Question

Molar conversion help please!

Answer 1

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

Answer 2

Whats your question to be exact?

Answer 3

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

Answer 4

Some direction or correction would be helpful

Answer 5

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.

Answer 6

.151 x 1 = .151, but could you mind checking my work of converting grams to moles? (like the process/formula)

Answer 7

sure

Answer 8

thank you

Answer 9

your math is correct but if you want the correct number of significant digits it should be .1514

Answer 10

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?

Answer 11

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

Answer 12

lemme know if you have any other questions

Answer 13

It does. So all that's left is NaCl mole -> gram conversion to find it's theoretical yield?

Answer 14

yes if you need to find the mass of NaCl produced that would be the only thing left to do in this situation

Answer 15

Ok thank you. I'll be back in like 30 seconds.

Answer 16

aight

Answer 17

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.

Answer 18

Alright glad to help :).