Question

How would your experimental mass of NaCl change if you did not add enough HCL for all the sodium carbonate to completely react?

Answer 1

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

Using that you get the balanced equation:
HCl + NaHCO₃ --> H₂O + 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.

For Example :

From the balanced equation it can be seen that 1 mole of NaHCO3 produces 1 mole of NaCl. Assume that 2.00 grams of NaHCO3 were used.

Moles of NaHCO3 =           grams         =          2.00 g          = 0.0238 moles                                     molecular mass         84.01 g/mole

Since the NaHCO3: NaCl mole ratio is 1:1, then 0.0238 moles of NaHCO3 would form 0.0238 moles of NaCl. The mass of this amount of NaCl would be:

Grams of NaCl = Number of moles * molar mass = 0.0238 mol * 58.44 g/mol = 1.39 grams of NaCl

The theoretical yield of NaCl expected is 1.39 grams.

Assume also that the student obtains 1.30 grams of NaCl. Their percentage yield would then be

what they actually obtained    = 1.30 grams * 100% = 93.4%                    what they should have obtained     1.39 grams