anonymous
  • anonymous
Molar conversion help please!
Chemistry
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The NaHCO3 is the limiting reactant and the HCl is the excess reactant in this experiment. Determine the theoretical yield of the NaCl product, showing all of your work in the space below. (5 points) So far I have 12.71 g NaHCO3 / 83.96 g per mole NaHCO3 = 0.151
sweetburger
  • sweetburger
Whats your question to be exact?
anonymous
  • anonymous
What is the theoretical yield of NaCl with the mass of NaHCO3 being 12.71 g and HCl being in excess. NaHCO3 +HCl → NaCl + CO2+ H2O

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anonymous
  • anonymous
Some direction or correction would be helpful
sweetburger
  • sweetburger
Alright so you found the number of moles of NaHCO3. Im assuming your math is correct (it looks correct). So from here we can find the number of moles of NaCl produced. |dw:1440617996106:dw|
anonymous
  • anonymous
.151 x 1 = .151, but could you mind checking my work of converting grams to moles? (like the process/formula)
sweetburger
  • sweetburger
sure
anonymous
  • anonymous
thank you
sweetburger
  • sweetburger
your math is correct but if you want the correct number of significant digits it should be .1514
anonymous
  • anonymous
Oh thank you, I appreciate it. So, we don't have to mess with HCl conversion because it's in excess, right? I only really have to convert .1514 mole NaCl to grams?
sweetburger
  • sweetburger
yea because HCl is in excess it would never be the limiting reagent this means that we are limited by the amount of NaHCO3 present and therefor whatever amount of NaHCO3 we have will directly affect the amount of products we will have when the reaction is completed
sweetburger
  • sweetburger
lemme know if you have any other questions
anonymous
  • anonymous
It does. So all that's left is NaCl mole -> gram conversion to find it's theoretical yield?
sweetburger
  • sweetburger
yes if you need to find the mass of NaCl produced that would be the only thing left to do in this situation
anonymous
  • anonymous
Ok thank you. I'll be back in like 30 seconds.
sweetburger
  • sweetburger
aight
anonymous
  • anonymous
OK. I'm resubmitting this and I was checking to see if I made any of the same mistakes. Well, all looks in order, you've been a great help thanks.
sweetburger
  • sweetburger
Alright glad to help :).

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