Find theoretical yield from a balanced equation

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SUMMARY

The discussion focuses on calculating the theoretical yield of nitrogen gas (N2) from the balanced chemical equation NaNO2 + HSO3NH2 → NaHSO4 + H2O + N2. The user determined that NaNO2 is the limiting reactant with a mass of 0.1120g, equating to 0.0016 moles. To find the theoretical yield of N2, the user needs to calculate the mass of N2 produced assuming 100% reaction efficiency. The final percent yield was calculated to be approximately 97% based on the actual yield of 0.00156 moles of N2 collected.

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I searched and wasn't able to find what I was looking for. I need to find the theoretical yield so I can find the percent yield, which I know how do to. I just can't remember how to find theoretical yield from a balanced equation. My equation which I believe to be correct is NaNO2+HSO3NH2=NaHSO4+H20+N2. Could you let me know what I need to do, thanks.
 
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NaNO_{2} + HSO_{3}NH_{2} \rightarrow NaHSO_{4}+H_{2}O+N_{2}

Find the number of moles of each starting material
Determine the limiting reactant
Calculate the moles of product expected if the yield was 100% based on the limiting reactant
Convert moles to grams
 
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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.
 
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If NaNO_{2} is the limiting reactant, than the theoretical yield of N_{2} is 0.0016 moles.
 
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I just need to find the theoretical yield of N2 in moles. I used the equation n=pv/rt to find mol N2, if that helps. We basically measured the volume of N2 gas by causing a reaction inside a test tub. When the gas was produced from the test tube the volume in the buret dropped and we moved a tube with a funnel and water down the buret to keep the pressure equal.
 
g-tar-man said:
I just need to find the theoretical yield of N2 in moles. I used the equation n=pv/rt to find mol N2, if that helps. We basically measured the volume of N2 gas by causing a reaction inside a test tub. When the gas was produced from the test tube the volume in the buret dropped and we moved a tube with a funnel and water down the buret to keep the pressure equal.
You've been given the theoretical yield of N2, and you know the actual yield from the experiment.

Percent yield = (Actual yield / Theoretical yield) x 100%

You can calculate this using moles.
 
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 NaNO2 and then, experimentally or because the book said so, you got a value for the amount of N2 created.

Now, just work out the mass of N2 created from 0.1120g, assuming the reaction is 100% efficient. This will give you the theoretical yield.

After this, take the value of N2 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 :biggrin:

I hope this helps.

The Bob (2004 ©)
 
Yes thanks, I was able to finish it this morning. I think I got the percent yield to be around 97%. So not to bad I guess. Thanks for your help. Sorry about posting in the wrong section. It just seemed like the right place to put it.
 

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