110 likes | 200 Views
Quiz 22 OCT 08. A process developed to treat minerals containing 4000 ppm of tetravalent uranium (UO 2 ) is based on the following conditions Consumption of H 2 SO 4 : 25 kg/t Consumption of NaClO 3 = 2 kg/t Volume of attacking solution/ mineral mass = 1 m 3 /t
E N D
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)
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
Distribution Ration KD Concentration (NH4)2SO4 or H2SO4 in Mol/L
CORRECTION • Write the oxidation equation of U(IV) by ClO3- ions OXIDATION REDUCTION
Determine the efficiency of the oxidizing treatment (1) • Number of moles of U solubilized 4000 ppm of U is equal to 4000 g/t of U Knowing 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 • Uranium Mass of U solubilized 4000 g * 96% = 3840g/t or 3.840g/L, knowing that the volume is 1 m3 = 1000L. [U] is 3.840/238 = 0.016M/L • H2SO4 25kg/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 Coefficient H2SO4 = 0.25M Kd 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