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

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.

To calculate the theoretical yield of nitroNapthalene using the given information, we need to first determine the balanced equation between naphthalene and nitroNapthalene. Assuming the equation is indeed balanced, as you mentioned (1:1), we can use the stoichiometry of the reaction to determine the amount of nitroNapthalene that should be produced from the given amount of naphthalene.

From the balanced equation, we can see that 1 mole of naphthalene reacts with 1 mole of nitroNapthalene. Therefore, the mole ratio between naphthalene and nitroNapthalene is 1:1.

Given that 3 g of naphthalene was used in the experiment, we have calculated that it corresponds to 0.023 mol of naphthalene.
Since the mole ratio is 1:1, we can conclude that the theoretical yield of nitroNapthalene should also be 0.023 mol.

To convert this theoretical yield from moles to grams, we can use the molar mass of nitroNapthalene, which you provided as 173.17 g/mol. Multiplying the molar mass by the number of moles gives:

0.023 mol * 173.17 g/mol = 3.98 g

Therefore, the theoretical yield of nitroNapthalene, based on the given information, is 3.98 grams.