how can I convert praseodymium oxide (Pr6O11) to praseodymium nitrate [Pr(NO3)3]
Hi Anand
This involves a redox reduction of Pr, thus you will need a reducing agent to supply 4 electrons
Pr6O11 + 22HNO3 + 4e- --> 6Pr(NO3)3 + 11H2O + 4NO3-
I guess water could act as the reductant and a proton source (meaning less HNO3 is required)
2H2O --> O2 + 4H+ + 4e-
Giving the overall equation
Pr6O11 + 18HNO3 --> 6Pr(NO3)3 + 9H2O + O2
This seems reasonable, since the attached paper describes the synthesis of Pr(NO3)3 from Pr6O11 by refluxing in 65% HNO3.
Best regards
Geoff
thank you so much..... but sir how can we calculate this ratio?
because according to my calculation that was 52%
Hi Anand
The equation below has a molar ratio HNO3:Pr6O11 of 18:1
Pr6O11 + 18HNO3 --> 6Pr(NO3)3 + 9H2O + O2
Thus, if you have 1 g of Pr6O11 (i.e. 1 g / 1021.44 g mol-1 = 0.000979 mol), then you need 18 x 0.000979 mol = 0.01762 mol HNO3. Multiply this by the molar mass of HNO3 (63.01 g mol-1) and you get 1.11 g of HNO3. Most nitric acid is 70 wt.%, thus the mass of 70 wt.% HNO3 you would add to your 1 g of Pr6O11 is 1.11 g/0.7 = 1.59 g. You probably want to use a slight excess of HNO3 to ensure complete conversion of the Pr6O11, as was also recommended in the paper I sent you.
Best regards
Geoff
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Materials Science
- Materials Chemistry