Calculating % Yield of NaN3 Experiment

[crybllrd]

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

 

[Borek]

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?

 

[p21bass]

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.

 

[Borek]

You are right. I have ignored it completely as it is irrelevant to the question, my fault.

 

[DeeNos]

Great job balancing the equation and converting the known mass to moles! Now, to calculate the percent yield, we need to compare the actual yield (in moles) to the theoretical yield (also in moles). The theoretical yield is the amount of product that should have been produced according to the balanced equation.

In this case, we know that 2 moles of NaN3 were used in the experiment. From the balanced equation, we can see that for every 2 moles of NaN3, we should get 3 moles of N2. So the theoretical yield of N2 would be 3 moles.

Now, we can use the equation for percent yield to calculate the percentage:

% yield = (actual yield/theoretical yield) * 100%
% yield = (2.25 mol/3 mol) * 100%
% yield = 75%

So the percent yield of the experiment is 75%. This means that only 75% of the expected amount of N2 was produced. There could be a variety of reasons for this, such as incomplete reactions or loss of product during the experiment. It is important to always strive for a high percent yield in experiments, as it indicates the efficiency and accuracy of the procedure.