Page 1 of 1

### Amount of solvent

Posted: Sat Nov 13, 2021 2:18 am
I have this balanced reaction:
Y2O3 + 6HNO3 = 2Y(NO3)3 + 3H2O
How I can determine the amount of HNO3 (solvent)?
Thanks

### Re: Amount of solvent

Posted: Sat Nov 13, 2021 5:47 am
You have a ratio 1:6 for Yttriumoxide and nitric acid. So you have to know how much Yttriumoxide you want to dissolve. You calculate the mole of it and know you need 6 times more of the nitric acid. According the concentration and specific gravity you can convert the mole of nitric acid to the volume.

### Re: Amount of solvent

Posted: Sat Nov 13, 2021 11:04 pm
Okay, So I have 0.002 mole of yttriumoxide that means I should have 0.012 mole of nitric acid, Please can you tell me how I can calculate the amount of nitric acid.

### Re: Amount of solvent

Posted: Sun Nov 14, 2021 1:43 am
You need m = n* M ( moles time molar mass) this gives 100% HNO3 , divided by the perecentage of it you get the Real mass and then V = m/σ. σ is specific gravity of nitric.

### Re: Amount of solvent

Posted: Sun Nov 14, 2021 2:25 am
Thanks