Little confused on how to get part B.

Question

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!

Answer 1

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.

Answer 2

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!!