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Chemistry 19 Online
OpenStudy (anonymous):

1 C8H9NO3 -> 4 C8H10NO3 to convert the moles of the reactant I would just multiply the moles of it by 4 and then multiply the moles by the molecular weight of C8H10NO3 right?

OpenStudy (anonymous):

I'm half asleep I know this but I just need to confirm so I can finish my lab

OpenStudy (anonymous):

so if I had 0.00309mol of reactant I would have 0.01236moles of product therefore I would have 2.1g of product?

OpenStudy (anonymous):

167g/mol is the molecular mass of the product

OpenStudy (sasogeek):

I'm not good at this but logically that makes sense

OpenStudy (anonymous):

I'm a pro at stochiometry but is 4 in the morning lol

OpenStudy (anonymous):

so im not a pro at the momment

OpenStudy (sasogeek):

lol k

OpenStudy (anonymous):

thanks :) wow I didnt get much of yield at all ugh

OpenStudy (sasogeek):

yw :)

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