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ankara55t · Dec 6, 2009, 04:01 PM

For the reaction:

NaHCO3 + HCl ---yields----NaCl + H20 + CO2

How can we determine the theoretical moles of NaCl produced?

I don't know where to begin.

As well, I need to find the theoretical mass of NaCl. How do I begin to find that?

Perito · Dec 6, 2009, 09:36 PM

$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$

This is a balanced reaction. For 1 mole of NaHCO3, you require 1 mole of HCL to produce 1 mole of NaCl. You start with the weight of NaHCO3. Then you determine the number of moles that that represents. Then, from the equation, this will be the number of moles of NaCl that are produced (if there's enough HCl around). Finally, from the molecular weight of NaCl, you can calculate the weight of NaCl that's produced.

Remember these relations:

$Molecular\,weight = \frac {grams}{mole}$

If you have a weight then

$moles = weight \,\div\, molecular\,weight$

If you have the number of moles

$weight\, (grams) = moles \,\times\, \frac {grams}{mole}$

ankara55t · Dec 7, 2009, 05:16 AM

Thank you so much!!
You're the best

ankara55t · Dec 7, 2009, 05:21 AM

Also could you look at how to calculate the moles and grams of HCl present in 6.0 ml. of 6.0M HCl solution used as a reactant.

do I start with 6.0ml. X 6.0 moles/1000ml. And get the # of moles then multiply the # of moles by the molectular weight to get the # of grams?

Unknown008 · Dec 7, 2009, 08:22 PM

That's it ankara55t! :)