Calculating % Yield of NaN3 Experiment

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To calculate the % yield of the NaN3 experiment, the balanced equation is 2NaN3 → 2Na + 3N2. The actual yield of nitrogen isolated is 21 g, which converts to 0.75 moles of N2 using the molar mass of 28 g/mol. The theoretical yield from 2 moles of NaN3 is 42 g of N2, since 3 moles of N2 are produced. The % yield is calculated as (21 g / 42 g) * 100%, resulting in a yield of 50%. Understanding the conversion and balancing of the equation is crucial for accurate yield calculations.
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Homework Statement


Use the unbalanced equation NaN3 → Na + N2 to answer the following question(s).
The % yield of an experiment in which two moles NaN3 was used and 21 g nitrogen was isolated is ________.

Homework Equations



percent yield=({actual yield}/{theoretical yield})100%

The Attempt at a Solution



First, I balanced:

2NaN3 → 2Na + 3N2

And I convert my known mass to moles:

21g of N_{2}* \frac{3 mol N_{2}}{28 g N_{2}}=2.25 mol N_{2}

...and that's where I'm stuck. I'm preparing for a comprehensive final exam, and it seems I have forgotten this piece from the beginning of the semester.

Edit:The LATEX ref is failing for some reason. it is 21g(3mol/28g)=2.25 mol
 
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LaTeX is OK, you were probably seeing cached image. That is a known problem.

What should be mass of nitrogen produced from 2 moles of NaN3?
 
Not to step on Borek's toes, or anything... And maybe I forgot something, but 21g(N2) does not equal 2.25mol(N2). There are 28.02g in one mol of nitrogen gas. Three mols N2 = 42.03g. N=14.01g/mol.
 
You are right. I have ignored it completely as it is irrelevant to the question, my fault.
 
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