g-tar-man said:
Ok I determined the moles of NaNO2 and determined that was the limiting reactant, I think. Because I was given the mass of NaNO2 to be .1120g I figured that to be .0016mol NaNO2. I'm not sure how to calculate the theoretical yield of N2 in mols. I also found mol N2 collected to be .00156 mol.
Firstly, ignore that you are trying to find out the percentage yield. This, it would seem, is changing the meaning of your question so much that you are not understanding.
I assume either you or a book reacted 0.1120g of NaNO
2 and then, experimentally or because the book said so, you got a value for the amount of N
2 created.
Now, just work out the mass of N
2 created from 0.1120g, assuming the reaction is 100% efficient. This will give you the theoretical yield.
After this, take the value of N
2 you actually got and divide it by the value you just calculated (but obviously with the same units). Then multiply by 100 to get your percentage yield.
I hope that all of the explanations have put a different light on the problem and that you start to understand it in your own way
I hope this helps.
The Bob (2004 ©)