Method for removing tungsten, vanadium, phosphor and arsenic from molybdate solution by deposition
A molybdate and solution technology, applied in chemical instruments and methods, molybdenum compounds, inorganic chemistry, etc., can solve the problems of limited tungsten removal depth, complicated operation, and small unit production capacity, and achieve environmental pollution and impurity removal depth High, easy-to-implement effect
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[0018] Example 1. 1L of sodium molybdate solution, containing Mo 100g / L, WO 3 1.0g / L, pH=9.6. Under the condition of room temperature and stirring speed of 150rpm, MnO is generated in situ 2 Is 50 times the theoretical amount (assuming MnO 2 With WO 3 The molar ratio is equal to 1 as the theoretical amount.) Add the required reagent, that is, add 13.6g of potassium permanganate, and after it is dissolved, slowly add 29.24g of manganese sulfate (prepared to 200-300g / L of manganese sulfate solution), Finally, 6.88g of NaOH was slowly added. Continue to stir for 1 hour and then filter, the filtrate contains WO 3 0.05g / L, Mo 97.5g / L.
Example Embodiment
[0019] Example 2. 1L of sodium molybdate solution, containing Mo80g / L, WO 3 1.0g / L, V 2 O 5 1.0g / L, P0.1g / L, As0.5g / L, pH=11. Under the condition of room temperature and stirring speed of 150rpm, MnO is generated in situ 2 Is 80 times the theoretical amount (assuming MnO 2 With WO 3 The molar ratio is equal to 1 as the theoretical amount) Add the required reagent, that is, add 21.76g of potassium permanganate, and after it is dissolved, slowly add 46.78g of manganese sulfate (prepared to 200-300g / L of manganese sulfate solution), Finally, 11.0 g of NaOH was slowly added. Continue to stir for 2 hours and then filter, the filtrate contains WO 3 0.03g / L, V 2 O 5 0.02g / L, P0.01g / L, As0.045g / L, Mo75g / L.
Example Embodiment
[0020] Example 3. 1L of sodium molybdate solution, containing Mo 100g / L, V 2 O 5 1.0g / L, pH=10. Under the condition of room temperature and stirring speed of 150rpm, MnO is generated in situ 2 Is 40 times the theoretical amount (assuming MnO 2 With V 2 O 5 The molar ratio is equal to 1 as the theoretical amount.) Add the required reagent, that is, add 13.88g of potassium permanganate, and after it is dissolved, slowly add 29.84g of manganese sulfate (prepared to 200-300g / L of manganese sulfate solution), Finally, 7.02g of NaOH was slowly added. Continue to stir for 1 hour and then filter, the filtrate contains V 2 O 5 0.06g / L, Mo 98g / L.
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