See what proportions materials produced in a chemical reaction constitute of the origional reactant?
I know the above sentance didn't make all that much sense, so I'll attempt to explain:
Say we have the reaction:
2NaN3 -> 2Na + 3N2
In this reaction, I know how much N2 is produced - is it somhow possible to calculate how much 2NaN3 there was origionaly present in the reaction?
Sorry for the question... Chemistry and Physics cross over, plus me not being a chemist = difficulty!
Pezzoni wrote:See what proportions materials produced in a chemical reaction constitute of the origional reactant?
I know the above sentance didn't make all that much sense, so I'll attempt to explain:
Say we have the reaction:
2NaN3 -> 2Na + 3N2
In this reaction, I know how much N2 is produced - is it somhow possible to calculate how much 2NaN3 there was origionaly present in the reaction?
Sorry for the question... Chemistry and Physics cross over, plus me not being a chemist = difficulty!
Thanks very much for an answer
Well if the reaction runs to completion as you wrote it, then for every 3 mols of N2 you produce, 2 moles of NaN3 would've been required for the reactants. If it's in equilibrium instead, you'd need to look up the equilibrium constant K for the reaction and calculate off of that.
The only minor complication is that you may need to convert the measurements to moles, look at the masses of the constituent atoms: N2 is 28.01g/mol and NaN3 is 65.01g/mol, so that one gram of N2 corresponds to 0.2873g of NaN3 if your measurements are by mass instead of by mole, plus whatever is left over that has not reacted. That varies by temperature but the equilibrium case can be calculated straightforwardly from the energy difference between the reactants and products or the equilibrium constant for that particular temperature (the two are really equivalent).
Kuroneko wrote:The only minor complication is that you may need to convert the measurements to moles, look at the masses of the constituent atoms: N2 is 28.01g/mol and NaN3 is 65.01g/mol, so that one gram of N2 corresponds to 0.2873g of NaN3 if your measurements are by mass instead of by mole, plus whatever is left over that has not reacted. That varies by temperature but the equilibrium case can be calculated straightforwardly from the energy difference between the reactants and products or the equilibrium constant for that particular temperature (the two are really equivalent).
I have the measurements in moles, as well as mass already. Forgot to mention that
