# Stoichiometry

How much iron (II) sulfide would be needed to prepare 15L of hydrogen sulfide? My balanced equation is 2HCl + FeS --> H2S + FeCl2. I have converted 15L H2S to mols, but I have no idea where to go from here.

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1. You're half way there IF the H2S is at STP. If not then an adjustment needs to be made.
But let me start over to make sure we don't miss something. Here are the general steps (assuming STP conditions for the H2S).

Step 1. Write the balanced equation. You have done that.
2HCl + FeS ==> H2S + FeCl2

Step 2. Convert what you have into mols. You have done that.
15L H2S x (1 mol H2S/34 g H2S) = 0.441 mol H2S.

Step 3. Using the coefficients in the balanced equation, convert mols of what you have (in this case H2S) to mols of what you want (in this case mols FeS).
mols FeS = 0.441 mols H2S x (1 mol FeS/1 mol H2S) = 0.441 mol x 1/1 = 0.441 mol FeS

Step 4. Now convert mols FeS to grams.
grams = mols FeS x molar mass FeS.

This procedure will work most stoichiometry problems.
Check my arithmetic. I estimated the molar mass of H2S so you need to redo that part and follow through.

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