Wastewater catalyst for treating small organic molecules and preparation method thereof
A catalyst and small molecule technology, which is applied in the field of wastewater catalysts for the treatment of small organic molecules and their preparation, can solve the problems of difficulty in degrading small organic molecules and the investigation of stability, and achieves low operating costs, low production costs, and catalytic stability. Good results
- Summary
- Abstract
- Description
- Claims
- Application Information
AI Technical Summary
Problems solved by technology
Method used
Image
Examples
Embodiment 1
[0029] Example 1: Ru / Ni 0.2 Zr 0.1 Ce 0.7 o 2 resolve resolution
[0030] 1) Weigh 0.23g of Ni(NO 3 ) 2 ·6H 2 O, 0.17g of Zr(NO 3 ) 4 ·5H 2 O and 1.22g of Ce(NO 3 ) 3 ·6H 2 O, dissolved in 100mL water together, stirred for 0.5h to obtain a mixed solution;
[0031] 2) According to the molar ratio of sodium hydroxide and the total molar number of metal ions of Ni, Zr and Ce in step 1) being 2 to 12:1, take 1.5 g of sodium hydroxide and dissolve it in 50 mL of water to obtain lye Wherein, step 1) in Ni, Zr, the metal ion total molar number of Ce and step 1) in the preparation mixed solution water volume, step 2) in the formula ratio of the water volume sum of the preparation lye used water volume sum is about 0.004mol: 150mL, the ratio of the volume of water used to prepare the mixed solution in step 1) to the volume of water used to prepare the lye in step 2) is 2:1; and the amount of sodium hydroxide can make the mixed solution of step 1) and the alkali of step 2) ...
Embodiment 2
[0036] Example 2: Ru / Ni 0.3 Zr 0.1 Ce 0.6 o 2 resolve resolution
[0037] 1) Weigh 0.35g of Ni(NO 3 ) 2 ·6H 2 O, 0.17g of Zr(NO 3 ) 4 ·5H 2O and 1.04g of Ce(NO 3 ) 3 ·6H 2 O, dissolved in 100mL water together, stirred for 0.5h to obtain a mixed solution;
[0038] 2) According to the molar ratio of sodium hydroxide and the total molar number of metal ions of Ni, Zr and Ce in step 1) being 2 to 12:1, take 1.5 g of sodium hydroxide and dissolve it in 50 mL of water to obtain lye Wherein, step 1) in Ni, Zr, the metal ion total molar number of Ce and step 1) in the preparation mixed solution use water volume, step 2) in the preparation ratio of the formula ratio of the use water volume sum of alkali lye is about 0.004mol: 150mL, the ratio of the volume of water used to prepare the mixed solution in step 1) to the volume of water used to prepare the lye in step 2) is 2:1; and the amount of sodium hydroxide can make the mixed solution of step 1) and the alkali of step 2)...
Embodiment 3
[0043] Example 3: Ru / Ni 0.4 Zr 0.1 Ce 0.5 o 2 resolve resolution
[0044] 1) Weigh 0.46g of Ni(NO 3 ) 2 ·6H 2 O, 0.17g of Zr(NO 3 ) 4 ·5H 2 O and 0.87g of Ce(NO 3 ) 3 ·6H 2 O, dissolved in 100mL water together, stirred for 0.5h to obtain a mixed solution;
[0045] 2) According to the molar ratio of sodium hydroxide and the total molar number of metal ions of Ni, Zr and Ce in step 1) being 2 to 12:1, take 1.5 g of sodium hydroxide and dissolve it in 50 mL of water to obtain lye Wherein, step 1) in Ni, Zr, the metal ion total molar number of Ce and step 1) in the preparation mixed solution use water volume, step 2) in the preparation ratio of the formula ratio of the use water volume sum of alkali lye is about 0.004mol: 150mL, the ratio of the volume of water used to prepare the mixed solution in step 1) to the volume of water used to prepare the lye in step 2) is 2:1; and the amount of sodium hydroxide can make the mixed solution of step 1) and the alkali of step 2...
PUM
| Property | Measurement | Unit |
|---|---|---|
| concentration | aaaaa | aaaaa |
Abstract
Description
Claims
Application Information
Login to View More 


