Little confused on how to get part B.

A) if 3.00g of hydrosulfuric acid is reacted with 3.00g of silver nitrate, calculate the mass (in g) of solid silver sulfide formed. Which is the limiting reactant? (Hint: the other product is nitric acid).
B) calculate the moles of EACH reactant remaining after the reaction is complete.

For part A, I got that AgNO3 (silver nitrate) was the limiting reactant and that 2.19g Ag2S (solid silver sulfide) was formed (Hope those are right). Not sure how to go about part B though if anyone could help please!

Your value of 2.19 g Ag2S appears to be right.

If you can calculate mols Ag2S why not use the same reasoning to calculate mols H2S used.
2AgNO3 + H2S ==> Ag2S + 2HNO3
mols AgNO3 = 3.00/170 = approx 0.00833
Convert mols AgNO3 used to mols H2S used the same way.
0.00833 mols Ag x (1 mol H2S/2 mols AgNO3) = 0.00833 x 1/2 = approx 0.0044
And subtract that from the initial mols H2S to find H2S left unreacted.

There won't be any AgNO3 left since that was the limiting regent and all of that was left. I don't know if your teacher wants to go the extra route since there will be SOME (not much) Ag^+ from the solubility of Ag2S (and that's complicated a little by the excess H2S present). I don't think your teacher expects that since there are no volumes listed so no way to calculate molarities.

Thank you so much for the help Dr. Bob222! I was thinking there wouldn't be anything left since it was the limiting reactant but just wanted to make sure . Thanks again!!

To calculate the moles of each reactant remaining after the reaction is complete, you need to use stoichiometry and the concept of limiting reactant.

First, we need to identify the limiting reactant. In this case, you correctly determined that silver nitrate (AgNO3) is the limiting reactant. This means that all the hydrosulfuric acid (H2S) will react completely, and any excess silver nitrate will not participate in the reaction.

Let's start by calculating the moles of the limiting reactant, which is silver nitrate (AgNO3).

1. Calculate the molar mass of AgNO3:
AgNO3 = 107.87 g/mol (Ag) + 14.01 g/mol (N) + (3 * 16.00 g/mol) (O)
= 169.87 g/mol

2. Convert the mass of AgNO3 to moles:
moles of AgNO3 = mass of AgNO3 (g) / molar mass of AgNO3 (g/mol)
= 3.00 g / 169.87 g/mol
≈ 0.0177 mol (rounded to four decimal places)

Since the stoichiometry of the balanced equation for the reaction is 1:1 (1 mol of AgNO3 reacts with 1 mol of H2S), we know that there were 0.0177 moles of H2S initially.

To calculate the moles of each reactant remaining, we need to determine the moles of the excess reactant. In this case, it is the hydrosulfuric acid (H2S).

3. Calculate the molar mass of H2S:
H2S = 1.01 g/mol (H) + 32.06 g/mol (S)
= 33.07 g/mol

4. Convert the mass of H2S to moles:
moles of H2S = mass of H2S (g) / molar mass of H2S (g/mol)
= 3.00 g / 33.07 g/mol
≈ 0.0907 mol (rounded to four decimal places)

Since all the AgNO3 is completely consumed in the reaction, the remaining moles of H2S will be the excess reactant.

Now, let's calculate the moles of each reactant remaining after the reaction is complete:

Moles of AgNO3 remaining = 0 mol (since it is the limiting reactant and completely consumed)
Moles of H2S remaining = 0.0907 mol (the excess reactant that didn't react completely)

Therefore, after the reaction is complete, there will be 0 moles of AgNO3 remaining and 0.0907 moles of H2S remaining.