1. So I am doing this lab and one of the conclusions asks me:

In our lab we are doing the double replacement of HCl + NaHCO3 yielding -> NaCl + H2O + CO2.

If you started with X moles of NaHCO3, how many moles of NaCl would you expect to be formed?

This was my response:

If you started with X moles of NaHCO3, there would be X moles of NaCl. The reason for this is because to find out part of the product (NaCl in HCl + NaHCO3 -> NaCl + H2O + CO2) of how many moles there are, you would have to know how many moles there are in NaHCO3 because sodium is part of the chemical compound, NaHCO3.  2.

3. Yes, but it's a little more than "you would have to know how many moles there are in NaHCO3 because sodium is part of the chemical compound, NaHCO3.". The ratio formed by components on one side to the other side comes into play.

Restating the reaction:

aHCl +
bNaHCO3 cNaCl + dH2O +eCO2

where a, b, c, d, and e = 1

You can use this molecular reaction in terms of moles. The moles of NaCl given the moles of NaHCO3 is c/b, which is 1/1 = 1. You began with 1 mole of NaHCO3, so you get 1 × 1 moles of NaCl, or 1 mole.

If you had a reaction equation that said ...

blah blah blah + 2∙NaHCO3 5∙NaCl + blah blah blah

... then the factor is 5/2 = 2.5. If you began with 1 mole of
NaHCO3, then you would get 1 × 2.5 moles of NaCl, or 2.5 moles.

Maybe you meant this, but just didn't complete the thought.  Bookmarks
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