Question

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.

Answer 1

HCl + NaHCO3 yields H2O + CO2 + NaCl

Answer 2

@bohotness

Answer 3

yes

Answer 4

I need help understanding this question. I'm not quite sure what exactly theoretical yield is. Help??

Answer 5

ookay

Answer 6

Me putting down the answer doesn't help me understand how I got the answer. I need to know for future reference.

Answer 7

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=6&ved=0CEoQtwIwBQ&url=https%3A%2F%2Fwww.khanacademy.org%2Fscience%2Fchemistry%2Fchemical-reactions-stoichiome%2Flimiting-reagent-stoichiometry%2Fv%2Flimiting-reactant-example-problem-1&ei=hfHoVO2GGIb5yAT4woLYDQ&usg=AFQjCNE15qWPAcHuZP9oXJU17Pkof0mHSQ&sig2=XizBfNyAOyXsKANKPtx6Kw&bvm=bv.86475890,d.aWw

Answer 8

In other words, you don't know how to do it?

Answer 9

Are you missing anything in this problem? It feels like it.

Answer 10

Given the information provided, the only theoretical yield I see is just the mass of NaCl.

Answer 11

Yes, one moment

Answer 12

i am not really good with this :P sorry

Answer 13

Empty Dish: 24.35 g
w/ NaHCO3: 37.06 g
after reaction: 40.06 g
after H2O Evaporated: 31.52 g

Answer 14

@dtan5457 The missing information is above ^^^

Answer 15

Yeah, I see it. I solved a problem similar to this a day ago, let me see if I can get it.

Answer 16

Ok, I got it. It's pretty confusing how I did it, a bit, long of an explanation. But I'll try to keep it simple.

Answer 17

When given a question that has a limiting and excess reactant, and asks for a yield of a product...

you need to make two ratios of moles of the reactants.

one for the problem, one for the equation...

the one of the problem says that the evaporating dish is 24.35 g, and that HCL is 40.06 g WITH the dish.

that means that the HCL is actually only 15.71.

now to get moles as you know..is given mass/formula mass

HCL is about 36.5 grams formulated..

15.71/36.5=0.43 moles (problem for HCL)

now calculate the NAHCO3 the same way, and it's moles is 0.15

problem ratio is 0.43/0.15 (about 2.8)

the equation ratio is just 1:1 as there is just one mole of HCL and one mole of NAHCO3

now take the limiting reactant of the PROBLEM RATIO (0.15), multiply it by the equation ratio ( with the excess on TOP, in this case it's 1/1 so it doesn't matter), and by the mass of NACL

0.15 x 1/1 x 59=roughly 8.85 grams theoretically

Answer 18

i'm sure there are easier ways and shorter ways..but this is how i learned it..

Answer 19

Thanks a lot. I'll try to break it down the best I can in my notes. Really appreciate it.

Answer 20

yw. feel free to asks any questions about it.