# If 0.130 g of p-aminophenol is allowed to react with excess acetic anhydride , what is the theoretical yield of acetaminophen in moles? In grams?

p-amino. - MW: 109.13g

aceta. - MW: 151.16 g

** I just forgot how to do this, I know it's probably REALLY simple. I took a stab at it, but if someone could just give me the basic steps so I can check my steps/answer.

THANKS!

In the interest of saving US time, we can be more productive if you post your work and let us check it. Thanks.

0.130g p-amino. x 1M / 109.13g p-amino. = 0.001 moles p-amino.

0.001 moles p-amino. x 151.16g aceta./1 moles aceta. = 0.15 g aceta.

then x100? If so, 15g aceta.

015g aceta. x 1 mole aceta. / 151.16 g aceta. = 9.92x10^-4 aceta.

then x100? If so, 0.099 moles aceta.

-----Let me know what went wrong..

You are allowed more places than you used.

0.130 x 1 mol/109.13 = 0.00119 mols.

0.00119 mols x 151.16 g/mol = 0.180 and not 0.15 grams.

So theoretical yield is 0.00119 mols or 0.180 grams.

The work you did with x 100 doesn't come into play until you calculate the percent yield. Thanks for using Jiskha.

Thank you!!