Question

thanks

Answer 1

yes, the actual yield is 1.57g since that's the actual result from the experiment

to get the theoretical yield, you have to take the reactant 2.36g NaHCO3, convert to moles NaHCO3, use the equation to convert to moles (H2CO3, I believe that's the product they're looking for), and convert that to grams of product

Answer 2

okay for the first part you said reactant 2.36 g NaHCO3 to moles NaHCO3
i think 1 mole = 84.006 g
do i multiply or divide that by 2.36 g?

Answer 3

to convert g to moles, divide by the molar mass

2.36/84.006

Answer 4

okay that gives me 0.0280

what do i do now?

Answer 5

you have moles NaHCO3

like I said earlier, use the equation (balance it first) to convert moles NaHCO3 to moles H2CO3

Answer 6

okay somebody else helped me with this part earlier but im not sure if they finished all the way through

2NaHCO3→Na2CO3+H2CO3

Answer 7

yeah that's the balanced equation

Answer 8

ok so i know 1 mol = 62.024 g of H2CO3
and 1 mol = 84.006 g of NaHCO3

but how do i go on fro here?

Answer 9

notice how the equation tells you that 2 mol NaHCO3 = 1 mol H2CO3
this lets you convert from moles NaHCO3 to moles H2CO3

Answer 10

ok i might just be confused here.. is it just 1 mol? im not sure at all

Answer 11

you are currently in moles of NaHCO3
if 2 mol NaHCO3 = 1 mol H2CO3

you need to divide moles of NaHCO3 by 2 to get moles of H2CO3.

Answer 12

are you saying to divide 0.0280 by 2? not sure

Answer 13

yes.

Answer 14

oh i see... i got a bit lost there.
it gives me 0.014

what after this?

Answer 15

you have moles of H2CO3 now
convert this to grams of H2CO3
end result should be your theoretical yield

Answer 16

will it give me 0.868?

Answer 17

yeah