The limiting reagent is Naphthalene, which means that the moles of NitroNapthalene will equal the moles of Napthalene.
Using your calculations, but with two extra-significant figures, moles of Napthalene=moles of NitroNapthalene.
0.02341 moles of Napthalene=0.02341 moles of NitroNapthalene
0.02341 moles of NitroNapthalene*(173.17 g/mol)= Theoretical Yield
***I used an extra digit because it usually makes a difference when calculating the answer that your text wants. Also, these types of questions usually contain three significant figures, and usually want an answer with three significant figures.
Okay, the reaction is this:
Naphthalene + HNO3 ---> NitroNapthalene
The limiting reagent is Naphthalene
The excess is the nitric acid
The equation is 1:1 balanced as far as I know
3 g of Nap used in the experiment x 1mol/128.17 g/mol = .023 mol Nap
12.69 g used in the experiment x 1 mol/63.01 g/mol = .201 mol NO3
How to do calculate the theoretical yield of nictroNapthalene using this info?? (Also, I looked up the mol weight of nitroNapthalene and it's 173.17 g/mol if that helps any)
Thanks
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