# Theoretical yield of sodium chloride from sodium bicarbonate

I need to calculate the theoretical yield of mass of sodium chloride which should have been produced in a reaction.

We used sodium bicarbonate and hydrochloric acid in the reaction.

I balanced the equation as Na4C2+4HCl --> 4NaCl+H4C2

I don't know if I even set that part of right, or if I even need it because I have no idea how to go about doing this.

Can someone give me some insight?

After I do this I have to get the percent yield, which I can do on my own, but I need this mass to be able to do it.

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Well, first off, you need to use the correct formula for sodium bicarbonate (NaHCO3)

Using that you get the balanced equation:

HCl + NaHCO3 --> H20 + CO2 + NaCl

I assume that you started your experiment by measuring the mass of the sodium bicarbonate, so you will need to take the mass of that and finding the number of moles of sodium bicarbonate.

The coefficients indicate that the mole ratio of NaCl to NaHCO3 is 1:1, therefore, the number of moles of NaCl formed will be the same as the number of moles of NaHCO3 you added.

Once you know the number of moles of NaCl produced you have to find the mass of NaCl by multiplying by the formula mass of the NaCl. That will be your theoretical yield.

spock

Thu, 2008-05-15 11:46

We used 2g of sodium bicarbonate, so I have to find the number of moles in that?

And once I get that I multiply what?

I really am trying to understand..

hollis

Thu, 2008-05-15 17:57

The formula mass of NaHCO3 is 84 g/mol

moles of NaHCO3 = 2 g / 84 g/mol = .0238 mol

Previous post: "The coefficients indicate that the mole ratio of NaCl to NaHCO3 is 1:1, therefore, the number of moles of NaCl formed will be the same as the number of moles of NaHCO3 you added."

.0238 mol of NaHCO3 will produce .0238 mol of NaCl

Convert .0238 mole of NaCl to grams by multiplying by the formula mass of NaCl.

spock

Thu, 2008-05-15 20:38

Okay. So:

.0238*58=1.4

And that's the theoretical yield?

In that case, is the yield set up:

1.5/1.4*100?

hollis

Thu, 2008-05-15 20:49

When I do that for the percentage yield I get 107%. Is there any way that's right?

hollis

Thu, 2008-05-15 20:53

I'm assuming that your experimental yield was 1.5 grams. If that is correct, then your percent yield was 107%.

Now that indicates a problem, since you should not be able to produce more than 100% of the theoretical yield. Based on my experience as a HS chem teacher that has students do this lab, I would guess that you did not add enough HCl to completely neutralize all of the NaHCO3. Since the left over NaHCO3 has a greater mass than the NaCl product, this would account for you getting a greater than 100% yield. Another possible error might be that the NaCl product was not completely dry. Any remaining water would make the product heavier than it should be, thus increasing your percent yield.

spock

Thu, 2008-05-15 22:46

Okay, I understand all of that. It makes since.

What should the percent yield be around?

Should it be closer to 100% or 0%?

I'm not entirely sure what the percent yield indicates.

hollis

Fri, 2008-05-16 11:28

sense* and I also see where I messed up the equation.

Sorry for all of these double posts. >.>

hollis

Fri, 2008-05-16 11:35