moles/t

Determine the efficiency of the oxidizing treatment (2)

• Number of moles of NaClO3 necessary to the process

After Eq 1, 1 mole of NaClO3 reacts with 3 moles of UO2

If we have 16.13 moles of U, we need 16.13/3 = 5.37 moles of NaClO3

The efficiency will be 5.37/ 18.78 = 0.286

Composition Solution

• UraniumMass of U solubilized4000 g * 96% = 3840g/t or 3.840g/L, knowing that

the volume is 1 m3 = 1000L. [U] is 3.840/238 = 0.016M/L• H2SO425kg/t = 25000g/t or 25 g/L, knowing that the

volume is 1 m3 = 1000L.M = 98g/mole[H2SO4] = 25/98= 0.255moles/L

Distribution CoefficientH2SO4 = 0.25MKd U(VI) = 200

Volume of the uraniferous solution

• Exchange capacity of the resin• 3 eq/kg * 100 kg of resin = 300 eq/kg in

monovalent ion• In uranium we will have 300/4 = 75 moles of

U/column• The saturation is 70% so 75*70% = 52.5 moles

of U

The total volume of the uraniferous solution is

52.5 moles/0.016 = 3254.8 liters

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