neutralization of nitric acid 2

[QualitymanPMF]

When nitric acid HNO3 is neutralized by sodium hydroxide NaOH the byproduct is H2O and NaHO3. Lets say you mix one pound of each what would be the amount in weight of the H2O and the NaHO3?

 

[orenda1168]

The reaction products are H2O and NaNO3, not NaHO3. For 1 pound each of the reactants, 1 pound each of the reaction products will be formed.

Orenda

 

[QualitymanPMF]

That’s what I thought but it is always good to check. Sorry the formula was a typo good catch. Now I know I have a good chance my EPA form R for N 511 nitric compounds is correct. Odd though because, H2O: H has an atomic weight of 1 and oxygen has an atomic weight of 16 water total atomic weight of 18. NaNO3: Na has an atomic weight of 23, N has an atomic weight of 14, and O has an atomic weight of 16, total atomic weight of 85. It must be in the balancing of the equation.
Thanks

 

[25362]

The stoichiometric balancing equation:

NaOH + HNO[sub]3[/sub] = NaNO[sub]3[/sub] + H[sub]2[/sub]O​

shows: 40 + 63 = 85 + 18 = 103

 

[Zoobie]

Am I missing something simple here? If I read orenda1168's post it would indicate that if you took a pound of caustic and a pound of nitric acid you would get a pound of water and a pound of sodium nitrate. That is absolutely absurd...high school chemistry is all you need for this one. You need .63 lbs of caustic to neutralize one pound of nitric acid. If you added one pound then you would have an excess of caustic (ie 0.37 lbs left over). This means you get 1.35 lbs of sodium nitrate, 0.29 lbs of additional water (to the water that these solutions are made up of), and 0.37 lbs of caustic.

Of course this only applies to strong acids and bases such as these...otherwise you have to take into account equilibrium constants (Ka's).

 

[orenda1168]

Zoobie:

Of course you are correct....I must have been daydreaming and didn't read beyond the original question of 1 pound each of the reactants, which (ignoring the actual H2O and NaNO3 product weights) still results in a TOTAL of 2 pounds of material on the right side of the equation.

Sorry about that!

Orenda

 

[25362]

Not so. Zoobie is right. On a stoichiometric basis and assuming all the acid is consumed, one lb of each reactant means 0.365 lb NaOH will remain unreacted, and the equation's right side products will amount to ~1.635 lb.

 

[QualitymanPMF]

Thank you for the updated posts I already figured that the outcome of the naturalizing process produced 32,028 pounds of Sodium Nitrite from 23,797 pounds of Nitric Acid when Sodium Hydroxide is used as the neutralizer and the acid was naturalized to a ph of 6. I was lazy and was looking for a quick conversion factor and thought someone might have it already, but they didn’t so I had to do the math. Thanks for your help.

 

[moltenmetal]

You get sodium nitrate, (NaNO3), not sodium nitrite (NaNO2) when neutralizing nitric acid.

 

[QualitymanPMF]

Ok so it has an a and not an i. a befor e except after c Your point is?

 

[moltenmetal]

Nitrate and nitrite are DIFFERENT MOLECULES with different properties, including different molecular weights. So yes, the "a" versus the "i" matters! I'm also sure it matters on an "EPA form R", too.

 

[chicopee]

Here is my stoichiometry
1.587HNO3+2.500NaOH= 2.044H2O+1.587NaNO3+.913Na+.456O
Do a mass balance and it checks out but instead of using 1lbs, I used 100lbs.. You get 36.8lbs of H2O and 134.9lbs of NaNO3

 

[QualitymanPMF]

Thank you!!!!

Great post. That is what I needed you are best today.

 

[ChemBuddy]

While most of what should be written was already written, let me add one link:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

Browse lectures using menu on the left.