Question

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

Answer 1

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.