Question

I'm working on a lab report where we have to find the percent yield of the reaction NaHCO3 + HCl yields NaCl + H2O + CO2. I believe the theoretical yield is 0.695 g NaCl, where NaHCO3 is the limiting reagent/reactant, and HCl the excess. In the beginning of the lab, we found that we had produced 7.170 g NaCl. I'm trying to find the ACTUAL yield of NaCl. I can't seem to set this up correctly. Please help; I'll award medals :)

Answer 1

I got as far as 7.170 g NaCl x 1 mol NaCl/58.44 g NaCl = 0.123 mol NaCl
but I need it in grams, I believe, to find the percent yield.

Answer 2

@Ashleyisakitty

Answer 3

@ashwinjohn3

Answer 4

if you made 7.17 g in the lab, then that's your actual yield.

Answer 5

i think your calculations for the theoretical yield might be off

Answer 6

But that gives me a percent yield of 10.32%. Isn't that a bit off??

Answer 7

Okay. Want me to post them here so you can have a look?

Answer 8

yeah, that would help.

Answer 9

also as it stands, your percent yield is \(\dfrac{7.17}{0.695}*100\%=1031.65\%\)

Answer 10

That's even more off, lol.

Answer 11

And for HCl, I got 1.603 g NaCl.

Answer 12

@aaronq Still there?

Answer 13

sorry, i had to step away from the computer, let me have a look

Answer 14

how much sodium bicarbonate did you start off with?